SLIDE 1
11/15/16 1
Conditional distributions, conditional expectation and the law of total expectation.
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Conditional distributions Let X and Y be discrete r.v.s. Conditional probability mass function of X given that Y=y
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pX|Y (x|y) = Pr(X = x|Y = y) = Pr(X = x, Y = y) Pr(Y = y) = p(x, y) pY (y) X
x
pX|Y (x|y) =? X
x
pX|Y (x|y) = X
x
p(x, y) pY (y) = pY (y) pY (y) = 1 Conditional distributions
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pX|Y (x|y) = Pr(X = x|Y = y) If X and Y are independent Poisson random variables with respective parameters and , calculate the conditional distribution of X, given that X + Y = n. λ1 λ2 P(X = k|X + Y = n) = P(X = k, X + Y = n) P(X + Y = n) = P(X = k, Y = n − k) P(X + Y = n) = P(X = k)P(Y = n − k) P(X + Y = n) Sum of Poisson random variables is Poisson
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X ∼ Poi(λ1) Y ∼ Poi(λ2) X + Y ∼ Poi(λ1 + λ2) Show that Proof:
P(X + Y = n) =
n
X
k=0
P(X = k, Y = n − k) =
n
X
k=0
P(X = k)P(Y = n − k) =
n
X
k=0
e−λ1 λk
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k! · e−λ2 λn−k
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(n − k)! = e−(λ1+λ2)
n
X
k=0
λk
1λn−k 2
k!(n − k)! = e−(λ1+λ2) n!
n
X
k=0
n! k!(n − k)!λk
1λn−k 2
= e−(λ1+λ2) n! (λ1 + λ2)n
Conditional distributions
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If X and Y are independent Poisson random variables with respective parameters and , calculate the conditional distribution of X, given that X + Y = n. λ1 λ2 P(X = k)P(Y = n − k) P(X + Y = n) = e−λ1 λk
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k! e−λ2 λn−k
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(n−k)!
e−(λ1+λ2) (λ1+λ2)n
n!
The conditional distribution of X, given that X+Y=n is: Binomial (n, ) λ1 (λ1 + λ2) = ✓n k ◆ λk
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(λ1 + λ2)k · λn−k
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(λ1 + λ2)n−k Conditional Expectation Expected value of random variable X given event A
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E(X|A) = X
x∈Range(X)