Let $X_1,X_2$ be a random variables with geometric distribution with $p$,

Find the probability function of the random variable $X_1+X_2$

My try:

$$X_1,X_2\sim G(p)$$

$$X_1+X_2 \in \{2,3,4,\dots\}$$





My attempt is correct?

  • $\begingroup$ What happened?- $\endgroup$ – Mankind Dec 16 '15 at 21:18
  • $\begingroup$ Nothing special $\endgroup$ – 3SAT Dec 16 '15 at 21:21
  • $\begingroup$ This is incomplete - for $P(X_1+X_2 = m),$ you need to account for all cases such that $(k+n) = m$. See this math.stackexchange.com/questions/548525/… $\endgroup$ – stochasticboy321 Dec 16 '15 at 21:32

The post contains some elements of a correct approach.

We cannot solve the problem without making some assumptions about the relationship between $X_1$ and $X_2$. We will assume that $X_1$ and $X_2$ are independent.

Let $S=X_1+X_2$. We want to find $\Pr(S=n)$. This is $0$ if $n\le 1$. So let $n\ge 2$.

We can have $S=n$ in several ways. For we could have $X_1=1$ and $X_2=n-1$. Or we could have $X_1=2$ and $X_2=n-2$. And so on up to $X_1=n-1$ and $X_2=1$.

For any $k$ from $1$ to $n-1$, we have $$\Pr(X_1=k\cap X_2=n-k)=(1-p)^{k-1}p(1-p)^{n-k-1}p=p^2(1-p)^{n-2}.$$ Adding up from $k=1$ to $k=n-1$ we find that $$\Pr(S=n)=(n-1)p^2 (1-p)^{n-2}.$$

Remark: We sketch another approach. We can think of $X_1$ as the number of trials until the first success, and of $X_2$ as the number of trials between the first success and the second success. So $S$ is the number of trials until the second success. The probability this is $n$ is the probability of exactly one success in the first $n-1$ trials times the probability of success on the $n$-th trial. That gives $$\Pr(S=n)=\binom{n-1}{1}p(1-p)^{n-2}p.$$

For related matters, please look under negative binomial distribution.


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