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Suppose I have $X \sim \mathrm{Geom}(p)$ and $Y=\mathrm{Pois}(\lambda)$. I want to create $Z = X + Y$, where the $X$ begins at $0$ rather than $1$.

Is this possible? Then I would calculate the mean and variance.

Lee David Chung Lin
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user1544953
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  • If you want only the expectation and the variance, then the answer by paw88789 below is all you need. Your way of phrasing the question leaves uncertainty about whether you want only that or something more. – Michael Hardy Apr 06 '15 at 18:31

2 Answers2

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Are $X$ and $Y$ independent?

For any random variables, $E[X+Y]=E[X]+E[Y]$; and for any independent random variables, $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$

paw88789
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  • This gives only the expectation and the variance. The question is vague, but one must suspect he wants something like the probability mass function. – Michael Hardy Apr 06 '15 at 18:28
  • @MichaelHardy: You may be right. I don't see a particularly nice way of expressing such a result. One would presumably be multiplying together various geometric and Poisson probabilities and adding up sums of these products. – paw88789 Apr 06 '15 at 18:35
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Something that I thought of to compute the pmf, then the expectation, is: $Pr(X=k)= p(1-k)^{k}$ $Pr(Y=k)=\frac{\lambda^k e^{\lambda}}{k!}$

Then: $Pr(Z=k)=p(1-p)^{k} \frac{\lambda^k e^{\lambda}}{k!}$

Finally, the expected value of $Z$ would be

$E[Z] = \sum_{k=0}^{\infty}k p(1-p)^k \frac{\lambda^k e^{\lambda}}{k!} $

$E[Z] = p \sum_{k=0}^{\infty}k (1-p)^k \frac{\lambda^k e^{\lambda}}{k!} $

Now we split into $k = 0$, and then remaining $k$:

$E[Z] = p\{0 \frac{\lbrack(1-p)\lambda\rbrack^0 e^{-\lambda}}{0!}\} + p\sum_{k=1}^{\infty}\frac{\lbrack(1-p)\lambda\rbrack^k e^{-\lambda}}{(k-1)!}$

$E[Z] = p \sum_{k=1}^{\infty}\frac{\lbrack(1-p)\lambda\rbrack^k e^{-\lambda}}{(k-1)!}$

$E[Z] = e^{-\lambda} p\lbrack(1-p)\lambda\rbrack \sum_{k=1}^{\infty}\frac{\lbrack(1-p)\lambda\rbrack^{k-1}}{(k-1)!}$

Let $j=k-1$

$E[Z] = e^{-\lambda} p\lbrack(1-p)\lambda\rbrack \sum_{j=0}^{\infty} \frac{\lbrack(1-p)\lambda\rbrack^j e^{-\lambda}}{j!}$

$E[Z] = e^{-\lambda} p\lbrack(1-p)\lambda\rbrack e^{(1-p)\lambda}$

$E[Z] = p\lbrack(1-p)\lambda\rbrack e^{-\lambda p}$

Then the variance is straightforward.

Thanks!

user1544953
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