# Geometric sum of symetric random variables

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.

• Maybe you could use the law of total expectation: $E(X)=E(E(X|Y))$ where in this case $Y=\nu_p$ and $X=\sum_{i=1}^{\nu_p}(Y_i+b)$ – gd1035 Nov 8 '18 at 17:57
• $\nu_p$ is independent of $Y_1, Y_2, \dots.$? – Daniel Camarena Perez Nov 10 '18 at 1:40
• Yes, $\nu_p$ si independent of $Y_1, Y_2, \dots$ – Waney Nov 10 '18 at 7:28
• – Daniel Camarena Perez Nov 10 '18 at 15:47

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:

$$\mathbb{E}\left[a_p\sum_{i=1}^{\nu_p}(Y_i+b) \right] = a_p(\mathbb{E}[Y]+b)\frac{1}{p} = a_p(\mathbb{E}[Y]+b)\mathbb{E}[\nu_p]$$.