Conditional Probability: The conditional probability that E occurs given that F has
already occurred and is denoted by
Sample space: The set of all possible outcomes of an experiment is known as the
sample space of the experiment and is denoted by S.
Examples:
1. If the experiment consists of flipping two coins, then the sample space consists
of the following four points:
S = {(h, h),(h, t),(t, h),(t, t)}
P(E|F) =
2. If the experiment consists of tossing two dice, then the sample space consists of
the 36 points S = {(i, j): i, j = 1, 2, 3, 4, 5, 6},
where the outcome (i, j) is said to occur if i appears on the leftmost die and j on
the other die.
3. If the experiment consists of measuring (in hours) the lifetime of a transistor,
then the sample space consists of all nonnegative real numbers; that is,
S = {x: 0 x < }
Example1: A coin is flipped twice. Assuming that all four points in the sample space S =
{(h, h),(h, t),(t, h),(t, t)} are equally likely, what is the conditional probability that both
flips land on heads, given that (a) the first flip lands on heads? (b) at least one flip lands
on heads?
Example2: Joe is 80 percent certain that his missing key is in one of the two pockets of
his hanging jacket, being 40 percent certain it is in the left-hand pocket and 40 percent
certain it is in the right-hand pocket. If a search of the left-hand pocket does not find
the key, what is the conditional probability that it is in the other pocket?
Example3: In the card game bridge, the 52 cards are dealt out equally to 4 players—
called East, West, North, and South. If North and South have a total of 8 spades among
them, what is the probability that East has 3 of the remaining 5 spades?
Event: Any subset E of the sample space is known as an event. In other words, an
event is a set consisting of possible outcomes of the experiment.
Examples:
1. If E = {(h, h),(h, t)}, then E is the event that a head appears on the first coin.
2. if E = {(1, 6),(2, 5),(3, 4),(4, 3),(5, 2),(6, 1)}, then E is the event that the sum of
the dice equals 7.
3. If E = {x: 0 x 5}, then E is the event that the transistor does not last longer
than 5 hours.
Mutually exclusive: If E and F are any two events such that E
are said to be mutually exclusive.
Example4: Celine is undecided as to whether to take a French course or a chemistry
course. She estimates that her probability of receiving an A grade would be 1 2 in a
French course and 2 3 in a chemistry course. If Celine decides to base her decision on
the flip of a fair coin, what is the probability that she gets an A in chemistry?
F = , then E and F
Example5: Suppose that an urn contains 8 red balls and 4 white balls. We draw 2 balls
from the urn without replacement. (a) If we assume that at each draw, each ball in the
urn is equally likely to be chosen, what is the probability that both balls drawn are red?
(b) Now suppose that the balls have different weights, with each red ball having weight
r and each white ball having weight w. Suppose that the probability that a given ball in
the urn is the next one selected is its weight divided by the sum of the weights of all
balls currently in the urn. Now what is the probability that both balls are red?
1. If E = {(1, 6),(2, 5), (3, 4),(4, 3),(5, 2),(6, 1)} is the event that the sum of the dice
is 7 and F = {(1, 5),(2, 4), (3, 3),(4, 2),(5, 1)} is the event that the sum is 6, then
the event E F does not contain any outcomes and hence could not occur, i.e.,
E F = Thus E and F are said to be mutually exclusive.
2. If E = {(h, h),(h, t),(t, h)} is the event that at least 1 head occurs and F = {(h, t),(t,
h),(t, t)} is the event that at least 1 tail occurs, then E F = {(h, t),(t, h)} is the
event that exactly 1 head and 1 tail occur. Thus E and F are not mutually
exclusive.
Axioms of Probability: The three axioms of probability
Axiom 1: 0 P(E) 1
Axiom 2: P(S) = 1
Axiom 3: For any sequence of mutually exclusive events
which
= Ø when i j),
(
provided if P(F) > 0.
... (that is, events for
)=
We refer to P(E) as the probability of the event E.
New Section 1 Page 1
Answer1.: (a). 1/2
Answer2.
2/3
Answer3.
(b). 1/3
Answer4.
P(C A) = P(C)P(A|C)=
Answer5.
(a). P(R1
R2) = P(R1)P(R2|R1)=
(b). P(R1
R2) =
Independent Events: Two events E and F are said to be independent if
= P(E)P(F).
Ex 1: Two coins are flipped, and all 4 outcomes are assumed to be equally likely. If E is
the event that the first coin lands on heads and F the event that the second lands on
tails, then E and F are independent, since P(E F) = P({(H,T)}) = 1/4 , whereas P(E) =
P({(H, H),(H,T)}) = 1/2 and P(F) = P({(H,T),(T,T)}) = 1/2 .
Ex 2: A card is selected at random from an ordinary deck of 52 playing cards. If E is
the event that the selected card is an ace and F is the event that it is a spade, then
E and F are independent. This follows because P(E F) = 1/52 , whereas P(E) = 4/52
and P(F) = 13/52 .
Ex 3: Suppose that we toss 2 fair dice.
(a). Let E denote the event that the sum of the dice is 6 and F denote the event that
the first die equals 4. Then
=
P(E)P(F) =
Hence, E and F are not independent. Intuitively, the reason for
this is clear because if we are interested in the possibility of throwing a 6 (with 2 dice),
we shall be quite happy if the first die lands on 4 (or, indeed, on any of the numbers 1,
2, 3, 4, and 5), for then we shall still have a possibility of getting a total of 6. If,
however, the first die landed on 6, we would be unhappy because we would no longer
have a chance of getting a total of 6. In other words, our chance of getting a total of 6
depends on the outcome of the first die; thus, E and F cannot be independent.
(b). Now, suppose that we let G be the event that the sum of the dice equals 7. Is G
independent of F? The answer is yes.
New Section 1 Page 2
Random variables: A random variable is any function that assigns a numerical value
to each possible outcome.
Remarks: Random variables are denoted by capital letters X,Y, and so on, to distinguish them from their possible
values given in lowercase x, y.
Probability mass function (pmf): A random variable whose set of possible values is
either finite or countably infinite is called discrete. If X takes the values x1, x2, . . .,
the function defined by
= P{X = }
(i = 1, 2, …)
is called the probability mass function of X. Moreover, p(x) = 0 if
for all i.
