Random Sums

• Let ${\xi }_1,{\xi }_2,\cdots$ be independent, identically, distributed RVs.
• Let $N$ be a discrete RV independent of ${\xi }_i$ that takes a value in $\mathbb{N}$. It is also pairwise independent to ${\xi }_i$.
• $X={\sum }_{i=1}^N{\xi }_i$ is called a random sum

Expectation of $X$

$$\begin{array}{rl}\text{E}\sum _{i=1}^N{\xi }_i& ={\mathbb{E}}_N\left[\text{E}\sum _{i=1}^N{\xi }_i\mid N\right]\\ & =\sum _{n=0}^{\infty }{\mathbb{P}}_N(n)\cdot \text{E}\sum _{i=1}^N{\xi }_i\mid N=n\\ & =\sum _{n=0}^{\infty }{\mathbb{P}}_N(n)\cdot n\cdot \text{E}{\xi }_i\\ & =\text{E}N\cdot \text{E}{\xi }_i\end{array}$$

1. Decompose to be conditional on $N$ since the summation depends on $N$
2. Since it is conditional on $N$, and we need to find the expectation, we can expand it out.
3. Independence of $N$ to ${\xi }_i$.

Variance of $X$

Denote $\text{E}{\xi }_i=m$ and $\text{Var}({\xi }_i)={\sigma }^2$

$$\begin{array}{rl}\text{Var}(X)& =\text{E}(X-\text{E}X{)}^2\\ & =\text{E}{\left(\sum _{i=1}^N{\xi }_i-\text{E}N\cdot m\right)}^2\\ & =\sum _{n=0}^{\infty }\text{E}{\left(\sum _{i=1}^N{\xi }_i-\text{E}N\cdot m\right)}^2\mid N=n\cdot {\mathbb{P}}_N(n)\\ & =\sum _{n=0}^{\infty }\text{E}{\left(\sum _{i=1}^N{\xi }_i-(n\cdot m)+m(n-\text{E}N)\right)}^2\cdot {\mathbb{P}}_N(n)\end{array}$$

Upon expansion, we get three terms:

$$\text{E}{\left(\sum _{i=1}^n{\xi }_i-nm\right)}^2$$

which is $n\cdot \text{Var}({\xi }_i)$. If we suppose $S$ is the summation, then $\text{E}S=mn$ as $\text{E}{\xi }_i=m$ and there are $n$ iid RVs. Then, we apply the definition of variance.

$$2\text{E}\left(\sum _{i=1}^n{\xi }_i-nm\right)\cdot m(n-\text{E}N)$$

which is $0$ since the left term of $\cdot$ is $0$.

$$\text{E}{m}^2\cdot {\left(n-\text{E}N\right)}^2=m^2\text{Var}(N)$$

is our final term. So,

$$\begin{array}{rl}& =\sum _{n=0}^{\infty }\left(n\text{Var}({\xi }_i)+m^2\text{Var}(N)\right)\cdot {\mathbb{P}}_N(n)\\ & =m^2\text{Var}(N)+\sum _{n=0}^{\infty }n{\mathbb{P}}_{N(n)}{\sigma }^2\\ \text{Var}(X)& ={\sigma }^2\text{E}N+m^2\text{Var}(N)\end{array}$$

In summary,

$$\begin{array}{rl}\text{E}X& =\text{E}N\cdot \text{E}{\xi }_i\\ \text{Var}(X)& =\text{E}N\cdot \text{Var}({\xi }_i)+(\text{E}{\xi }_i{)}^2\text{Var}(N)\end{array}$$