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Expectations
Expectations. (See also Hays, Appendix B; Harnett, ch. 3).
A. The expected value of a random variable is the arithmetic mean of that variable, i.e. E(X) = μ. As Hays notes, the idea of the expectation of a random variable began with probability theory in games of chance. Gamblers wanted to know their expected long-run winnings (or losings) if they played a game repeatedly. This term has been retained in mathematical statistics to mean the long-run average for any random variable over an indefinite number of trials or samplings.
B. Discrete case: The expected value of a discrete random variable, X, is found by multiplying each X-value by its probability and then summing over all values of the random variable. That is, if X is discrete,
μ (^) X AllX
E(X)= ∑ xp(x)=
C. Continuous case: For a continuous variable X ranging over all the real numbers, the expectation is defined by
μ (^) X
-
E(X) = ∫ xf(x)dx=
∞
∞
D. Variance of X: The variance of a random variable X is defined as the expected (average) squared deviation of the values of this random variable about their mean. That is,
V(X)= E[(X- μ =E X − μ = σ^2 x )^2 ] (^2 )^2
In the discrete case, this is equivalent to
AllX
V ( X ) σ^2 ( x μ)^2 P ( x )
E. Standard deviation of X: The standard deviation is the positive square root of the variance, i.e.
SD ( X )=σ = σ^2
F. Examples.
1. Hayes (p. 96) gives the probability distribution for the number of spots appearing on two fair dice. Find the mean and variance of that distribution.
x p(x) xp(x) (x - μx) 5 (x - μx) 5 p(x)
2 1/36 2/36 25 25/
3 2/36^ 6/36^16 32/
4 3/36 12/36 9 27/
5 4/36 20/36 4 16/
6 5/36 30/36 1 5/
7 6/36 42/36 0 0
8 5/36^ 40/36^1 5/
9 4/36 36/36 4 16/
10 3/36 30/36 9 27/
11 2/36 22/36 16 32/
12 1/36 12/36 25 25/
Σ xp(x) = 252/36 = 7 = μx. The variance σ 5 = 210/36 = 35/6 = 5 5/6. (NOTE: There is a simpler solution to this problem, which takes advantage of the independence of the two tosses.)
2. Consider our earlier coin tossing experiment. If we toss a coin three times, how many times do we expect it to come up heads? And, what is the variance of this distribution?
x p(x) xp(x) (x - μx) 5 (x - μx) 5 p(x)
0 1/8 0 2.25 2.25/
1 3/8 3/8 0.25 0.75/
2 3/8 6/8 0.25 0.75/
3 1/8 3/8 2.25 2.25/
Σ xp(x) = 1.5. So (not surprisingly) if we toss a coin three times, we expect 1.5 heads. And, the variance = 6/8 = 3/4.
PROBLEMS: HINT. Keep in mind that μX and σX are constants.
- Prove that V(X) = E[(X - μX) 5 ] = E(X 5 ) - μX 5. HINT: Rules 4, 5, and 8 are especially helpful here.
Solution.
Equation Explanation
E[(X - μ (^) X)^2 ] = Original Formula for the variance.
E( (^) X^2 - 2X μ (^) X+ μ^2 X) = Expand the square
E( (^) X^2 )-E(2 μ (^) XX)+E( μ^2 X) = Rule 8: E(X + Y) = E(X) + E(Y). That is, the expectation of a sum = Sum of the expectations
E( (^) X^2 )- 2 μ (^) XE(X)+ μ^2 X = Rule 5: E(aX) = a * E(X), i.e. Expectation of a constant times a variable = The constant times the expectation of the variable; and Rule 4: E(a) = a, i.e. Expectation of a constant = the constant
E( (^) X^2 )- u^2 X Remember that E(X) = μX, hence 2μXE(X) = 2μX 5. QED.
- Prove that V(aX) = a 5 * V(X). HINT: Rules 3 and 5 are especially helpful.
Solution. Let Y = aX. Then,
Equation Explanation
V(Y) =E(Y^2 ) - E(Y )^2 = Rule 3: V(X) = E[(X - E(X))^5 ] = E(X^5 ) - E(X)^5 = σ (^5) X, i.e. Definition of the variance
E( (^) a^2 X^2 ) − E(aX )^2 = Substitute for Y. Since Y = aX, Y^5 = a^5 X^5
a 2 E(X^2 )^ - a^2 E(X )^2^ = Rule 5: E(aX) = a * E(X), i.e. Expectation of a constant times a variable = The constant times the expectation of the variable
a 2 (E(X^2 )^ - E(X )^2 )^ = Factor out a^5
a 2 V(X) Rule 3: Definition of the variance, i.e. V(X) = E(X 5 ) - E(X) 5. QED.
- Let Z = (X - μX)/σX. Find E(Z) and V(Z). HINT: Apply rules 7b and 14.
Solution. In this problem, a = -μX, b = 1/σX.
Equation Explanation
X-
E(Z) =E
X
X ⎟⎟ ⎠
μ Definition of Z
E(X)-
X
X
μ Rule 7b: E[(a^ ±^ X) * b] = (a^ ±^ E(X)) * b.
Remember, a = -μX, b = 1/σX.
0 Remember E(X) = μX, so the numerator = 0. QED
Intuitively, the above makes sense; subtract the mean from every case and the new mean becomes zero. Now, for the variance,
Equation Explanation
X-
V(Z)= V
X
X ⎟⎟ ⎠
μ Definition of Z
*V(X) =
2 σ X
Rule 14: V(a ± bX) = b 5 * V(X) = σ (^5) bX. Remember, b = 1/σX
1 Remember, V(X) =^ σX^5 , hence^ σX^5 appears in both the numerator and denominator. QED.
NOTE: This is called a z-score transformation. As we will see, such a transformation is extremely useful. Note that, if Z = 1, the score is one standard deviation above the mean.