Further, the probability distribution always satisfies the conditions
(i)
≥0
and (ii)
Ex: Check whether the following can serve as probability distributions:
(a) f(x) = (x − 2)/2 for x = 1, 2, 3, 4
(b)
25 for x = 0, 1, 2, 3, 4
Sol:
(a) This function cannot serve as a probability distribution because f (1) is negative.
(b) The function cannot serve as a probability distribution because the sum of the
five probabilities is 6/5 and not 1.
If we let X denote the number of heads that appear, then X is a random variable taking
on one of the values 0, 1, 2 and 3 with respective probabilities 1/8, 3/8, 3/8 and 1/8.
Ex: A life insurance agent has 2 elderly clients, each of whom has a life insurance
policy that pays $100,000 upon death. Let Y be the event that the younger one dies
in the following year, and let O be the event that the older one dies in the following
year. Assume that Y and O are independent, with respective probabilities P(Y) = .05
and P(O) = .10. If X denotes the total amount of money (in units of $100, 000) that
will be paid out this year to any of these clients’ beneficiaries, then X is a random
variable that takes on one of the possible values 0, 1, 2 with respective probabilities.
Find the pmf for X?
Ex2: Letting X denote the random variable that is defined as the sum of two fair dice,
then
P{X = 2} = P{(1, 1)} = 1/36
P{X = 3} = P{(1, 2),(2, 1)} = 2/36
P{X = 4} = P{(1, 3),(2, 2),(3, 1)} = 3/36
P{X = 5} = P{(1, 4),(2, 3),(3, 2),(4, 1)} = 4/36
P{X = 6} = P{(1, 5),(2, 4),(3, 3),(4, 2),(5, 1)} = 5/36
P{X = 7} = P{(1, 6),(2, 5),(3, 4),(4, 3),(5, 2),(6, 1)} = 6/36
P{X = 8} = P{(2, 6),(3, 5),(4, 4),(5, 3),(6, 2)} = 5/36
P{X = 9} = P{(3, 6),(4, 5),(5, 4),(6, 3)} = 4/36
P{X = 10} = P{(4, 6),(5, 5),(6, 4)} = 3/36
P{X = 11} = P{(5, 6),(6, 5)} = 2/36
P{X = 12} = P{(6, 6)} = 1/36
Distribution function[or Cumulative distribution function
(CDF)]. For a random variable X, the function F defined by
F(x) = P(X ≤ x), −∞ < x < ∞,
is called the distribution function of X.
Proposition. If F is the distribution function of a random variable X, then the
following properties hold:
1. F(x) ≤ F(y) if x < y. (i.e., non-decreasing)
2.
3.
4. F is right continuous.
New Section 1 Page 3
5 defective pieces of ABC units which are accidently mixed with 15
good pieces and looking at them. Assume that 4 pieces are picked up
at random. Construct a probability mass function and probability
distribution function of random variable corresponding to number of
defective pieces.
A bag contains 6 red and 4 white balls. Three balls are drawn at random.
Obtain the probability mass function and probability distribution function
of the number of white balls drawn.
Suppose a coin is tossed two times. Construct the probability
mass function and probability distribution function of random
variable corresponding to number of Heads.
Pb 4. The probability mass function of a random variable X is given by p(i) =
some positive value. Find (a) c (b) P{X = 0} and (c) P{X > 2}.
, i = 0, 1, 2, ... , where λ is
New Section 1 Page 4
Continuous Random Variable and Probability density function
Expectation or Expected value of a random variable:
A random variable X is continuous if its distribution function is given by an integral
F(x) =
whose integrand f(x), called the density of the distribution, is nonnegative, and is
continuous, perhaps except for finitely many x-values.
The expected value of a discreate random variable is a weighted
average of all possible values of the random variable, where the
weights are the probabilities associated with the corresponding values.
Note. If f(x) is the probability density function of a continuous random variable, then
(a) f(x) ≥ 0 for all x.
(b) f is piecewise continuous
(c)
= 1.
(d) P{
}=
The mathematical expression for computing the expected value of a
discreate random variable X with probability mass function (p.m.f.) f(x)
is given below:
E(X) =
= F(b) - F(a)
(for discrete r.v.)
or
,.…..}
If X= {
E(X) =
{ provided
}
In words, the expected value of X is a weighted average of the possible
values that X can take on, each value being weighted by the probability
that X assumes it. For instance, on the one hand, if the probability mass
function of X is given by
p(0) = 1/2 = p(1) then E[X] = 0 (1/2) + 1 (1/2) = 1/2
is just the ordinary average of the two possible values, 0 and 1, that X
can assume. On the other hand,
if p(0) = 1/3 p(1) = 2/3 then E[X] = 0 (1/3) + 1(2/3) = 2/3
is a weighted average of the two possible values 0 and 1, where the
value 1 is given twice as much weight as the value 0, since p(1) = 2p(0).
Pb1. Find E[X], where X is the outcome when we roll a fair die.
Ans: 7/2
Pb.2 A school class of 120 students is driven in 3 buses to a symphonic
performance. There are 36 students in one of the buses, 40 in another, and
44 in the third bus. When the buses arrive, one of the 120 students is
randomly chosen. Let X denote the number of students on the bus of that
randomly chosen student, and find E[X]?
Ans: 40.2667
Pb.3 Let X= {0, 1,2,…}, P{X=k} =
Ans:
New Section 1 Page 5
where
. Find E[X]?
The expected value of a continuous random variable X with probability density
function (p.d.f) f(x) is:
E(x) =
Pb.1 Let X be a
{ provided
random ariable, if pdf is i en by
f(x) =
Find CDF and E[X]?
ns
X
}
Some properties of Expectation:
1. If X and Y are random variable, then E(X+Y) = E(X) + E(Y), provided all
expectations exist.
Generalization: E(
) = E( ) + E( ) +.…+ ( )
2. If X and Y are independent random variable, then E(XY) = E(X)E(Y),
provided all expectations exist.
Generalization: E(
) = E( )E( )...E( )
3. If X is a random variable and a is a constant, then
(a). E[ah(X)] = aE[h(X)]
(b). E[h(X)+a] = E[h(X)]+a
Where h(X), a function of X, is a r.v. and all the expectations exist.
4. If X is a random variable and a & b are constant, then
E[aX+b] = aE(X)+b, provided all expectations exist.
