Random Variables
Experiment :
Let X be a number chosen at random from 1 to 10. X is called a random
variable. The range of X (Range(X)) is {1,2,3,4,5,6,7,8,9,10}.
Experiment:
Flip a coin. Suppose that you win $1 for heads and lose $1 for tails. Let Y
be the amount that you win or lose. Y, too, is a random variable.
Range(Y) = { -1,1}
Experiment:
Flip a coin 100 times. Let Z be the number of times out of 100 that heads
appears. Z is a random variable. Range(Z)= {0,1,2,3.., 97,98,99,100}
In general,
A random variable X is a variable that can assume various values
with various probabilities.
A more technical definition might be:
A random variable associated with a sample space S is a function whose
domain is S and whose range is a set real numbers.
So for example: Flip a coin and win/lose $1. Sample space { H, T}. The random
variable Y is simply the function:
H1
T  -1.
For our purposes, the first, more intuitive, informal definition will suffice.
There are two distinct types of random variables: discrete and continuous
A discrete random variable is one whose
1. range is finite e.g. {1,2,3,4,5,6} or
2. range is infinite but can be explicitly listed. e.g. {1,2,3,4….} or
{2,4,6,8,10……} or perhaps { -1,1,-2,2,-3,3,-4,4,….}
A continuous random variable is one whose range contains an entire interval
eg. (-  , +  ) or (-1,1).
A the range of a discrete random variable contains no complete interval of real
numbers.
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For now, we will be concerned with discrete random variables.
Example:
A card is selected from a deck. If the card is an Ace you lose $100. If it is
a picture card you win $20. If it is any other card you win $5. Let X be the
amount you win or lose.
X is a discrete random variable.
The Range(x) = { -100, 5, 20} (finite)
Would you play this game?
Example:
Let X be the number of hits on some website wich occur in a given month.
Range(X) = {0,1,2,3,4,5,6,………..} ( infinite, no limit)
X is a discrete random variable.
Notation:
Uppercase letters are usually used to denote random variables.
Every discrete random variable has:
1. a probability distribution
2. an expected value or mean value
3. a variance and standard deviation
Definition:
Let X be a discrete random variable with range { x1,x2,x3,…}
The probability distribution (or probability mass function [pmf] or probability
function [pf] ) of X is a function p such that
p(x) = Prob(X = x) for each xi in Range(X)
We often describe the probability distribution of a random variable using a table
or graph. An example should clarify the definition.
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Example: Roll a die. Let X be the number of spots. Range(X) = {1,2,3,4,5,6}
Probability function:
x
1
2
3
4
5
6
p(x)
1/6
1/6
1/6
1/6
1/6
1/6
Graphically:
Or
p(x) = 1/6 for x = 1,2,3,4,5,6
The probability distribution of a random variable tells us what to “expect” of the
random variable. Theoretically, we expect to roll a 1 one-sixth of the time; we
expect to roll 2 one-sixth of the time etc. You might think of the probability
distribution of a random variable as a description of its theoretical behavior.
Expected value or mean of a random variable:
Suppose that we roll a die many times and compute the average of all the
values rolled Theoretically, what would we expect?
We would expect to roll:
1 – 1/6th of the time
2 – 1/6th of the time
3 – 1/6th of the time
4 – 1/6th of the time
5 – 1/6th of the time
6 – 1/6th of the time
So the “average value” or “expected value” might be:
1*(1/6) + 2*(1/6) + 3*(1/6) + 4*(1/6) + 5*(1/6) + 6*(1/6) = 3.5
This is a weighted average.
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For example, if we rolled the die 60000 times, in theory we expect
1—10000 times,
2—10000 times,
3—10000 times,
4—10000 times,
5—10000 times,
6---10000 times
The average of these 60000 numbers is 3.5.
Thus, if random variable X is the number of spots that appear on a roll of a
single die, we say that
The Expected Value or mean of X is 3.5
i.e E(X) = 3.5 or μ = 3.5 or μX = 3.5
.
You can think of the expected value (i.e. mean ) of a random variable as a
theoretical average or a long run average. If we roll a die 100000000 times , in
theory the values should average out to 3.5.
Definition:
If X is a discrete random variable with
probability function p(x) and
range(X) = ( x1,x2, x3, x4,…}
then E(X) defined as
E(X) = μ = x1p(x1) + x2p(x2) + x3p(x3) + x4p(x4)…
Notice that each value in Range(X) is weighted by its probability.
