• A random variable is a variable whose values
are numerical outcomes of a random
experiment. That is, we consider all the
outcomes in a sample space S and then
associate a number with each outcome
• Example: Toss a fair coin 4 times and let
X=the number of Heads in the 4 tosses
We write the so-called probability distribution of
X as a list of the values X takes on along with
the corresponding probabilities that X takes on
those values. We know X is B(4,.5)
• Recall the figure below showing the probability
distribution of X. Eachindividual outcome has
prob=1/16, and using dbinom(0:4,4,.5) you can
find the P(X=x), where x=0,1,2,3,4
• There are two types of r.v.s: discrete and
continuous. A r.v. X is discrete if the number of
values X takes on is finite (or countably infinite).
In the case of any discrete X, its probability
distribution is simply a list of its values along with
the corresponding probabilities X takes on those
values.
Values of X: x1 x2 … xk
P(X):
p1 p2
pk
NOTE: each value of p is between 0 and 1 and all
the values of p sum to 1. We display probability
distributions for discrete r.v.s with so-called
probability histograms. The next slide shows the
Binomial probability histogram for X=# of Hs in 4
tosses of a fair coin.
The next slide gives a similar example...
• Toss a fair coin until you get the first occurrence of "H".
Let X = the number of the toss on which the first "H"
appears.
• What are the possible values of X? What are the
corresponding probabilities?
Values of X: 1
P(X)
2
3
4
5
6 …
:
X is called a geometric r.v. and in R is computed
with the dgeom, pgeom, qgeom, rgeom
functions - the d, p, q, and r stand for the same
functions we've seen before… what would the
probability histogram look like?
• A continuous r.v. X takes its values in an interval
of real numbers. The probability distribution of a
continuous X is described by a density curve,
whose values lie wholly above the horizontal
axis, whose total area under the curve is 1, and
where probabilities about X correspond to
areas under the curve.
• The first example is the random variable which randomly
chooses a number between 0 and 1 (perhaps using a
spinner). This r.v. is called the uniform random variable
and has a density curve that is completely flat!
Probabilities correspond to areas under the curve... use
the punif(x) = P(X <= x) to get the areas under the
uniform r.v.; e.g., P(.3 < X < .7) = punif(.7) - punif(.4)
A continuous random variable X takes all values in an interval.
Example: There are an infinite number of values between 0 and 1 (e.g., 0.001, 0.4,
0.0063876).
How do we assign probabilities to events in an infinite sample space?
 We use density curves and compute probabilities for intervals.
 The probability of any event is the area under the density curve
for the values of X that make up the event.
This is a uniform density curve for the variable X.
The probability that X falls between 0.3 and 0.7 is
the area under the density curve for that interval
(base x height for this density):
P(0.3 ≤ X ≤ 0.7) = (0.7 – 0.3)*1 = 0.4
X
The probability of a single point is meaningless for a
continuous random variable. Only intervals can have a
non-zero probability, represented by the area under the
density curve for that interval.
The probability of a single point is zero since
there is no area above a point! This makes
the following statement true:
Height
=1
The probability of an interval is the same whether
boundary values are included or excluded:
P(0 ≤ X ≤ 0.5) = (0.5 – 0)*1 = 0.5
P(0 < X < 0.5) = (0.5 – 0)*1 = 0.5
X
P(0 ≤ X < 0.5) = (0.5 – 0)*1 = 0.5
P(X < 0.5 or X > 0.8) = P(X < 0.5) + P(X > 0.8) = 1 – P(0.5 < X < 0.8) = 0.7
(You may use either the “OR” Rule or the “NOT” Rule...)
• The other example of a continuous r.v. that
we’ve already seen is the normal random
variable. See the next slide for a reminder
of how we’ve used the normal and how it
relates to probabilities under the normal
curve...
Continuous random variable and population distribution
The shaded area under a
density curve shows the
proportion, or %, of individuals
in a population with values of X
between x1 and x2.
Because the probability of
drawing one individual at
random depends on the
frequency of this type of
individual in the population, the
probability is also the shaded
area under the curve.
% individuals with
X such that x1 < X
< x2
Mean of a random variable
•The mean x bar of a set of observations is their arithmetic average.