Proof: If X is a discreate random variable, then
Expectation of a function of a Random Variable:
Suppose that we are given a random variable X along with its probability mass
function or probability density function and that we want to compute the
expected value of some function of X, say, h(X).
random ariable
E(h(x)) =
If X is a continuous random variable, then
random ariable
Pb.2 Let X denote a random variable that takes on any of the values −1, 0,
and 1 with respective probabilities:
P{X = −1} = .2 P{X = 0} = .5 P{X = 1} = .3
Compute E[ ] and ( X )
Ans:
( X)
Pb.3 The density function of X is given by
5. The expected value of a random variable X, E[X], is also referred to as the
mean or the first moment of X about origin. The quantity E[ ], n ≥ 1, is
called the nth order moment of X about origin. That is
f(x) =
Find E[
Ans:
E[
].
]=
6. Expectation of a linear combination of random variables:
Let
be any n linear combination of random variables and if
are any n constants, then
E(
)=
7. If X ≥ 0 then E(X) ≥ 0.
8. If X and Y are two random variables such that Y X then E(Y)
.
9. |E(X)|
10. If X and Y are independent random variable, then
E[h(X).k(Y)] = E[h(X)].E[k(Y)], where h(.) is a function of X alone and k(.) is
a function of Y alone, provided expectations on both sides exist.
New Section 1 Page 6
Moments:
1. The
moment of a variable x about any point x=A is:
(
,
where
2. If x= (mean)
(
New Section 1 Page 7
Variance:
Let f(x) be the p.d.f. of a random variable X, where X is defined from a to b:
Definition: If X is a random ariable with mean μ, then the ariance
of X, denoted by Var(X), is defined by
Var(X) =
= called the 2nd order moment of X
about mean.
An alternative formula for Var(X) is derived as follows:
A.M =
In words, the variance of X is equal to the expected value of
minus the square of its expected value. In practice, this formula
frequently offers the easiest way to compute Var(X).
Pb1. Calculate Var(X) if X represents the outcome when a fair die
is rolled.
Ans: Var(X) = 35/12
Properties of Variance:
1. If X is a random variable and a & b are constant, then V[aX+b] =
Proof:
V(X).
2. The square root of the Var(X) is called the standard deviation of X, and we
denote it by SD(X). That is, SD(X) =
ar(X)
3. Analogous to the means being the center of gravity of a distribution of
mass, the variance represents, in the terminology of mechanics, the
moment of inertia.
New Section 1 Page 8
Moment Generating Function (MGF)
How to calculate the Mean and Variance from MGF ?
Note: In some cases, MGF does not exist, since the
Or the series
does not converge absolutely for real
values of t for some distribution.
Uniqueness Theorem of Moment Generating function: The MGF
of a distribution, if it exists, uniquely determines the distribution.
This implies that corresponding to given probability distribution,
there is only one MGF (provided it exists) and corresponding to a
given MGF, there is only one probability distribution.
Hence
X and Y are identically distributed.
New Section 1 Page 9
The Bernoulli and Binomial Random Variables
Suppose that a trial, or an experiment, whose outcome can be classified as
either a success or a failure is performed. If we let X = 1 when the outcome is
a success and X = 0 when it is a failure, then the probability mass function of
X is given by
Thus, only 0.4 percent of the packages will have to be replaced.
Pb. 3 : The following gambling game, known as the wheel of fortune (or
chuck-a-luck), is quite popular at many carnivals and gambling casinos:
A player bets on one of the numbers 1 through 6. Three dice are then
rolled, and if the number bet by the player appears i times, i = 1, 2, 3,
then the player wins i units; if the number bet by the player does not
appear on any of the dice, then the player loses 1 unit. Is this game fair
to the player? (Actually, the game is played by spinning a wheel that
comes to rest on a slot labelled by three of the numbers 1 through 6,
but this variant is mathematically equivalent to the dice version.)
where p, 0 p 1, is the probability that the trial is a success. A random
variable X is said to be a Bernoulli random variable (after the Swiss
mathematician James Bernoulli) if its probability mass function is given as
above for some p ∈ (0, 1).
A random variable X is said to follow binomial distribution if it assumes only
non-negative values and its pmf is given by:
where n is independent Bernoullian trails (n being finite) in which the
probability 'p' of success in any trail is constant for each trail, then q = 1-p is
the probability of failure in any trail.
The two independent constant n and p in the distribution are known as the
parameters of the distribution. 'n' is also sometimes, known as the degree of
the binomial distribution.
Any random variable which follows binomial distribution is known as
binomial variate.
Notation:
Pb. 1 : Five fair coins are flipped. If the outcomes are assumed independent,
find the probability mass function of the number of heads obtained.
Pb. 2 : It is known that screws produced by a certain company will be
defective with probability .01, independently of one another. The company
sells the screws in packages of 10 and offers a money-back guarantee that at
most 1 of the 10 screws is defective. What proportion of packages sold must
the company replace?
Sol : If X is the number of defective screws in a package, then X is a binomial
random variable with parameters (10, .01). Hence, the probability that a
package will have to be replaced is
New Section 1 Page 10
Moment Generating function of binomial distribution:
Moments of binomial distribution:
where Y is a binomial r. v. with
parameters (n−1) and p.
New Section 1 Page 11
Geometric Distribution:
Suppose that independent trials, each having a probability p, 0 < p < 1, of
being a success, are performed until a success occurs. If we let X equal the
number of trials required, then
In order for X to equal n, it is necessary and sufficient that the first n − 1
trials are failures and the nth trial is a success. Here the outcomes of the
successive trials are assumed to be independent. Since
it follows that with probability 1, a success will eventually occur. Any random
variable X whose probability mass function as above is said to be a geometric
random variable with parameter p.
Pb 1. Find the M. G. F. of the geometric distribution and hence
find its mean and variance. Does
(t) exist for all values of t ?
New Section 1 Page 12
Poisson Random Variable:
A random variable X that takes on one of the values 0, 1, 2, ... is said to
be a Poisson random ariable with parameter λ if, for some λ > 0,
The above function is called probability mass function, since
Pb2. Suppose that the probability that an item produced by a certain machine will be
defective is 0.1. Find the probability that a sample of 10 items will contain at most 1
defective item.
Sol.
whereas the Poisson approximation yields the value
Poisson approximation of Binomial distribution.
The Poisson random variable has a tremendous range of applications in
diverse areas because it may be used as an approximation for a
binomial random variable with parameters (n, p) when n is large and p
is small enough so that np is of moderate size. To see this, suppose that
X is a binomial random ariable with parameters (n, p), and let λ = np.
Then
Find the M. G. F. of the Poisson distribution
and hence find its mean and variance. Does
(t) exist for all values of t ?
