Download Lecture 16: Expected Value, Variance (Geometric Dist., Independent Random Variables) and more Study notes Linear Algebra in PDF only on Docsity!
Jim Lambers Math 6A Spring Quarter 2003- Lecture 16 Notes
These notes correspond to Section 5.3 in the text.
Expected Value and Variance, cont’d
We now continue our analysis of the behavior of random variables.
The Geometric Distribution
The following probability distribution is useful in applications in which outcomes correspond to sequences of experiments that are, in some sense, “failures” until a single “success” occurs.
Definition (Geometric Distribution) A random variable X has a geometric distribution with parameter p if p(X = k) = (1 − p)k−^1 p for each positive integer k.
Theorem If X is a random variable that has a geometric distribution with parameter p, then
E(x) =
p
Example Suppose that a coin has a 1/10 chance in coming up heads. We let X be a random variable with a geometric distribution, with parameter p = 1/10, on the sample space of sequences of coin flips that end with a result of heads. Then, for each positive integer k, p(X = k) is equal to the probability that a sequence of k coin flips will have k − 1 tails and one head. Because of our choice of sample space, X(s) = k if and only if s is the sequence of k flips that ends with a result of heads. From the preceding theorem, E(X) = 1/(1/10) = 10, so we expect that 10 flips are necessary before the coin comes up heads. 2
Independent Random Variables
Definition (Independent Random Variables) Let X and Y be random variables on a sample space S. We say that X and Y are independent if
p(X(s) = r 1 ∧ Y (s) = r 2 ) = p(X(s) = r 1 ) · p(Y (s) = r 2 ),
for all s ∈ S.
The following result provides a simple way of determining whether two random variables are independent.
Theorem If X and Y are independent random variables on a sample space S, then
E(XY ) = E(X)E(Y ).
Proof Since every value r in the range of X(s)Y (s) is obtained by multiplying two values r 1 ∈ X(S) and r 2 ∈ Y (S), we have
E(XY ) =
r∈X(s)Y (s)
p(XY = r)r
r 1 ∈X(s)
r 2 ∈Y (s)
p(X(s) = r 1 ∧ Y (s) = r 2 )r 1 r 2
r 1 ∈X(s)
r 2 ∈Y (s)
p(X(s) = r 1 )r 1 · p(Y (s) = r 2 )r 2
r 1 ∈X(s)
p(X(s) = r 1 )r 1
r 2 ∈Y (s)
p(Y (s) = r 2 )r 2
r 1 ∈X(s)
p(X(s) = r 1 )r 1 E(Y )
= E(Y )
r 1 ∈X(s)
p(X(s) = r 1 )r 1
= E(X)E(Y ).
Example Let X 1 and X 2 be a random variables on a sample space S that have the same values for each outcome in S; that is, X 1 (s) = X 2 (s) for all s ∈ S. We will show that if we redefine the domain of X 1 and X 2 to be S 2 = S × S, so that X 1 ((a, b)) = a and X 2 ((a, b)) = b for each ordered pair (a, b) ∈ S 2 , then X 1 and X 2 must be independent. Let i ∈ X 1 (S), j ∈ X 2 (S), and let p 1 = p(X 1 = i), while p 2 = p(X 2 = j). It follows that p(X 1 = i ∧ X 2 = j) = p 1 p 2 = p(X 1 = i)p(X 2 = j), which implies that X 1 and X 2 are independent. 2
Variance
The expected value, by itself, is not always a useful indicator of a random variable’s behavior on its sample space, because it only indicates what number the values of the variable tend to cluster around. To have a clearer picture of the variable’s behavior, it is helpful to have a measure of how much the variable’s values deviate from the expected value. We now state a precise definition of this concept.
s∈S
(X(s)^2 − 2 E(X)X(s) + E(X)^2 )p(s)
s∈S
X(s)^2 p(s) −
s∈S
2 E(X)X(s)p(s) +
s∈S
E(X)^2 p(s)
= E(X^2 ) −
s∈S
2 E(X)X(s)p(s) + E(X)^2
s∈S
p(s)
= E(X^2 ) − 2 E(X)
s∈S
X(s)p(s) + E(X)^2 · 1
= E(X^2 ) − 2 E(X) · E(X) + E(X)^2
= E(X^2 ) − E(X)^2.
Example Let S = { 1 , 2 , 3 , 4 , 5 , 6 } and let S 2 = S × S be a sample space that represents the outcomes of rolling two six-sided dice, where each number is equally likely to be rolled for each die. Let X : S 2 → N be a random variable that assigns to each ordered pair (a, b) ∈ S 2 the number a + b, the sum of the two rolls. Then E(X) = 7, as computed in the Lecture 15 notes. We also have
E(X^2 ) =
s∈S 2
X(s)^2 p(s)
s 1 ∈S
s 2 ∈S
(s 1 + s 2 )^2 p((s 1 , s 2 ))
∑^6
s 1 =
∑^6
s 2 =
(s 1 + s 2 )^2
∑^6
s 1 =
∑^6
s 2 =
s^21 + 2s 1 s 2 + s^22
[ 6
s 1 =
s^21
s 2 =
∑^6
s 1 =
s 1
s 2 =
s 2
∑^6
s 1 =
∑^6
s 2 =
s^22
]
[
∑^6
s 1 =
s^21 + 2
∑^6
s 1 =
s 1
∑^6
s 1 =
∑^6
s 2 =
s^22
]
[
∑^6
s 1 =
s 1
]
[
]
It follows from the preceding theorem that
V (X) = E(X^2 ) − E(X)^2 =
Theorem If X and Y are independent random variables on a sample space S, then
V (X + Y ) = V (X) + V (Y ).
Proof Since X and Y are independent, E(XY ) = E(X)E(Y ). It follows that
V (X + Y ) =
s∈S
(X(s) + Y (s) − E(X + Y ))^2 p(s)
s∈S
(X(s) + Y (s) − E(X) − E(Y ))^2 p(s)
s∈S
(X(s) − E(X))^2 p(s) +
s∈S
(Y (s) − E(Y ))^2 p(s) −
s∈S
(X(s) − E(X))(Y (s) − E(Y ))p(s)
= V (X) + V (Y ) − 2
s∈S
(X(s) − E(X))(Y (s) − E(Y ))p(s)
= V (X) + V (Y ) − 2
s∈S
[X(s)Y (s) − E(X)Y (s) − E(Y )X(s) + E(X)E(Y )]p(s)
= V (X) + V (Y ) − 2
s∈S
X(s)Y (s)p(s) + 2E(X)
s∈S
Y (s)p(s) +
2 E(Y )
s∈S
X(s)p(s) − 2 E(X)E(Y )
s∈S
p(s)
= V (X) + V (Y ) − 2
[
s∈S
X(s)Y (s)p(s) − E(X)E(Y ) − E(Y )E(X) + E(X)E(Y )
]
= V (X) + V (Y ) − 2 [E(XY ) − E(X)E(Y )]
= V (X) + V (Y ).
Remark This result can be generalized to n independent random variables X 1 , X 2 ,... , Xn on a sample spaces S. Specifically,
V (X 1 + X 2 + · · · Xn) = V (X 1 ) + V (X 2 ) + · · · + V (Xn).
