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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Review en.wikipedia.org/wiki/Wald%27s_equation $\endgroup$ – 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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.