Example: Flip a fair coin.
Win $1 for H, lose $1 for tails. Let X be the amount that you win or lose.
Range(X) = { -1,1}
Probability distribution of X:
E(X) = μ = 1*(1/2) + (-1)*(1/2) = 0
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So, in the long run we expect to break even.
Example: Roulette
Rules: Bet $1 on Red. If a red number appears you win $1 otherwise lose $1.
Let X be the amount you win or lose. Range(X) = {-1,1}
Note: There are 18 Red numbers, 18 Black numbers and also 0 and 00
so, on average, you win 18 times out of 38 and lose 20 times out of 38
Probability function:
E(X) = 1*(18/38) + (-1)*(20/38) = -2/38 ~ -.05
The game is not fair. On the AVERAGE you would expect to lose 5 cents/game.
So the casino take a nickel for every dollar bet in this way.
A fair game is one for which the expected value is 0.
Example: A card is selected from a deck. If the card is an Ace you lose $100. If
it is a picture card you win $20. If it is any other card you win $5. Let X be the
amount you win or lose.
X is a discrete random variable. The Range(x) = { -100, 5, 20}
The probability function:
E(X) = -100*4/52 + 20*12/52 +5*36/52 = -400/52 + 420/52 = 20/52 = .38
The expected value is positive. The game favors the player.
(If, instead, you lost $110 for an Ace, E(X) would be -.38 ---unfavorable)
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Variance of a random variable:
The variance of a random variable X is a measure of the average amount of
(squared) deviation form the mean μ ( or E(X)).
If X is a random variable with E(X) = μ and Range(X) = {x1,x2,x3,….} then
Var(x)= (x1- μ)2*p(x1) + (x2- μ)2*p(x2) + (x3- μ)2*p(x3) + (x4- μ)2*p(x4)…
The standard deviation of a random variable X is the square root of Var(X).
Notation : Var(X) : σ2
Standard deviation :σ
Another way to denote variance is Var(X) = σ2 = E[ (X- μ)2]
Example: Roll a die. Let X be the number of spots.
Range(X) = {1,2,3,4,5,6}
Probability function:
x
1
2
3
4
5
6
p(x)
1/6
1/6
1/6
1/6
1/6
1/6
E(X) = μ = 3.5
Var(X) = σ2 = (1-3.5)2*1/6 + (1-3.5)2*1/6+ (1-3.5)2*1/6+ (1-3.5)2*1/6+ (1-3.5)2*1/6+ (1-3.5)2*1/6
= 2.92
σ=
2.92 = 1.707
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Shortcut formula for variance:
With some ordinary algebra you can derive the following shortcut formula for σ 2:
Var(X) = σ2 = E(x2) - [E(x)]2
Example using this version :
X
X2
p(x)
1
2
3
4
5
6
1
4
9
16
25
36
1/6
1/6
1/6
1/6
1/6
1/6
E(X) = 1*(1/6) + 2*(1/6) + 3*(1/6) + 4*(1/6) + 5*(1/6) + 6*(1/6) = (19)/6 = 3.5
E(X2) = 1*(1/6) + 4*(1/6) + 9*(1/6)+16*(1/6)+25*(1/6)+36*(1/6) = (91)/6 = 15.167
Var(X) = σ2 = E(x2) - [E(x)]2 = 15.167 – (3.5)2 = 2.92
Summary:
If X is a discrete random variable with
o Range(X) = {x1, x2, x3,…} and
o probability function p(x)
then
E(X) = μ = x1p(x1) + x2p(x2) + x3p(x3) + x4p(x4)…
Var(x)= σ2 = (x1- μ)2*p(x1) + (x2- μ)2*p(x2) + (x3- μ)2*p(x3) + (x4- μ)2*p(x4)…
or
Var(X) = σ2 = E(x2) - [E(x)]2
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Example:
Roll 2 dice. X is the number of Spots.
