Given $X\sim\text{NB}(r,p)$ and $Y\sim\text{NB}(r+1,p)$, how do I prove the above identity?
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Your hint would be to couple $X$ and $Y$ : make one dependent on the other. This can be done in an obvious manner. – Sarvesh Ravichandran Iyer Sep 27 '20 at 14:01
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I thought of equating X=Y-1 but I could not see why P(X=k)=P(Y-1=k) @TeresaLisbon – Sep 27 '20 at 14:07
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That is not correct, right? What you need is : given $X \sim NB(r,p)$, you need to wait for another success to get $r+1$ successes i.e. $Y$,and each success is given by a geometric random variable independent of $X$. So take $Y = X+G$ where $G$ is a geometric random variable (with probability $p$), and now see what happens. – Sarvesh Ravichandran Iyer Sep 27 '20 at 16:28
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Ohhh I see, does that mean I will have to solve for the pmf of the joint probability X+G? @TeresaLisbon – Sep 28 '20 at 09:20
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Exactly. Basically, the idea is that $Y = X+G$ in distribution, therefore $E[Y^k] = E[(X+G)^k]$ for any $k$. This means that you only need to know what the moments of the geometric distribution are, and then use the binomial theorem in the expansion of $(X+G)^k$. – Sarvesh Ravichandran Iyer Sep 28 '20 at 09:21
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1I see okay thank you for your help! @TeresaLisbon – Sep 28 '20 at 10:12