Download Expectation and Discrete Random Variables - Mr-Kuijpers-Math

Expectation and Probability Distributions
This is the concept of the expected mean of a random variable. It is
closely related to the arithmetic mean or average.
x
n
 x = E(X) =
i=1
i
P(X = x i)
A table or formula listing all possible values that a discrete variable can
take on, together with associated probabilities is called a discrete
probability distribution
e.g. Calculate the expected value of X, E(X) where X denotes the
outcomes on a die.
x
1
2
3
4
5
6
P(X = x)
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 = 3.5
e.g. Find E(T) for the probability distribution for T, where T counts the
sum when 2 dice are thrown.
t
1
2
3
4
5
6
7
8
9
10 11 12
0/36
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36 2/36 1/36
P(T=t)
E(T) = 1×0/36 + 2×1/36 + 3×2/36 + ... + 12×1/36 = 7
Note:
1.
The probability distribution lists all of the possible outcomes.
2.
The sum of the probabilities must be 1 if it is a probability
distribution
Games of Chance
A fair price for a ticket in a lottery would be one which returned the same
amount from ticket sales as was distributed in prizes
Gain = payoff – entry price
e.g. In a gambling game a man is paid $5 if he gets all heads or all tails
when three coins are tossed, and he pays out $3 if either 1 or 2
heads occur. What is his expected gain?
x
P(X = x)
$5
1/4
-$3
3/4
E(X) = 5 × ¼ + -3 × ¾ = -1
Therefore: on average man will lose $1 per toss. Game is not fair.
Variance of a Random Variable
- gives us an indication of the spread of results every time we conduct
the experiment
- the variance is the average of the squared distances from the mean.
i.e. VAR(X) = E[(X –  )2]
The formula can be simplified to:
2
2
VAR(X) = E(X ) – [E(X)]
Note: to find E(X2) square all x-values and multiply by original
probabilities.
e.g. Using the previous distribution tables, calculate VAR(X) for the die
experiment, and VAR(T) for the sum of two dice.
VAR(X) = 12×1/6 + 22×1/6 + 32×1/6 + 42×1/6 + 52×1/6 + 62×1/6 – (3.5)2
= 2.92 (2 d.p.)
VAR(T) = 12×0/36 + 22×1/36 + … + 112×2/36 + 122×1/36 – (7)2
= 5.83 (2 d.p.)
Note: The standard deviation is the square root of the variance
PROOF:

VAR(X) = E  (X – )












2


2
2
= E X – 2X + 
2


2


2


=E X



2
– 2E(X) + E  
2


=E X
– 2(E(X)) + (E(X))
=E X
– (E(X))
2
2
Linear Functions of a Random Variable
Consider the random variable:
x
1
P(X = x)
0.2
2
0.3
3
0.4
4
0.1
VAR(X) = 6.6 – 2.42
= 0.84
E(X) = 2.4
Each data value is doubled then one is subtracted (i.e. 2X – 1).
Calculate the new expected mean and variance
x
P(X = x)
E(X) = 3.8
1
0.2
3
0.3
5
0.4
7
0.1
VAR(X) = 17.8 – 3.82
= 3.36
By taking the original mean and variance, we could have obtained the
same results by using:
E(aX + b) = aE(X) + b
2
VAR(aX + b) = a VAR(X)
where a and b are constants and VAR(b) = 0
i.e. E(X) = 2×2.4 – 1
VAR(X) = 22×0.84
= 3.8
= 3.36
Note: SD(aX) = aSD(X) as the standard deviation is the square root of
the variance
e.g. X is a random variable with mean 20 and variance 4. Find the
mean and standard deviation of:
the mean
the variance
the standard
(VAR)
deviation (SD)
X+2
22
4
2
3X
60
36
6
X–1
19
4
2
2X – 1
39
16
4
-X
-20
4
2
e.g. A soup recipe requires 1200mL of milk (available in two 600mL
cartons). The standard deviation of the volume of milk in a 600mL
carton is 3mL. Calculate the standard deviation of the total volume
purchased.
Let C = C1 + C2
VAR(C) = VAR(C1 + C2)
= VAR(C1) + VAR(C2)
= 32 + 32
= 18
Therefore standard deviation = √18
Note: treat as separate containers C1 and C2 not 2C
PROOF (see workbook)
Sums of Random Variables
Consider the random variables X and Y
x
1
2
3
P(X = x)
1/3
1/3
1/3
E(X) = 2
2/3
y
P(Y = y)
VAR(X) = 2/3
2
1/3
E(Y) = 3
3
1/3
4
1/3
VAR(Y) =
Let Z = X + Y, then the distribution of Z is:
y
2
3
4
x
2
4
5
6
1
3
4
5
3
5
6
7
and the probability distribution for Z is:
z
3
4
P(Z = z)
1/9
2/9
5
3/9
E(Z) = 5
VAR(Z) = 4/3
i.e. E(Z) = E(X) + E(Y)
VAR(Z) = VAR(X) + VAR(Y)
6
2/9
7
1/9
RESULTS:
1.
E(aX + bY) = aE(X) + bE(Y)
2.
VAR(aX  bY) = a VAR(X) + b VAR(Y)
2
2
Note: You still add the variances when the variables are subtracted.
Square root the variance to get the standard deviation
You generally only square a/b terms when dealing with non-context
questions or contextual questions dealing with money (or if variables are
independent).
e.g. The “second-hand” bookstore purchases magazines and books. It
pays $5 for books and $2 for magazines.
The mean number of books purchased each day is 50 with a
standard deviation of 9 books
The mean number of magazines purchased each day is 40 with a
standard deviation of 4 books.
Calculate the mean and standard deviation of the amount they pay
each day. Let P = amount paid each day, B = books and M =
magazines.
P = 5B + 2M
PROOF: see workbook
E(P) = 5×50 + 2×40
= $330
VAR(P) = VAR(5B + 2M)
= 52×VAR(B) +22×VAR(M)
= 2089
SD(P) = $45.71