Some examples:
1. The number of misprints on a page (or a group of pages) of a book
2. The number of people in a community who survive to age 100.
3. The number of wrong telephone numbers that are dialled in a day.
4. The number of packages of dog biscuits sold in a particular store each day.
5. The number of customers entering a post office on a given day.
6. The number of vacancies occurring during a year in the federal judicial system.
7. The number of α-particles discharged in a fixed period of time from some radioactive
material
Pb1. Suppose that the number of typographical errors on a single page of this
book has a Poisson distribution with parameter λ = 1/2 . Calculate the
probability that there is at least one error on this page.
Sol. Letting X denote the number of errors on this page, we have
New Section 1 Page 13
Problems based on Binomial, Poisson and Geometric random variables
Q1: An urn contains N white and M black balls. Balls are
randomly selected, one at a time, until a black one is
obtained. If we assume that each ball selected is replaced
before the next one is drawn, what is the probability that
(a) exactly n draws are needed? (b) at least k draws are
needed?
Q2: Two cricket players, Markus and Dean, decide to throw balls
at a wicket, in alternate fashion, starting with Markus. The
winner is the player who is first to hit the wicket. The probability
that Markus hits the wicket is 0.2 for any of his throws. The
probability that Dean hits the wicket is p for any of his throws. If
Markus throws first, the probability he wins the game is 5/13.
Determine the value of p.
Q3: Suppose the number of hits a web site receives in any time interval is a Poisson random variable. A particular site
gets on average 5 hits per second.
a) What is the probability that there will be no hits in an interval of two seconds?
Solution: X
Poisson(2 × 5)
P(X = k) =
P(X = 0) =
=
b) What is the probability that there is at least one hit in an interval of one second?
Solution: = 5, X
Poisson(5), P(X ≥ 1) = 1 - P(X = 0) = 1 -
New Section 1 Page 14
=1-
Q3: A company sells LED bulbs in packages of 20 for $25. From past records, it known's that a bulb will
be defective with probability 0.01. The company agrees to pay a full refund if a customer finds more
than 1 defective bulb in a pack. If the company sells 100 packs, how much should it expect to refund ?
New Section 1 Page 15
Uniform distributions
Pb.1 If X is uniformly distributed over (0, 10), calculate the probability that
(a) X < 3, (b) X > 6, and (c) 3 < X < 8.
X is a uniform random ariable on the inter al (α, β) if the probability density
function of X is given by
Pb.2 Buses arrive at a specified stop at 15-minute intervals starting at 7 A.M. That is,
they arrive at 7, 7:15, 7:30, 7:45, and so on. If a passenger arrives at the stop at a time
that is uniformly distributed between 7 and 7:30, find the probability that he waits
(a) less than 5 minutes for a bus; (b) more than 10 minutes for a bus.
Sol. Let X denote the number of minutes past 7 that the passenger arrives at the stop.
Thus X is a uniform random variable over the interval (0, 30) and hence its pdf is given
by
It follows that the passenger will have to wait less than 5 minutes if (and only if) he
arrives between 7:10 and 7:15 or between 7:25 and 7:30. Hence, the desired
probability for
(a)
Similarly, he would have to wait more than 10 minutes if he arrives between 7 and
7:05 or between 7:15 and 7:20, so the probability for
(b)
New Section 1 Page 16
Exponential distributions
Pb2. Suppose that the length of a phone call in minutes is an exponential
random ariable with parameter λ = 1/10 . If someone arri es immediately
ahead of you at a public telephone booth, find the probability that you will
have to wait (a) more than 10 minutes; (b) between 10 and 20 minutes
A continuous random variable whose probability density function is given, for some
λ > 0, by
is said to be an exponential random variable (or, more simply, is said to be
exponentially distributed) with parameter λ.
The cumulative distribution function F(x) of an exponential random variable is
given by
Note that:
Pb1. Let X be an exponential random ariable with parameter λ.
Calculate (a) E[X] and (b) Var(X) (c) MGF
One of the most important joint distributions is the bivariate normal
distribution. We say that the random variables X, Y have a bivariate
normal distribution if, for constants , , σx > 0, σy > 0, −1 <ρ< 1,
their joint density function is given, for all −q < x, y < q, by
We note from this derivation that for the exponential distribution, M(t) is
defined only for alues of t less than λ. Differentiation of M(t) yields
New Section 1 Page 17
Gamma distributions
random ariable is said to ha e a amma distribution with parameters (α, λ),
λ > 0, α > 0, if its density function is i en by
where
Pb 1. Let X be a amma random ariable with parameters α and λ.
Calculate (a) E[X] (b) Var(X) and (c) MGF.
, called the gamma function, is defined as
For inte ral alues of α, say, α = n, we obtain
Note that when n = 1, this distribution reduces to the exponential distribution.
New Section 1 Page 18
Normal distribution, mean and variance of normal distribution
A random variable is said to have a Normal distribution, with parameters μ
(mean) and (variance) if the probability density function of X is given by
This normal density function is a bell-shaped curve that is symmetric about
μ.
(a) μ = 0, σ = 1
(b) arbitrary μ,
The normal distribution was introduced by the French mathematician
Abraham DeMoivre in 1733, who used it to approximate probabilities
associated with binomial random variables when the binomial parameter n
is large. This result was later extended by Laplace and others and is now
encompassed in a probability theorem known as the central limit theorem,
which we will discuss later. The central limit theorem, one of the two most
important results in probability theory, gives a theoretical base to the often
noted empirical observation that, in practice, many random phenomena
obey, at least approximately, a normal probability distribution. Some
examples of random phenomena obeying this behaviour are the height of a
man or woman, the velocity in any direction of a molecule in gas, and the
error made in measuring a physical quantity.
Important Observation:
An important fact about normal random variables is that
if X is normally distributed with parameters μ and , then
Y = aX + b is normally distributed with parameters aμ + b and
.
To prove this statement, suppose that a > 0. (The proof when a < 0 is similar.)
Let
denote the cumulative distribution function of Y. Then
To prove that f(x) is indeed a probability density function, we need to show
that
where
is the cumulative distribution function of X. By differentiation, the
density function of Y is the
Proof:
which shows that Y is normal with parameters aμ + b and
New Section 1 Page 19
.
• An important implication of the preceding result is that if X is normally
distributed with parameters μ and , then Z = (X − μ)/σ is normally
distributed with parameters 0 and 1. Such a random variable (Z) is
said to be a standard, or a unit, normal random variable.
