1.2 Conditional Expectation
For discrete random variables \(X\) and \(Y\) with \(P(Y = y)>0\), the conditional distribution of \(X\) given \(Y = y\) is \(P(X = x\mid Y = y) = \frac {P(X = x, Y = y)}{P(Y = y)}\), and the conditional expectation is \[E(X\mid Y = y) = \sum _x x\,P(X = x\mid Y = y) .\] For continuous variables the sums are replaced by integrals of the conditional density \(f_{X\mid Y}(x\mid y) = \frac {f(x, y)}{f_Y(y)}\).
Example 1.2.1. Let \(X\) and \(Y\) be independent Poisson variables with means \(\lambda _1\) and \(\lambda _2\). Find the conditional distribution of \(X\) given \(X + Y = n\), and \(E(X\mid X + Y = n)\).
Solution. \(X + Y\) is Poisson with mean \(\lambda _1 + \lambda _2\). For \(r = 0, 1, \dots , n\), \begin {align*} P(X = r\mid X + Y = n) &= \frac {P(X = r)P(Y = n - r)}{P(X + Y = n)} = \frac {\frac {e^{-\lambda _1}\lambda _1^r}{r!}\cdot \frac {e^{-\lambda _2}\lambda _2^{n - r}}{(n - r)!}}{\frac {e^{-(\lambda _1 + \lambda _2)}(\lambda _1 + \lambda _2)^n}{n!}}\\ &= \binom nr\left (\frac {\lambda _1}{\lambda _1 + \lambda _2}\right )^r\left (\frac {\lambda _2}{\lambda _1 + \lambda _2}\right )^{n - r} . \end {align*}
So given the total, \(X\) is binomial with \(n\) trials and success probability \(p = \frac {\lambda _1}{\lambda _1 + \lambda _2}\), and \(E(X\mid X + Y = n) = \frac {n\lambda _1}{\lambda _1 + \lambda _2}\).
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