Convolution of two geometric random variables starting at zero

Question

I'm asked to compute the probability function of $X + Y$ given that $P(X=k) = P(Y=k) = p(1-p)^k$ for $k \in \mathbb{N}_0$, where $X$ and $Y$ are independent random variables. This is my attempt:

$$P(X + Y = k) = P(\bigsqcup_{i=0}^k\{X = i, Y = k-i\}) = \sum_{i=0}^kP(X=i,Y=k-i)$$ $$=\sum_{i=0}^kp(1-p)^ip(1-p)^{k-i} = p^2\sum_{i=0}^k(1-p)^k = p^2(k+1)(1-p)^k$$

Is this computation correct? Can it be further be simplified/is this probability function known?

Answer 1

It looks ok. As a sanity check, we could see that the pmf sums up to $1$. Instead of computing the sum explicity, let's recall that $X$ is a geometric random variable with mean $E[X]=(1-p)/p$. Now

$$ \sum _{k=0}^\infty p^2(k+1)(1−p)^k = p \sum _{k=0}^\infty k \, p(1−p)^k + p \sum _{k=0}^\infty p(1−p)^k =p E[X]+ p=1$$

Looks fine.

Answer 2

Your calculation is correct.   I'd tweak the presentation a bit, for clarity, but there is no actual problem with the working.

$$\begin{align}\mathsf P(X + Y = k) ~&=~ \mathsf P(\bigcup_{i=0}^k\{X = i, Y = k-i\}) \\ &=~ \sum_{i=0}^k\mathsf P(X=i)\,\mathsf P(Y=k-i)\\&=\sum_{i=0}^k p\,(1-p)^i\cdot p\,(1-p)^{k-i}\\& = p^2\,(1-p)^k\sum_{i=0}^k 1\\&= (k+1)\,p^2\,(1-p)^k\end{align}$$

$X, Y$ are geometrically distributed random variables. Each is the count of failures before the first success in a sequence of independent and identically distributed Bernoulli trials.

$X+Y$ is therefore the count of failures before the second success in a sequence of independent and identically distributed Bernoulli trials.   Thus the event $\{X+Y=k\}$ is the event of some arrangement of $k$ failures and one success, followed by the second success.   Thus from first principles: we multiply the count of ways to make the arrangement, by the probability for each outcome in the arrangement.$$\mathsf P(X+Y=k)=\binom{k+1}{k} \,(1-p)^k\, p^2~\mathbf 1_{k\in\Bbb N_0}$$

This is a Negative Binomial Distribution, with parameters $2, 1-p$. [Negative Binomial Random Variables count the successes before the stopping amount of failures, so we complement the success parameter.]

$$Z\sim\mathcal{NegBin}(r, q)~~\iff~~\mathsf P(Z=k)~=~\binom{k+r-1}k (1-q)^r q^k~\mathbf 1_{k\in\Bbb N_0}$$

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A Geometric Random Variable is also a Negative Binomial Random Variable.$$X\sim\mathcal{Geo}_0(p)\quad\iff \quad X\sim\mathcal{NegBin}(1, 1-p)$$

The sum of two independent Negative Binomial random variable with identical failure rates, is a Negative Binomial Random Variable with the same failure rate and a stopping amount equal to the sum of their stopping amounts. $$X\sim\mathcal{NegBin}(n,1-p),Y\sim\mathcal{NegBin}(m,1-p), X\perp Y~~\vdash~~X+Y\sim\mathcal{NegBin}(n+m,1-p)$$