Conditional Expectation of random sum of independent random variables

Question

Let $Y=X_1+X_2+\dots+X_N$ where $X_1,X_2,\dots,N$ are jointly independent random variables, $X_1,X_2 ...$ identically distributed continuous random variables with finite expectation, and $N$ a discrete random variable with finite expectation.

What is the definition of conditional density $f(y|n)$ of $Y$ given $N=n$ (if it exists). I need it to compute $E(Y|N=n)$.

Is there another approach to compute $E(Y|N=n)$ or even $E(Y|N)$?

For simplicity we can assume that the random variable $N$ is bounded by a positive integer M

Answer 1

The expected value is relatively easy for this case as long as the expectation for $X_i$ exists. Since $N$ is known, you are looking to find $$E[Y|N=n] = E[X_1+X_2+ \cdots + X_N|N=n]= E[X_1 + X_2+ \cdots + X_n].$$ Since expectation is a linear operator, you can write this as $$E[Y|N=n] = E[X_1]+E[X_2]+ \cdots + E[X_n].$$ From there we have $$ E[Y|N=n]=nE[X_1] $$

Answer 2

In actuarial science, this is the basic form of a collective risk model. There is a discrete distribution from which an annual loss frequency is drawn, and each of those losses has a severity drawn from some continuous distribution. The sum total of those losses is the expected aggregate loss the insurer/reinsurer/homeowner/etc. expects.

Somewhat more formally, if $N$ is a discrete random variable, and $X$ is a continuous random variable, we can define $S$ as the aggregate loss random variable where $S = X_1 + X_2 + \ldots + X_N, X \in {0, 1, 2, \ldots}$.

Given some basic independence assumptions, simply the $X's$ (jointly or singly) do not depend on $N$ and vice versa, it can be shown that: $$ E(S) = E(N)E(X)\\ Var(S) = E(N)Var(X) + Var(N)E(X)^2 $$ The proof is based on convolution of the probability generating functions. See (Klugman et al. 1998, pp.295–298).

As soakley showed below, the first expectation is relatively simple to derive. The second can be understood knowing the law of total variance which is: $$ Var(X) = E_Y[Var(X|Y)] + Var_Y[E(X|Y)] $$ or, in English, the total variance is the expected value of the conditional variance plus the variance of the conditional expected value. See (Heckman & Meyer 1989, p.30) for this derivation.

Adding derivation of Variance (edit)

Given the law of total probability, we can say that $Var(S) = E[Var(S|N)] + Var[E(S|N)]$. Now, $S|N$ is merely $X_1 + X_2 + \ldots + X_N$, and we are given the value of $n$. So let's rewrite. $$ Var(S) = E[Var(X_1 + X_2 + \ldots + X_N)] + Var[E(X_1 + X_2 + \ldots + X_N)] $$ One of our assumptions in compound variance, which you also mentioned in your question, is that the X's are iid. Therefore, in the first term, $Var(X_1 + X_2 + \ldots + X_N)$ collapses to $N\cdot Var(X)$. Similarly, the second term $E(X_1 + X_2 + \ldots + X_N)$ simply becomes $N*E(X)$, so we now have: $$ Var(S) = E[N\cdot Var(X)] + Var(N\cdot E(X)] $$ To complete the conditioning, we need to take the expectations and variances over $N$. But the $X$'s are constants with respect to $N$, so we get: $$ Var(S) = E[N]Var(X) + Var(N)E[X]^2 $$ since the $E(X)$ component in the variance comes outside squared and the $Var(X)$ component in the expectation comes out directly, which demonstrates the second relationship.

Heckman, Philip E., and Glenn G. Meyers. "The calculation of aggregate loss distributions from claim severity and claim count distributions." Proceedings of the Casualty Actuarial Society. Vol. 70. 1983.

Klugman, S. A.; Panjer, H. H. & Willmot, G. E. Loss models: from data to decisions John Wiley & Sons, Inc., 1998

Answer 3

Recall that the definition of conditional probability is:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$, and so $P(A|B)P(B) = P(A \cap B)$

For your problem, the second form is applicable:

$P(Y|N=n)P(N=n) = P(Y \cap N=n)$

You now take expectations across the probability distributions. The key insights are:

1) When you consider $E(Y|N=n)$, you have temporarily fixed the value of $N$, so you treat it as the non-random value $n$. Then as soakley notes, $E(Y|N=n)=nE(X_1)$ by the definition of $Y$ and that the $X_i$'s are iid. You "unfix" $N$ by then taking the expectation of $N$ across all its possible values

2) Since the $X_i$'s and $N$ are independent you can use the product rule from the conditional probability definition. Taking expectations over all possible $N$'s, you convert

$E(Y|N=n)=E(X_1)n$ to $E(Y|N)E(N)=E(X_1)E(N)$, which is Avraham's answer. Avraham's answer is quite thorough, by the way, and the references are worth checking.

In general, $f(y|n)$ might not have a closed form, so if all you want is the expectation (and variance), you are better off simply computing those quantities directly, as the answers here have shown.

EDIT: Let's try this: $E(Y)=E_N(E_X(Y|N=n))$

This iterated expectation formula is valid by definition. It says that first we condition $N=n$, and then look at the inner expectation on $Y$, which is now a fixed sum of $n$ iid random variables, each with expectation $E(X_1)$. Now we take a second expectation across $N$, the outer expectation. So the total expectation is the product of the expectation of $N$ and the expectation of $X_1$, given that these are independent.