• Given any normal r.v. X, we can scale it and look at it as a r.v. being
symmetry with respect to zero.
Observation 2. Cumulative distribution function of a standard normal
random variable is:
Pb1. Find X and ar(X) when X is a normal random ariable with parameters μ
and .
Sol1. Let us start by finding the mean and variance of the standard normal
random variable Z = (X − μ)/σ. We have
If Z is a standard normal random variable, then
Integration by parts (with u = x and v =
) now gives
Since Z = (X − μ)/σ is a standard normal random variable
whenever X is normally distributed with parameters μ and , it
follows that the distribution function of X can be expressed as
Pb 2. If X is a normal random ariable with parameters μ = 3 and
= 9, find (a) P{2 < X < 5}; (b) P{X > 0}; (c) P{|X − 3| > 6}.
Since Z = (X − μ)/σ which implies X = μ + σZ, the precedin yields the results
New Section 1 Page 20
DeMoivre-Laplace limit theorem:
If Sn denotes the number of successes that occur in n independent
trails with the success probability p, then for any a < b,
Note that we now have two possible approximations to binomial
probabilities:
1. The Poisson approximation, which is good when n is large and p is
small, and
2. The normal approximation, which can be shown to be quite good
when npq is large.
Pb1. Let X be the number of times that a fair coin that is flipped 40
times lands on heads. Find the probability that X = 20. Use the normal
approximation and then compare it with the exact solution.
Sol. Note that Binomial is discrete r. v. and Normal is continuous r.v.,
it is best to write
before applying the normal approximation (this is called the
continuity correction). Since
The Normal approximation to the Binomial Distribution:
An important result in probability theory known as the DeMoivreLaplace limit theorem states that when n is large, a binomial random
variable with parameters n and p will have approximately the same
distribution as a normal random variable with the same mean and
variance as the binomial. This result was proved originally for the
special case of p = 1/2 by DeMoivre in 1733 and was then extended to
eneral p by Laplace in 1812. It formally states that if we “standardize”
the binomial by first subtracting its mean np and then dividing the
result by its standard deviation np(1 − p), then the distribution function
of this standardized random variable (which has mean 0 and variance 1)
will converge to the standard normal distribution function as n→ .
New Section 1 Page 21
Pb 2. The ideal size of a first-year class at a particular college is 150
students. The college, knowing from past experience that, on the
average, only 30 percent of those accepted for admission will actually
attend, uses a policy of approving the applications of 450 students.
Compute the probability that more than 150 first-year students attend
this college.
Sol.
If X denotes the number of students who attend, then X is a binomial
random variable with parameters n = 450 and p = 0.3. Using the
continuity correction, we see that the normal approximation yields
Pb 4. An examination is frequently regarded as being good (in the sense
of determining a valid grade spread for those taking it) if the test scores of
those taking the examination can be approximated by a normal density
function. (In other words, a graph of the frequency of grade scores should
have approximately the bell-shaped form of the normal density.) The
instructor often uses the test scores to estimate the normal parameters μ
and and then assigns the letter grade A to those whose test score is
reater than μ + σ, B to those whose score is between μ and μ + σ, C to
those whose score is between μ − σ and μ, D to those whose score is
between μ − 2σ and μ − σ, and F to those getting a score below μ − 2σ.
(This strategy is sometimes referred to as grading “on the curve.”)
Since
Hence, less than 6 percent of the time do more than 150 of the first
450 accepted actually attend.
Pb 3. To determine the effectiveness of a certain diet in reducing the amount of
cholesterol in the bloodstream, 100 people are put on the diet. After they have
been on the diet for a sufficient length of time, their cholesterol count will be taken.
The nutritionist running this experiment has decided to endorse/certify the diet if at
least 65 percent of the people have a lower cholesterol count after going on the
diet. What is the probability that the nutritionist endorses the new diet if, in fact, it
has no effect on the cholesterol level?
Sol.
Assume: Diet has no effect on the cholesterol count.
By chance: ach person’s count will be lower than it was before the diet with
probability 1/2 .
R.V. : X is the number of people whose count is lowered.
then the probability that the nutritionist will endorse the diet when it actually has
no effect on the cholesterol count is
it follows that approximately 16 percent of the class will receive an
A grade on the examination, 34 percent a B grade, 34 percent a C
grade, and 14 percent a D grade; 2 percent will fail.
New Section 1 Page 22
Markov’s inequality
Note: The importance of Marko ’s and Chebyshe ’s inequalities is
that they enable us to derive bounds on probabilities when only the
mean or both the mean and the variance of the probability
distribution are known. Of course, if the actual distribution were
known, then the desired probabilities could be computed exactly and
we would not need to resort to bounds.
Proposition 1.: If X is a random variable that takes only nonnegative
values, then for any value a > 0
Pb 1. Suppose that it is known that the number of items produced in a
factory during a week is a random variable with mean 50.
a) What can be said about the probability that this week’s production
will exceed 75?
b) If the ariance of a week’s production is known to equal 25, then
what can be said about the probability that this week’s production
will be between 40 and 60?
Sol. Let X be the number of items that will be produced in a week.
a) By Marko ’s inequality
b) By Chebyshe ’s inequality,
Chebyshev’s inequality (Corollary of Markov’s inequality)
Proposition 2. If X is a random ariable with finite mean μ and ariance
,
then for any value k > 0,
Since (X μ) is a nonnegative random variable, we can apply
Marko ’s inequality (with a = ) to obtain
New Section 1 Page 23
Joint Distribution Functions
Thus far, we have concerned ourselves only with probability distributions
for single random variables. However, we are often interested in
probability statements concerning two or more random variables. In
order to deal with such probabilities, we define, for any two random
variables X and Y, the joint cumulative probability distribution function
of X and Y by
All joint probability statements about X and Y can, in theory, be answered
in terms of their joint distribution function.
Case1. When X and Y are discrete random variables, with X taking on
one of the values , i 1, and Y one of the values , j 1, it is convenient
to define the joint probability mass function of X and Y by
Pb. 1. Suppose that 3 balls are randomly selected from an urn
containing 3 red, 4 white, and 5 blue balls. If we let X and Y denote,
respectively, the number of red and white balls chosen, then find
a) the joint probability mass function of X and Y
b) the probability mass function of X (marginal probability mass
functions of X)
c) the probability mass function of Y (marginal probability mass
functions of Y)
Sol.
