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I'm trying to find:

$$\Pr(B = 0)$$

Where:

$$B = \sum_{i=0}^N b_i$$

And:

\begin{align} N &\thicksim \mathrm{Poisson}(\lambda=10) \\ b_i &\thicksim \mathrm{Geometric}(p=0.8) \end{align}

NOTE: the Geometric distribution being used here is the one which models the number of failures until first success, not including the first success, that is:

$$\Pr(Y=k) = (1-p)^kp.$$

Here's my thinking so far: When all the $b_i = 0$, their sum would just be 0 + 0 + ... + 0, which subsequently would make our summation $B = 0$. This is simple enough to find if $N$ is fixed because we know $\Pr(b_i = 0) = 0.8$ from our Geometric distribution, so...

$$\Pr(B=0) = 0.8^N$$

However, this is where the problem begins! $N$ is not fixed, it's a random variable, so how do I go about calculating this summation when the length of the summation is completely dependent on the value of a random variable?

As stated to begin with, the main goal is to find $\Pr(B = 0)$, I think my thinking might be close but I'm unsure how to work with these random variables and it's doing my head in.

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    $\begingroup$ Add the self-study tag and read its wiki. $\endgroup$ – StubbornAtom Apr 15 '18 at 13:11
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    $\begingroup$ en.wikipedia.org/wiki/Law_of_total_probability $\endgroup$ – Mark L. Stone Apr 15 '18 at 13:37
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    $\begingroup$ Try to apply the representation$$\mathbb{P}(B=0)=\mathbb{E}^N[\mathbb{P}(B=0|N)]$$ $\endgroup$ – Xi'an Apr 15 '18 at 13:39
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    $\begingroup$ The 3 secrets to real estate are location, location, location. The 3 secrets to probability calculation are condition, condition, condition. In this case, once ought to be enough. $\endgroup$ – Mark L. Stone Apr 15 '18 at 13:53
  • $\begingroup$ @MarkL.Stone I don't mean to get spoon fed, but I really have no clue how to even phrase this in conditional probability. I think the AND bit of conditional would just be what I worked out $0.8^4 = Pr(B = 0 and N = 4)$, that's about as far as I can get and I'm not even sure if that's even following the right path. $\endgroup$ – Troy Apr 15 '18 at 14:18
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Suppose $X_i\sim\text{Geometric}(p)$ and $Y=\sum_{i=1}^NX_i$ where $N\sim\text{Poisson}(\lambda)$.

As mentioned in the comments, to derive the pmf of $Y$, you need to apply the total probability theorem by conditioning on the random variable $N$.

$\begin{align} \Pr(Y=0)&=\Pr\left(\sum_{i=1}^NX_i=0\right)\\ &=\sum_{n=0}^\infty\Pr\left(\sum_{i=1}^nX_i=0\mid N=n\right)\Pr(N=n)\\ \end{align}$

You haven't mentioned if $N$ is independent of the $X_i$'s, in which case that last expression can be further simplified as the conditional distribution of $\sum_{i=1}^nX_i\mid N=n$ will be the same as the unconditional distribution of $\sum_{i=1}^nX_i$. Otherwise you need the conditional distribution $\sum_{i=1}^NX_i\mid N$. Nor have you mentioned whether the $X_i$'s are independent or not. If they are, then the sum of independent and identically distributed Geometric random variables has another known distribution. Try to think of that, and you have a sum left to calculate.

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  • $\begingroup$ You were correct in mentioning that both $N$ and $X_i$ could be assumed to be independent. And I figured out what you mentioned; that the sum of an independent, identical Geometric random variables is known to be Negative Binomially distributed. But how does this help reduce everything and find some finite solution? $$\sum_{n=0}^\infty\Pr(\sum_{i=1}^nX_i=0)\Pr(N=n)$$ Aren't we still left with this, but all we know is the first part can be negatively binomially distributed? Am I missing something? $\endgroup$ – Troy Apr 16 '18 at 3:48
  • $\begingroup$ @Troy You know both the probabilities in that expression you wrote. Have you tried simplifying from that? If you don't write the probabilities out explicitly how can you tell if the sum has a closed form or not? $\endgroup$ – StubbornAtom Apr 16 '18 at 5:04
  • $\begingroup$ @SturbbornAtom Figured it out I think. By approaching it logically, or by actually using a Binomial Distribution where k = N, either way the $\Pr(\sum_{i=1}^nX_i=0)$ part simplifies down to $p^n$. And then the $\Pr(N=n)$ is of course just the PMF of the Poisson Distribution. Then everything is in terms of $n$, which means we can take the sum to a reasonably large number (I did 1000000) to approximate $\Pr(B=0)$; which when $p = 0.8$ and $\lambda = 10$ evaluates to roughly 0.1353 - what my simulated value was before any of these conditional calculations! Thanks for all your help guys! $\endgroup$ – Troy Apr 16 '18 at 13:47

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