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Theorem : Expectation and Variance of the Sum of a Random Number of Independent Random Variables

Consider the sum, where $N$ is a random variable that takes nonnegative integer values, and $X_1, X_2, …, X_n$ are identically distributed random variables. Assume that $N, X_1, X_2, …, X_n$ are independent.

$$Y = X_1, X_2, …, X_N$$

Then

$$E[Y] = E[N] E[X] $$

$$\var{Y} = E[N] \var{X} + (E[X])^2 \var{N} $$

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Fix a $n > 0$. The random variable $X_1 + … + X_n$ is independent of $N$ and, therefore, independent of $\set{ N = n } $. Hence,

$$\begin{align*} E [ Y | N = n ] &= E [ X_1 + … + X_N | N = n] \br &= E [ X_1 + … + X_n | N = n ] \br &= E [ X_1 + … + X_n ] \br &= n E [ X ] \end{align*}$$

This is true for all $n > 0$, so

$$E [ Y | N ] = N E [ X ] $$

Using the law of iterated expectations, we obtain

$$E[Y] = E[E[Y|N]] = E[NE[X]] = E[N]E[X]$$

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Similarly,

$$\begin{align*} \var{ Y | N = n } &= \var{ X_1 + … + X_N | N = n } \br &= \var{ X_1 + … + X_n } \br &= n \var{ X } \end{align*}$$

This is true for all $n > 0$, so

$$\begin{align*} \var{ Y } &= E [ \var{ Y | N } ] + \var{ E [ Y | N ] } \br &= E [ N \var{ X } ] + \var{ N E[X] } \br &= E [ N ] \var{ X } + \pare{E [ X ]}^2 \var{ N } \end{align*}$$

Properties: The Sum of a Random Number of Independent Random Variables

Let $X_1, X_2, …, X_n$

Example: Sum of a geometric number of independent geometric random variables

We let $N$ be geometrically distributed with parameter $p$. We also let each random variable $X_i$ be geometrically distributed with parameter $q$. We assume that all of these random variables are independent. Let $Y = X_1, X_2, …, X_n$. We have

$$M_N(s) = \frac{ p e^s }{ 1 - (1 - p) e^s }, M_X(s) = \frac{ q e^s }{ 1 - (1 - q) e^s }$$

To determine $M_Y(s)$, we start with the formula for $M_N(s)$ and replace each occurrence of $e^s$ with $M_X(s)$. This yields

$$M_Y(s) = \frac{ p M_X(s)}{ 1 - (1 - p) M_X(s)} $$

and, after some algebra,

$$M_Y(s) = \frac{ pq e^s }{ 1 - (1 - pq) e^s } $$

We conclude that $Y $ is geometrically distributed, with parameter $pq$.