The joint probability mass function of X and Y, p(i, j) = P{X = i, Y = j}, is
obtained by noting that X = i, Y = j if, of the 3 balls selected, i are red, j
are white, and 3 − i − j are blue. Because all subsets of size 3 are
equally likely to be chosen, it follows that
Using that the event {X = x} is the union of the mutually exclusive events
{X = x, Y = }, j 1, it follows that the probability mass function of X
(marginal probability mass functions of X) can be obtained from the
joint probability mass function by
These probabilities can most easily be expressed in tabular form
Similarly, the probability mass function of Y (marginal probability mass
functions of Y) is obtained from
marginal probability
mass functions of Y
New Section 1 Page 24
marginal probability
mass functions of X
Pb.2. Suppose that 15 percent of the families in a certain community
have no children, 20 percent have 1 child, 35 percent have 2 children,
and 30 percent have 3. Suppose further that in each family each child is
equally likely (independently) to be a boy or a girl. If a family is chosen at
random from this community, then B, the number of boys, and G, the
number of girls, in this family will have
a) the joint probability mass function of B and G
b) the probability mass function of B (marginal probability mass functions
of B)
c) the probability mass function of G (marginal probability mass functions
of G)
Pb 3. The joint density function of X and Y is given by
Compute (a) P{X > 1, Y < 1}, (b) P{X < Y}, and (c) P{X < a}.
Sol.
Case2. We say that X and Y are jointly continuous if there exists a
function f(x, y), defined for all real x and y,
The function f(x, y) is called the joint probability density function of X
and Y.
Here, the probability density function of X (marginal density functions
of X) is given by:
Similarly, the probability density function of Y (marginal density
functions of Y) is given by
New Section 1 Page 25
Independent Random Variables
Pb4. The joint density of X and Y is given by
The random variables X and Y are said to be independent if, for any
two sets of real numbers A and B,
P{X ∈ A, Y ∈ B} = P{X ∈ A}P{Y ∈ B}
In other words, X and Y are independent if, for all A and B, the events
= {X ∈ A} and
= {Y ∈ B} are independent.
Find
a) the density function of the random variable X/Y.
b) the density function of the random variable X.
c) the density function of the random variable Y.
It can be shown by using the three axioms of probability that the above
equation will follow if and only if, for all a, b,
P{X
Sol. a) We start by computing the distribution function of X/Y. For a > 0
a, Y
b} = P{X
a}P{Y
b}
Hence, in terms of the joint distribution function F of X and Y, X and Y
are independent if
F(a, b) =
(a) (b) for all a, b
X and Y are joint discrete
random variables
X and Y are joint continuous
random variables
Differentiation shows that the density function of X/Y is given by
Proposition The continuous (discrete) random variables X and Y are
independent if and only if their joint probability density (mass)
function can be expressed as
b)
(x, y) = h(x) g(y)
−
<x<
,−
<y<
Pb 1: If the joint density function of X and Y is
f(x, y) = 6e−2xe−3y
if 0 < x < , 0 < y <
and is equal to 0 outside this region, are the random variables independent?
What if the joint density function is
f(x, y) = 24xy
if 0 < x < 1, 0 < y < 1, 0 < x + y < 1
and is equal to 0 otherwise?
c)
New Section 1 Page 26
Pb 2 Suppose that n + m independent trials having a common probability
of success p are performed. If X is the number of successes in the first n
trials, and Y is the number of successes in the final m trials, then X and Y
are independent? If yes, find P{X = x, Y = y} . If Z is the total number of
successes in the n + m trials, then X and Z are independent?
Sol. Yes, X and Y are independent since knowing the number of successes
in the first n trials does not affect the distribution of the number of
successes in the final m trials (by the assumption of independent trials).
In fact, for integral x and y,
In contrast, X and Z are dependent.
Pb 3. A man and a boy decide to meet at a certain location. If each of
them independently arrives at a time uniformly distributed between 12
noon and 1 P.M., find the probability that the first to arrive has to wait
longer than 10 minutes.
Sol. If we let X and Y denote, respectively, the time past 12 that the man
and the boy arrive, then X and Y are independent random variables, each
of which is uniformly distributed over (0, 60). Then
The desired probability, P{X + 10 < Y} + P{Y + 10 < X}, which, by symmetry,
equals 2P{X + 10 < Y}, is obtained as follows:
New Section 1 Page 27
Conditional Distributions:
For any two events E and F, the conditional probability of E given F is defined, provided that P(F) > 0, by
If X and Y are discrete random variables
If X and Y are Joint continuous random variables
1. The conditional probability mass function of X given that Y = b, by
1. The conditional probability density function of X given that Y = y, by
2. The conditional probability distribution function of X given that Y = b is
defined, for all b such that (b) > 0
Notes:
1. In other words, the definitions are exactly the same as in the unconditional
case, except that everything is now conditional on the event that Y = b.
2. If X is independent of Y, then the conditional mass function and the
distribution function are the same as the respective unconditional ones.
This follows because if X is independent of Y, then
New Section 1 Page 28
2. The conditional probability distribution function of X given that Y =
y is defined by,
Notes:
1. If X and Y are independent continuous random variables, the
conditional density of X given that Y = y is just the unconditional
density of X. This is so because, in the independent case,
Pb. 1. Suppose that p(x, y), the joint probability mass function of X and Y,
is given by p(0, 0) = 0.4, p(0, 1) = 0.2, p(1, 0) = 0.1, p(1, 1) = 0.3.
Calculate
a). the conditional probability mass function of X given that Y = 1.
b). the conditional probability distribution function of X given that Y=1.
Sol. a)
Pb. 2 The joint density of X and Y is given by
Compute
a). the conditional density of X given that Y = y, where 0 < y < 1.
b). the conditional probability distribution function of X given that Y=y,
where 0 < y < 1.
Sol. a). For 0 < y < 1, we have
b).
b).
New Section 1 Page 29
Pb 3. Suppose that the joint density of X and Y is given by
Conditional expectation:
Find P{X > 1|Y = y}.
Sol. We first obtain the conditional density of X given that Y = y
Discrete
Continuous
1. If X and Y are jointly discrete
random variables, then the
conditional probability mass
function of X, given that Y = b, is
defined for all b such that P{Y =
b} > 0, by
1. If X and Y are jointly continuous
with a joint probability density
function f(x, y), then the
conditional probability density
of X, given that Y = y, is defined
for all values of y such that
(y) > 0 by
Then the conditional expectation
of X given that Y = b, for all values
of b such that (b) > 0, by
Then the conditional expectation
of X, given that Y = y, by
provided that
(y) > 0.
