# How to evaluate a summation equation containing a random variable?

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.

• Add the self-study tag and read its wiki. – StubbornAtom Apr 15 '18 at 13:11
• en.wikipedia.org/wiki/Law_of_total_probability – Mark L. Stone Apr 15 '18 at 13:37
• Try to apply the representation$$\mathbb{P}(B=0)=\mathbb{E}^N[\mathbb{P}(B=0|N)]$$ – Xi'an Apr 15 '18 at 13:39
• 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. – Mark L. Stone Apr 15 '18 at 13:53
• @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. – Troy Apr 15 '18 at 14:18

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.
• 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? – Troy Apr 16 '18 at 3:48
• @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! – Troy Apr 16 '18 at 13:47