Let $Y_1,Y_2,\dots$ be nondegenerate and symetirc independent, identically distributed random variables with finite variance $\sigma^2>0$ and let $\nu_p$ be a geometric random variable with mean $\frac{1}{p}$, independent of the $Y_1, Y_2, \dots$.
$$E\left[a_p\sum_{i=1}^{\nu_p}\left(Y_i+b\right)\right]=E[\nu_p]E[a_p(Y_i+b)]$$
where $a_p$ and $b$ are constants.
I appreciate any help, I have never before worked with this type of $\sum$ so I do not uderstand why this equation is true. Thank you again.
Best Answer
You can write the LHS of the equation as:
$\mathbb{E}\left[a_p\sum_{i=1}^{\nu_p}(Y_i+b) \right] = \sum_{k=1}^\infty \mathbb{E}\left[a_p\sum_{i=1}^{k}(Y_i+b) \mathbb{P}(\nu_p=k) \right]$
$=\sum_{k=1}^\infty a_p (\mathbb{E}[Y]+b)k(1-p)^kp$
$=a_p(\mathbb{E}[Y]+b)p\underbrace{\sum_{k=1}^\infty k(1-p)^{k-1}}_S$.
Let's evaluate the infinite summation $S$ in the expression above.
$S = 1 + 2(1-p) + 3(1-p)^2 + ...$
$(1-p)S = (1-p) + 2(1-p)^2 + ...$
Subtracting the two equations, we get:
$pS = 1+(1-p)+(1-p)^2+... = \frac{1}{p}$, implying $S = \frac{1}{p^2}$.
Substituting back in our original expression, we get:
$\mathbb{E}\left[a_p\sum_{i=1}^{\nu_p}(Y_i+b) \right] = a_p(\mathbb{E}[Y]+b)\frac{1}{p} = a_p(\mathbb{E}[Y]+b)\mathbb{E}[\nu_p]$.