Remark: Just as conditional probabilities satisfy all of the properties of
ordinary probabilities, so do conditional expectations satisfy the
properties of ordinary expectations. For instance, such formulas as
in the discrete case
in the continuous case
and
remain valid. As a matter of fact, conditional expectation given that Y = y
can be thought of as being an ordinary expectation on a reduced sample
space consisting only of outcomes for which Y = y.
New Section 1 Page 30
Computing Expectations by Conditioning
Pb 1. If X and Y are independent binomial random variables with
identical parameters n and p, calculate the conditional expected value of
X given that X + Y = m.
Sol. Let us first calculate the conditional probability mass function of X
given that X + Y = m. Hence, X + Y is a binomial random variable with
parameters 2n and p. For k min(n, m),
Let us denote by E[X|Y] that function of the random variable Y whose
value at Y = y is E[X|Y = y]. Note that E[X|Y] is itself a random variable. An
extremely important property of conditional expectations is given by the
following proposition.
Proposition: The Conditional Expectation Formula
E[X] = E[E[X|Y]]
From the above equation:
Discrete R.V
Continuous R.V.
Pb.1 A miner is trapped in a mine containing 3 doors. The first door leads
to a tunnel that will take him to safety after 3 hours of travel. The second
door leads to a tunnel that will return him to the mine after 5 hours of
travel. The third door leads to a tunnel that will return him to the mine
after 7 hours. If we assume that the miner is at all times equally likely to
choose any one of the doors, what is the expected length of time until he
reaches safety?
Pb2. Suppose that the joint density of X and Y is given by
Sol. Let X denote the amount of time (in hours) until the miner reaches
safety, and let Y denote the door he initially chooses. Now,
Compute E[X|Y = y].
Sol. We start by computing the conditional density
Hence, the conditional distribution of X, given that Y = y, is just the
exponential distribution with mean y. Thus
To understand why Equation (1) is correct, consider, for instance, E[X|Y =
2] and reason as follows: If the miner chooses the second door, he spends
5 hours in the tunnel and then returns to his cell. But once he returns to
his cell, the problem is as before; thus, his expected additional time until
safety is just E[X]. Hence, E[X|Y = 2] = 5 + E[X]. The argument behind the
other equalities in Equation (4) is similar. Hence,
New Section 1 Page 31
Transformation of random variables
Objective of this topic is to find the pmf of the variables using Transformation
Method such as:
Pb. 4. Let X be the number of heads when two coins are tossed. Find the
probability distribution of Y =
Sol.
Pb1. Let X be a r.v with Prob. distribution
Find the Prob. Distribution of Y = 2X-1?
Pb2. Let X be the number of heads when three coins are tossed. Find the
probability distribution of Y =
Pb3. Let X be a binomial r. v. with the density function
Find the Prob. Distribution of Y =
Sol 3.
?
Transformation Method of one Variables:
1. Let X be a discrete r.v with pmf p(x). If the transformation y = g(X) is
defined for all values within the range of X, then the pdf of Y = g(X) is
obtained by consisting two cases:
Case1. When the relation Y = g(X) is one-to-one.
Case2. When the relation Y = g(X) is not one-to-one.
Pb. 5. Let X be a geometric r. v. with the probability distribution
Case1. When the relation Y = g(X) is one-to-one.
New Section 1 Page 32
Case2. When the relation Y = g(X) is not one-to-one.
Pb6. Let X be the number of heads when two coins are tossed. Find the
probability distribution of Y =
Pb 2. If pdf of r.v X is given by
Pb7. Let X be the number of heads when three coins are tossed. Find the
probability distribution of Y =
2. Let X be a continuous r. v. with pdf f(x). If the transformation y = v(x) is
• continuous differentiable and
• either non-decreasing or non-increasing
for all values within the range of X for which f(x) 0, then the ppdf of Y = v(X)
is given by
Remark: When the equation y =v(x) has a finite number of solutions, say
, .…
we can find the pdf of Y by
,
?
Pb 1. If X is a uniform r. v. over (0, 1). Find the distribution function of Y =
Sol.
Here the transformation Y =
is continuous differentiable and increasing.
New Section 1 Page 33
Find the pdf of Y =
Sol.
?
Pb 3. If X is normal r.v with mean 0 and variance
, find the pdf of Y =
?
Pb 6. If X is a continuous r.v with pdf
then find the pdf of Y = 2X-3 ?
Ans.
Pb 7. If X is a r.v uniformly distributed in (0, 1), then
find the pdf of Y= -2
?
Ans.
Pb 8. If X is a r.v uniformly distributed in (-1, 1), then
find the pdf of Y= sin ( )?
Ans.
Pb 4. If X is a r.v uniformly distributed in (0, 1), then find the pdf of Y = sin X?
Ans.
Pb 9. If X is a r.v with pdf
then find the pdf of Y=
?
Ans.
Pb 5. If X is a r.v with CDF as F(x), show that the r. v. Y = F(X) is uniformly
distributed in (0, 1).
Sol. As y = F(x) is continuously differentiable and non-decreasing for x
(properties of CDF), so we apply transformation method.
Pb 10. If X has exponential distribution with parameter , find the
distribution of Y =
Ans.
Hence, Y follows the uniform distribution in (0, 1).
New Section 1 Page 34
Joint probability distribution of functions of random variables
(Transformation Method of two Variables:)
Pb 1. Consider two r.v X and Y such that their joint densities are defined as,
Let and
be jointly continuous random variables with joint probability
density function
. It is sometimes necessary to obtain the joint
distribution of the random variables and , which arise as functions of
and . Specifically, suppose that = ( , ) and = ( , ) for
some functions and .
Assume that the functions and satisfy the following conditions:
1. The equations = ( , ) and =
solved for and in terms of and
= ( , ), = ( , ).
Find (i) Marginal distribution function of X and Y
(ii) Conditional distribution function of X given Y = 1.
( , ) can be uniquely
, with solutions given by, say,
2. The functions and have continuous partial derivatives at all points
( , ) and are such that the 2 by 2 determinant
Marginal distribution function of X
Marginal distribution function of Y
at all points ( ,
).
Under these two conditions, it can be shown that the random variables
and are jointly continuous with joint density function given by
where
=
( ,
),
=
( ,
).
