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Suppose we flip a coin a random number $N$ of times, where $$P(N = n) = (e^{-r})\frac{r^n}{n!}\qquad n = 0,1,2,\ldots$$ that is, $N$ has a Poisson distribution. The flips are independent, and the probability of seeing heads is equal to p. Hence if we know that $N = n$, the number of heads is simply the number of heads in $n$ coin flips with success probability $p$. We write $X$ for the total number of heads, and $Y$ for the total number of tails, so $X +Y = N$.

Question: Why is $$P(X=x) = \sum_{n\geq x} P(X=x|N=n)P(N=n)\quad ?$$ Also, how does one calculate this sum?

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This is the law of total probability. The event $\{X = x\}$ can occur only if the coin is tossed $x, x+1, x+2, \ldots $ times. That is, $N$ must necessarily take on value $n \geq x$, and if $N=n$, then $P\{X=x|N=n\}$ is a binomial probability. So, $$P\{X=x\}=\sum_{n = x}^\infty P\{X=x|N=n\}P\{N=n\}.$$ It is good to write out the formula explicitly, cancel $n!$, and sum the series (yes it can all be done quite straightforwardly). You will get a nice result if you do this. –  Dilip Sarwate Feb 8 '12 at 1:54
    
Can you please also explain its solution. I already understood what you are explaining, but stuck at the sum part. –  user669083 Feb 8 '12 at 2:02

2 Answers 2

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What you describe is called the law of total probabilities. The proof goes like this $$P(X = x) = \sum_{n=0}^\infty P(X = x\; {\rm and}\; N=n) = \sum_{n= 0}^\infty P(X = x|N=n)P(N=n)$$ Now $X$ is a sum of $n$ binomials, so if $n < x$ we cannot have $X = x$; in this case $P(X = x | N = n) = 0$.

Since we have independent trials, $P(X = x | N = n) = {n\choose x}p^x(1-p)^{n-x}$ giving the summation.

Now it is time to compute the summation. We have $$P(X=x) = \sum_{n=x}^\infty P(X=x|N=n)P(N=n) = \sum_{n=x}^\infty {n\choose x}p^x(1-p)^{n-x}{r^ne^{-r}\over n!}.$$ Regrouping and cleaning up we have $$P(X=x) = e^{-r}\sum_{n=x}^\infty {1\over x!(n-x)!}(rp)^x(r - rp)^{n-x} $$ Now reindex to obtain $$P(X=x) = e^{-r}\sum_{n=0}^\infty {1\over n!x!} (rp)^x(r-rp)^n = {(rp)^xe^{-r}\over x!}\sum_{n=0}^\infty {(r-rp)^n\over n!}= {(rp)^xe^{-r}\over x!} e^{r-rp} = {(rp)^xe^{-rp}\over x!}.$$ What you get is distrubuted Poisson(rp).

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You can add an intermediate step as follows:

$\begin{align} P[X = x] &= \sum_{n=1}^\infty P[X = x, N = n]\\ &= \sum_{n=1}^\infty P[X = x | N = n] P[N = n], \end{align}$

where we used Bayes' formula in the second step. Of course all terms with $n < x$ are zero because if $n < x$, it's impossible to get $x$ heads after $n$ flips.

As to how to calculate the sum, you should fill in the probabilities and see what you can do. Note that P[X = x | N = n] is the same as the probability of flipping $n$ coins and getting $x$ heads, which you can calculate.

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