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If $X$ and $Y$ are independent geometric random variables, $X \sim G(p)$ and $Y \sim G(q)$ then if $Z = X+Y$ I need to show that:

$$P[Z=z] = \frac{pq}{p-q}[(1-q)^{z-1}-(1-p)^{z-1}]$$

My attempt:
\begin{align} P[Z=z] &= \sum_{k=1}^z P[X=k]P[Y=z-k] && \text{(not sure of my summation limits here)} \\ &= \sum_{k=1}^z p(1-p)^{k-1} \cdot q(1-q)^{z-k-1} \\ &= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(1-q)}\cdot\sum_{k=1}^z (\frac{1-p}{1-q})^k \\ &= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(1-q)} \cdot \frac{1-(\frac{1-p}{1-q})^k}{1-\frac{1-p}{1-q}} \\ &= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(p-q)} \left[1 -(\frac{1-p}{1-q})^k\right] \\ &= \frac{pq}{p-q}\left[\frac{(1-q)^z}{1-p} - \frac{(1-p)^z}{1-p}\right] \end{align}

The result is pretty close to the answer. The only discrepancy is the first expression in the parentheses, $\frac{(1-q)^z}{1-p}$ which should be $\frac{(1-q)^z}{1-q}$.

Could someone please have a look at my working and show me where I have gone wrong? Or perhaps a neater calculation? Thanks!

PS
I did check similar questions answered here, but couldn't find anything relevant to my problem.
Is the sum of two independent geometric random variables with the same success probability a geometric random variable?
How to compute the sum of random variables of geometric distribution

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I think you made a mistake in the calculation. First geometric random variable is at least 1, hence $z\ge 2$.

\begin{align*} P[Z=z] &= \sum_{k=1}^{z-1} P[X=k]P[Y=z-k]\\ &=\sum_{k=1}^{z-1} p(1-p)^{k-1} \cdot q(1-q)^{z-k-1}\\ &=\frac{pq}{(1-p)} (1-q)^{z-1}\sum_{k=1}^{z-1} (\frac{1-p}{1-q})^k \\ &=\frac{pq}{(1-p)} (1-q)^{z-1} \cdot \frac{1-p}{1-q}\frac{1-(\frac{1-p}{1-q})^{z-1}}{1-\frac{1-p}{1-q}}\\ &=\frac{pq}{p-q} (1-q)^{z-1} \cdot [1 -(\frac{1-p}{1-q})^{z-1}]\\ &=\frac{pq}{p-q}[(1-q)^{z-1} - (1-p)^{z-1}], \end{align*} which is the desired result.

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  • $\begingroup$ Thanks for the helpful answer. You lost me in the fourth line: $=\frac{pq}{(1-p)} (1-q)^{z-1} \cdot \frac{1-p}{1-q}\frac{1-(\frac{1-p}{1-q})^{z-1}}{1-\frac{1-p}{1-q}}$ How did you get the $\frac{1-p}{1-q}$ part before the summation? $\endgroup$ – abruzzi26 Feb 8 '17 at 15:22
  • $\begingroup$ @abruzzi26 This follows for the formula for the summation of geometric series with finite terms $\endgroup$ – John Feb 8 '17 at 15:26
  • $\begingroup$ Ah, yes. I got confused there for a sec. Thanks again $\endgroup$ – abruzzi26 Feb 8 '17 at 16:55
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I like using the probability generating function for this: you know the pgf for the sum is the product of the pgfs for the two random variables. So

$$p_X(t)p_Y(t)={pqt^2\over (1-t(1-p))(1-t(1-q))}=$$

Applying partial fractions to just the reciprocal terms and setting consecutively $t=(1-p)^{-1}$ then $t=(1-q)^{-1}$ gives

$$A(1-t(1-q))+B(1-t(1-p))=1\iff A={1-p\over q-p}, B = -{1-q\over q-p}$$

leaving us with

$$p_{X+Y}(t) = {pqt^2\over q-p}\left({(1-p)\over 1-t(1-p)}-{1-q\over 1-t(1-q)}\right)$$ $$=\sum_{k=0}^\infty {pq\over q-p}((1-p)^{k+1}-(1-q)^{k+1})t^{k+2}$$

and reindexing appropriately gives

$$p_{X+Y}(t) = \sum_{i=2}^\infty {pq\over p-q}\Big((1-q)^{k-1}-(1-p)^{k-1}\Big)t^k$$

as desired.

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