Let $X_1,X_2,\dots$ be independent and identically distributed random variables. Furthermore, consider the sum $$ Y = X_1 + X_2 + \dots + X_N $$ where the number of terms $N$ is itself a random variable, independent of the $X_i$, all defined on the sampe probability space.

Given this preamble, the text I am reading claims the following

\begin{align} E[Y|N=n] &= E[X_1+X_2+\dots+X_N|N=n] \\ &=E[X_1+X_2+\dots+X_n|N=n] \\ &=E[X_1+X_2+\dots+X_n] \end{align}

While I understand the second and third equalities from an intuitive perspective (we know $N=n$, so this information can be incorporated into the number of terms in the sum), how can this be derived mathematically, or in a more pedantic/rigorous fashion using the laws of probability? Do we need to consider the joint distribution of the $X_i$ and $N$? Any ideas would be appreciated.


Let $I_k(n) = \begin{cases} 1 & : k\in \{1.. n\}\\ 0 & : \text{elsewhere}\end{cases}$

Then $Y = \sum_{k=1}^\infty X_k\; I_k(N)$

$$\begin{align} \mathsf E(Y\mid N=n) & = \mathsf E( \sum_{k=1}^\infty X_k \;I_k(N) \mid N=n) \\ & = \sum_{k=1}^\infty \mathsf E(X_k \;I_k(N)\mid N=n) & \text{linearity of expectation} \\ & = \sum_{k=1}^\infty \mathsf E(X_k \;\mathsf E(I_k(N)\mid N=n)) & \text{independence of r.v.} \\ & = \sum_{k=1}^\infty \mathsf E(X_k I_k(n)) \\ & = \mathsf E(\sum_{k=1}^\infty X_k I_k(n)) \\ & = \mathsf E(\sum_{k=1}^n X_k) \\ \end{align}$$

  • $\begingroup$ This is a great answer. In particular, expressing the random variable $g(N) = I_k(N)$, and the equality in step 3, were profoundly insightful. $\endgroup$ – jII Feb 20 '15 at 17:32

Let $S(N)=\sum_{i=1}^{\infty} X_i\mathbf{1}_{\leq N}(i)\implies E[S(N)|N]=N\mu$, where $\mu=E[X_i]$

Therefore, $E[S(N)|N]$ is simply a function of $N$. Once we set $N=n$, we've completely specified the function, just like $f(x)|_{x=2}$ has a definite value.

  • $\begingroup$ I think you have an error in the first line, since $E[S(N)] = E[N]E[X]$. Although I agree that $E[S(N)|N]=N\mu$. $\endgroup$ – jII Jan 26 '15 at 2:10
  • $\begingroup$ @jesterII I was treating $E[S(N)]$ as a random variable. $E[S(N)|N]$ would be more accurate. Ill correct. I hope my definition of $S(X)$ shows how you can arrive at your expectation. $\endgroup$ – user76844 Jan 26 '15 at 2:18

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