Pb. 1. Let and
be jointly continuous random variables with
probability density function
. Let = + , = - . Find the
joint density function of and
in terms of
.
Sol. Let ( , ) = + and ( , ) = - . Then
New Section 1 Page 35
Conditional distribution function of X given Y = 1.
Pb 2. Consider two r.v X and Y such that their joint densities are defined as, Pb 3. Consider two r.v X and Y such that their joint densities are
defined as,
Find (i) Marginal distribution function of X and Y
(ii) Conditional distribution function of Y given X = 3.
Find the means of X, Y, XY and X+Y.
Marginal distribution function of X
Marginal distribution function of Y
Conditional distribution function of X given Y = 1.
New Section 1 Page 36
Covariance, correlation coefficient
• In this topic, we will discuss the term Covariance, which is an important
statistical metric, for comparing the relationships between the multiple
variables.
• In probability theory and statistics, covariance is a measure of the joint
variability(variance/spread/dispersion) of the random variables.
• It is about the degree to which random variables vary similarly.
Def: For two jointly distribution real-valued r.v X and Y with finite second
moments i.e., E[ ]< the covariance is defined as
Cov(X, Y) = E[(X-
(Y-
] = E[(X- (X) Y- ( ) ]
The covariance is also sometimes denoted by
or
• The normalized version of the covariance, called as correlation
coefficient. However, shows by its magnitude the strength of the
linear relation.
What is the difference between Covariance and Correlation ?
• Covariance measures the direction of a relationship between two
variables, while correlation measures the strength of that
relationship.
• Both Covariance and Correlation are positive when the variables
move in the same direction, and negative when the variables move
in the opposite directions.
• A correlation coefficient must always be between -1 and 1, with the
extreme values indicating a strong relationship. On the other hand, a
covariance lies between - to
.
2nd Form: Cov(X, Y) = E(XY) - E(X)E(Y).
Intuitively, the covariance between X and Y indicate how the values of X
and Y move relative to each other.
• If larger the values of X tend to happen with larger values of Y then
(X- (X) Y- ( ) is positive on average. In this case, the covarince is
positive and we say X and Y are positively correlated.
• If X tends to small when Y is large, then (X- (X) Y- ( ) is negative on
average. In this case, the covarince is negative and we say X and Y are
negatively correlated.
Difference between Covariance and Variance ?
• Covariance and variance are both used to measure the distribution
of points in a data set.
• Variance is typically used in the data sets with only one variable, and
indicates how closely those data points are clustered around the
average. Whereas covariance used with more than one variable.
• Covariance measures the direction of the relationship between two
variables.
- A positive measure means that both variables tend to be high or low
at the same time.
- A negative measure means that when one variable is high, the other
tends to be low.
What does a Covariance of 0 mean ?
• A covariance of zero indicates that there is no clear directional
(linear) relationship between the variables being measured. In other
words, a high X value is equally likely to be paired with a high or low
value for Y.
• The sign of the covariance therefore shows the tendency in the linear
relationship between the variables.
• The magnitude of the covariance is not easy to interpret because it is not
normalized and hence depends on the magnitudes of the variables.
Note: It is possible that the variables have a strong curvilinear
relationship (non-linear relationship).
New Section 1 Page 37
Pb 1. Consider the series
In this case the relationship between X and Y is non-linear.
Note: If two variables are jointly normally (Bivariate) distributed
(but not if they are merely individually normal distributed),
uncorrelatedness does imply independence.
(No linear relation between X and Y)
Properties:
1. Covariance with itself.
For a random variable X;
Cov (X, X) = Var(X)
Variance is a special case of Covariance.
2. For a real-valued random variable X and a is real constant, then
Cov (X, a) = 0.
3. For two real-valued random variables X and Y,
Cov (X, Y) = Cov (Y, X).
4. If X, Y, Z are real-valued random variables then
Cov (X+Y, Z) = Cov (X, Z) + Cov (Y, Z)
5. If X and Y are real-valued random variables and a & b are real
constants then
Cov (aX, bY) = abCov (X, Y) and Cov (X+a, Y+b) = Cov (X, Y)
Uncorrelatedness and Independence
• Random variables whose covariance is zero are called uncorrelated.
• If X and Y are independent random variables, then Cov(X, Y) = 0.
This implies E(XY) = E(X)E(Y)
However, the converse is not generally true. For example see the Pb1,
where Cov(X, Y) = 0 but X and Y are not independent. That is, Y =
Pb 2. Let X be a uniform distribution in [-1, 1] and Y =
. Is the Cov(X, Y) = 0.
6. Cov (aX+bY, cX+dY) = ac Var(X) + bd Var(Y) + (ad+bc) Cov(X, Y).
7. Cov (aX+bY, cW+dV) = ac Cov(X, W) + ad Cov(X, V) + bc Cov(Y, W) + bd Cov(Y, V).
New Section 1 Page 38
Two Steps Approach:
Cov(X, Y) = E(XY) - E(X)E(Y)
Step 1.
Find Marginal distribution function of X and Y.
Discrete r. v.
Continuous r. v.
Step 2.
Find Expectation of X, Y and XY.
Discrete r. v.
Continuous r. v.
Pb1. Let X and Y be two r. v. each taking three values -1, 0, 1 and having the
joint probability distribution is
Pb2. The joint probability mass function of X and Y is given below:
Prove that X and Y have different expectations and they are uncorrelated.
Marginal distribution function of X
Find the Cov(X, Y) ?
Marginal distribution function of Y
New Section 1 Page 39
Pb3. Let X and Y be two r. v. each taking three values -1, 0, 1 and having
the joint probability distribution is
Pb6. Let X and Y be jointly distributed with correlation coefficient 1/2.
The variance of X and Y is 4 and 9. Find the Var(2X-4Y+3)?
Prove that X and Y are uncorrelated.
Pb4. If X and Y are two independent r. v. with means 5 and 10 and
standard deviations 2 and 3, respectively. Find the covariance between
3X+4Y and 3X-Y.
Pb7. Two independent r. v. X and Y have pdf's defined by
Show that X+Y and X-Y are uncorrelated.
Pb5. If X, Y and Z are uncorrelated r. v. with 0 means and standard
deviations5, 12 and 9, respectively. Let A = X+Y and B = Y+Z. Find the
covariance between A and B?
Ans. Cov(A, B) = 144.
New Section 1 Page 40
New Section 1 Page 41