I'm having trouble finding the PMF of a sum of pascal random variables.
The problem stated is: Let $X \sim Pascal(m,p)$ and $Y \sim Pascal(l,p)$ where X and Y are independent of each other. Let $Z = X + Y$. Find PMF of Z.
The textbook answer states that since the two random variables are independent we can think of Z as a pascal random variable where we want to see $m + l$ heads. So $Z \sim Pascal(m+l, p)$
The book doesn't give us an exact distribution, but Pascal(m+l, p) using the negative binomial formula has distribution $${k-1 \choose m+l-1}p^{m+l}(1-p)^{k-m-l}$$
However when I approached it before looking at the given solution, I didn't get the same same binomial coefficient.
My approach:
Let a = m+l where a is a fixed constant$$P(Z = a) = P(X + Y = a)$$$$=\sum_m P(X + Y = a - m|X = m)P(X = m)$$$$=\sum_m P(Y = a - m|X = m)P(X = m)$$
Since X and Y are independent of each other$$=\sum_m P(Y = a-m)P(X = m)$$Now applying the negative binomial formula, $$\sum_m {r-1 \choose a-m-1}p^{a-m}(1-p)^{r-a+m}{k-r-1 \choose m-1}p^m(1-p)^{k-r-m}$$$$\sum_m {r-1 \choose a-m-1}p^a(1-p)^{k-a}{k-r-1 \choose m-1}$$$$p^a(1-p)^{k-a}\sum_m {r-1 \choose a-m-1}{k-r-1 \choose m-1}$$By Vandermonde's Identity,$${k-2 \choose a-2}p^a(1-p)^{k-a}$$
Which makes more sense to me as the distribution since we choose the last place for the last heads to end for pascal random variable X and we choose the last place for the last heads to end for pascal random variable Y. Hence we have ${k-2 \choose a-2}$. However, using the variable a the textbook is saying our binomial coefficient is ${k-1 \choose a-1}$. What's the flaw in my logic? How can I go about understanding why the binomial coefficient is ${k-1 \choose a-1}$?
$\endgroup$1 Answer
$\begingroup$Thank you for a very well-formulated question! We can reality-check that the book's answer is correct by, for example, taking $m=0$. So we should definitely seek an error in your calculation. The problem is that you used the variable $m$ for two different things, and they got confused with each other.
The formula we want to apply is$$ P(Pascal(t,p)=k) = \binom{k-1}{t-1} p^t (1-p)^{k-t}. $$Therefore, if $X = Pascal(m,p)$ and $Y=Pascal(l,p)$, then$$ P(X=n) P(Y=a-n) = \binom{n-1}{m-1} p^m (1-p)^{n-m} \binom{a-n-1}{l-1} p^l (1-p)^{a-n-l}; $$and we want to sum this expression over $n$; this will require a different tool from Vandermonde's identity, since the variable of summation is in the "numerators" of the binomial coefficients.
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