Range(X) = {2,3,4,5,6,7,8,9,10,11,12}
Probability function:
x
p(x)
2
3
4
5
6
7
8
9
10
11
12
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
Picture:
8
x
x2
p(x)
2
3
4
5
6
7
8
9
10
11
12
4
9
16
25
36
49
64
81
100
121
144
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
E(X) = 2*1/36+3*2/36+4*3/36*5*4/36+6*5/36+7*6/36+8*5/35+9*4/36+10*3/36+11*2/36+12*1/36
=7
E(X2) = 4*1/36+9*2/36+16*3/36+25*4/36+36*5/36+49*6/36+…..+144*1/36 = 1974/36 = 54.83
Var(X) = E(X2) – E(X)2 = 54.83 - 72 = 5.83 (using the shortcut formula)
Thus μX = 7 and σX2 = 5.83 and σX =
583
. = 2.41
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An important property of E(X):
Suppose X is a number chosen at random between 1 and 10 inclusive.
Range(X) = {1,2,3,4,5,6,7,8,9,10}
Probability distribution: p(x) = 1/10 for all x in Range(X) i.e
X
p(x)
1
2
3
4
5
6
7
8
9
10
1/10
1/10
1/10
1/10
1/10
1/10
1/10
1/10
1/10
1/10
So
E(X) = 1*.1 + 2*.1 + 3*.1 + 4*.1 + 5*.1 + 6*.1 + 7*.1 + 8*.1 + 9*.1 + 10*.1 = 5.5
Now define another random variable Y = X + 5.
The Range(Y) = { 6,7,8,9,10,11,12,13,14,15,16}
Probability function for Y is
X
p(x)
6
7
8
9
10
11
12
13
14
15
1/10
1/10
1/10
1/10
1/10
1/10
1/10
1/10
1/10
1/10
Calculate E(Y),:
E(Y) = 6*.1 + 7*.1 + 8*.1 + 9*.1 + 10*.1 + 11*.1 + 12*.1 + 13*.1 +14*.1 + 15*.1
= 10.5
Notice that E(Y) = E(X + 5) = 10.5 -- which is E(X) + 5
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Similarl, let Z = 10X . Then, Range(Z) = {10, 20, 30,40,50,60,70,80, 90,100}
The probabilities are still 1/10 for each value in the range.
E(Z) = 10*.1 + 20*.1 +30*.1 +…+ 100*.1 = 55
Thus E(Z) =E(10X) = 55 which is 10*E(X)
In general:
If X is a random variable and a and b are real numbers then
E(aX + b) = aE(X) + b
Example: Roll two dice. X is the number of spots.
Range(X) = { 2,3,4,5,6,7,8,9,10,11,12};
E(X) = 7.
Let Y = 2X + 4. (Roll the dice, multiply by 2 and add 4)
Range(Y) = { 8, 10,12,14,16,18,20,22,24,26,28}
E(Y) = E(2X + 4) = 2E(X) + 4 = 1*7 + 4 = 18
Variance:
Var(aX + b) = |a|Var(X)
Example: Let X be a number chosen at random from 1 to 6 (roll a die)
We saw E(X) = 3.5 and Var(X) = 2.92.
Let Y = 3X + 5 ( Roll a die, multiply by 3 and add 5)
Range(Y) = { 8,11,14,17,21,24}
E(Y) = E(3X + 5) = 3E(X) + 5 = 3*(3.5) + 5 = 15.5
Var(Y) = Var(3X + 5) = |3|Var(X) = 3*2.92 = 8.76
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Example:
A family has 4 children. Let X be the number of girls .
Find:
a. Range(X),
b. Distribution of X
c. E(X)
d. Var(X)
a. Range(X) = {0,1,2,3,4}
Probability distribution:
Here is the sample space:
BBBB
BBBG
BBGB
BGBB
GBBB
BBGG
BGBG
BGGB
GGBB
GBGB
GBBG
GGGB
GGBG
GBGG
BGGG
GGGG
E(X) = 0*(1/16) +1*(4/16) +2*6/16 + 3*4/16 + 4* 1/16 = 32/16 = 2
To Calculate Var(X), get E(X2):
E(X2) = 0*(1/16) +1*(4/16) +4*6/16 + 9*4/16 + 16* 1/16 = 80/16 = 5
Var(X) = E(X2) – E(X)2 = 5 – 22 = 1 i.e. σ2 = 1
σ = 5 = 2.236
Summary:
If Y = aX = b
E(Y) = E(aX + b) = aE(X) + b
Var(Y) = Var(aX + b) = |a|Var(X)
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