Consider the following random variables in $(\Omega, \mathfrak{F}, P)$.

  1. $X_1,X_2, X_3,...$ where $\forall n \in \mathbb{N}, \mu_n = E(X_n), \sigma_n^2 = Var(X_n) < \infty$.
  2. $N$, a discrete RV whose range is $\mathbb{N}$. Let $A_n = ({\omega: N(\omega) = n})$ and $p_n = P (N = n) = P(A_n)$. Also, $N$ is independent of each $X_n$.
  3. $\mu_N$ defined as $\mu_N (\omega) = \mu_{N(w)}, \sigma_N^2 (\omega) = \sigma^2_{N (\omega)}$, and $X_N$ defined as $X_N (\omega) = X_{N(\omega)} (\omega)$.

Interpretation: if we choose an index n at random (so $N=n$), then $X_N$ is the RV $X_n$ and $\mu_N$ is the mean of $X_n$, and $\sigma_N^2$ is the variance of $X_n$.

Prove that $E[X_N|N] = E[X_N|\sigma(N)] = \mu_N$ and $Var[X_N|N] = Var[X_N|\sigma(N)] = \sigma_N^2.$ Use these to find formulas for $E[X_N]$ and $Var[X_N]$ in terms of $\mu_n, \sigma^2_n$ and $p_n$.

My attempt:

Edit: I already proved the conditional expectation stuff. I just want to see if my formulas for $E[X_N]$ and $Var[X_N]$ in terms of $\mu_n, \sigma^2_n$ and $p_n$ are correct:

$Var[X_N] = E[Var[X_N|N]] + Var[E[X_N|N]]$

$=E[\sigma_N^2] + Var[\mu_N]$

$=E[\sigma_N^2] + E[\mu_N^2] - E[\mu_N]^2$

$=\sum_{n=1}^{\infty} \sigma_n^2 p_n + \sum_{n=1}^{\infty} \mu_n^2 p_n - (\sum_{n=1}^{\infty} \mu_n p_n)^2$

Explanation of sums and stuff: (For $E[\mu_N]$)

$E[X_N|N]$ is $\sigma(N)$-measurable. Thus, $\exists$ Borel-measurable function g such that $g(N) = E[X_N|N]$

$\to E[E[X_N|N]] = E[g(N)]$

$\to E[X_N] = E[g(N)]$

$\to E[X_N] = \sum_{n=1}^{\infty} g(n) p_n$

where $g(n) = E(X_N | N = n) = E[X_n 1_{A_n}]/p_n = E[X_n]p_n/p_n = E(X_n) = \mu_n$

$E[X_N] = E[\mu_N] = \sum_{n=1}^{\infty} \mu_n p_n$

Similarly, I came up with expressions for $E[X_N^2]$ and $Var[X_N]$.

Are these expressions correct?


2 Answers 2


Since $$ X_N=\sum_{n\geqslant 1} X_n\mathbf{1}_{N=n} $$ holds pointwise, we have $$ {\rm E}[X_N]=\sum_{n\geqslant 1}\mu_np_n $$ agreeing with your expression. Similarly, $$ {\rm E}[X_N^2]=\sum_{n\geqslant 1}{\rm E}[X_n^2]p_n=\sum_{n\geqslant 1}(\sigma_n^2+\mu_n^2)p_n $$ and hence $$ {\rm Var}(X_N)=\sum_{n\geqslant 1} (\sigma_n^2+\mu_n^2)p_n-\left(\sum_{n\geqslant 1} \mu_np_n\right)^2 $$ also agreeing with your expression.

  • $\begingroup$ I like this even better than my own solution. Thanks Stefan! $\endgroup$
    – BCLC
    Oct 4, 2014 at 10:50

This question is another clear case of applying identity: $E[f(X,Y)|Y=y]=E[f(X,y)|Y=y]$. $$E[X_N|N=n]=E[X_n|N=n]=E[X_n]=\mu _n$$. In the same way, for the variance we have: $$Var[X_N|N=n]=E[(X_N-\mu_N)^2|N=n]=E[(X_n-\mu_n)^2|N=n]=E[(X_n-\mu_n)^2|=\sigma_n^2$$


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