•The mean µ of a random variable X is a weighted average of the
possible values of X, reflecting the fact that all outcomes might not be
equally likely.
A basketball player shoots three free throws. The random variable X is
the number of baskets successfully made (“H”).
MMM
HMM
MHM
MMH
HHM
HMH
MHH
HHH
Value of X
0
1
2
3
Probability
1/8
3/8
3/8
1/8
The mean of a random variable X is also called expected value of X.
What is the expected number of baskets made? Do the computations...
• We’ve already discussed the mean of a density curve as
being the “balance point” of the curve… to establish this
mathematically requires some higher level math… So
we’ll think of the mean of a continuous r.v. in this way.
For a discrete r.v., we’ll compute the mean (or expected
value) as a weighted average of the values of X, the
weights being the corresponding probabilities. E.g., the
mean # of Hs in 4 tosses of a fair coin is computed as:
(1/16)*0 + (4/16)*1 + (6/16)*2 + (4/16)*3 + (1/16)*4 =
(32/16) = 2.
• In either case (discrete or continuous), the
interpretation of the mean is as the long-run average
value of X (in a large number of repetitions of the
experiment giving rise to X).
• We've used mean(rbinom(1000, 5, .1)) for example to
simulate the mean of a binomial r.v. (n=5, p=.1).
• Look at the Pick 3 Lottery, like the old numbers
game…you pay $1 to play (pick a 3 digit number),
and if your number comes up, you win $500;
otherwise, you win nothing.
• What is the probability that you win (i.e., that your 3
digits match the ones chosen that night)?
• What is the probability that you lose?
• Define X = your winnings when you play "Pick
3"
possible values of X:
P(X)
:
So what is your expected winnings?
• There's also a discrete uniform r.v.: Like Table B that I've
handed out… sample(0:9, 1, replace=T), chooses 1
number from Table B; sample(0:9,100,replace=T)
chooses 100.
• Let X = the digit chosen at random from Table B
• What are the values of X? What are the corresponding
probabilities? What is the mean?
• Use the sample function to simulate this and see if you can
tell what the mean and s.d. are…
HINT: Try this:
z=numeric(1000); zz=numeric(1000)
for (i in 1:1000) {
x=sample(0:9,20,replace=T); m=mean(x); s=sd(x)
z[i] = m ; zz[i] = s }
par(mfrow=c(1,2))
hist(z) ; hist(zz)
mean(z) ; mean(zz)
• Now what if we look at means of samples of size n=20 of
these 10 digits (0,1,…,9) - what does the distribution look
like then?
Try this R code:
par(mfrow=c(1,1)) #a single plot per page
z=numeric(1000)
for (i in 1:1000) {
x=sample(0:9,20,replace=T); m=mean(x);
z[i] = m}
hist(z) ; mean(z) ; sd(z)
What's going on here? Notice especially the standard
deviation - compare the sd of this simulation of means of
samples of 20 with the sd of samples of 20… why is the sd
of the means ~ .6 while the sd of the digits is ~2.9 ?
• Is there an intuitive reason why this might be happening?
• The mathematical reason is called the Central Limit
Theorem.
•
• Central Limit Theorem: Suppose we take a large sample of
size n from a population with mean =  and sd =  (call the
sample X1, X2, … , Xn).
If Xbar = mean(X1, X2, … , Xn),
then the distribution of Xbar will look Normal with mean = 
and sd = /sqrt(n) .
• All the situations we've been looking at prior to this are
examples of the CLT:
– Binomial(n, p) with large n tends to look Normal (np, sqrt(np(1-p))
– p-hat with large n tends to look Normat( p , sqrt(p(1-p)/n) )
– means of samples of size n=20 from the digits 0:9, tend to look
Normal ( 4.5, 2.87/sqrt(20) )
• Now you try these (Hand in next time):
– Choose n=25 numbers at random from the interval [0,1] (see runif)
• simulate this choice of 25 numbers enough times (1000) to convince you
that the mean of this population of numbers from 0 to 1 is ~ .5 and the sd
is ~ .29
• so what would the CLT say about the mean of the samples of 25?
• simulate this and verify by looking at a histogram and by computing the
mean and sd of the means…