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you could help me with this problem I would really appreciate it. This is not a task just an exercise to understand the subject. Let the random variable $X \sim \mathrm{Bernoulli}(p)$. Suppose that given an $X = i$, $Y$ is a random variable of type $\mathrm{Poisson} (3 (i + 1))$. Find $\mathbb{E} \left[(X +1) Y^2\right]$.

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    $\begingroup$ Weirdest title I've seen here in a while. $\endgroup$ – Asaf Karagila Nov 20 '17 at 20:55
  • $\begingroup$ In some languages, the word for "expectation" could be translated to "hope". Of course, this is a bad translation (in this context). $\endgroup$ – madprob Nov 20 '17 at 20:56
  • $\begingroup$ @AsafKaragila: I guess the OP is Italian or French. We use the term speranza / ésperance (i.e. hope) to denote the expected value. $\endgroup$ – Jack D'Aurizio Nov 20 '17 at 21:05
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    $\begingroup$ @Jack: Well, I expect that it's not unreasonable to hope that people use the correct English term regardless. Or maybe I hope it's not unreasonable to expect. I'm not which one, and for Italian people, I guess that would be interchangeable anyway. :P $\endgroup$ – Asaf Karagila Nov 20 '17 at 21:07
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\begin{align*} E[(X+1)Y^2] &= E[E[(X+1)Y^2|X]] \\ &= E[(X+1)E[Y^2|X]] & \text{X+1 constant given X} \\ &= E[(X+1)(\mathbb{V}[Y|X]+E[Y|X]^2)] & E[Y^2|X]=\mathbb{V}[Y|X]+E[Y|X]^2 \\ &= E[(X+1)(3(X+1)+(3(X+1))^2)] & Y|X \sim \text{Poisson}(3(X+1)) \\ &= (1-p)((0+1)(3(0+1)+(3(0+1))^2) \\ &+ p((1+1)(3(1+1)+(3(1+1))^2) & X \sim \text{Bernoulli}(p) \\ &= 12(1-p) + 84p = 12+72p \end{align*}

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HINT Let $V\sim\mathcal{P}(3)$ and $W\sim \mathcal{P}(6)$. Then, by the Law of Iterated Expectation (aka Law of Total Expectation, Tower Property of Expectation):

$$ \begin{split} \mathbb{E}\left[(X+1)Y^2\right] & = \mathbb P[X=0]\,\Bbb E\left[V^2\right] +\mathbb P[X=1]~\mathbb E\left[2W^2\right] \\ & = (1-p)\,\mathbb{E}\left[V^2\right] + 2p\, \mathbb{E}\left[W^2\right]\\ \end{split} $$

Can you finish this?

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  • $\begingroup$ I do not understand, I think I can not finish it. :( $\endgroup$ – Nick Jr. Nov 20 '17 at 20:54
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    $\begingroup$ @NickJr. can you please ask which steps you do not understand and I will explain more. The 1st step uses conditional expectations, the 2nd - just evaluates the probabilities. $\endgroup$ – gt6989b Nov 20 '17 at 21:18

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