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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.

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    $\begingroup$ Maybe you could use the law of total expectation: $E(X)=E(E(X|Y))$ where in this case $Y=\nu_p$ and $X=\sum_{i=1}^{\nu_p}(Y_i+b)$ $\endgroup$ – gd1035 Nov 8 '18 at 17:57
  • $\begingroup$ $\nu_p$ is independent of $Y_1, Y_2, \dots. $? $\endgroup$ – Daniel Camarena Perez Nov 10 '18 at 1:40
  • $\begingroup$ Yes, $\nu_p$ si independent of $Y_1, Y_2, \dots$ $\endgroup$ – Waney Nov 10 '18 at 7:28
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    $\begingroup$ Review en.wikipedia.org/wiki/Wald%27s_equation $\endgroup$ – Daniel Camarena Perez Nov 10 '18 at 15:47
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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:

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

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