relating poisson and exponential distributions

Asked Jul 03 '22 at 23:32

Active Jul 03 '22 at 23:32

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Here is the entire question:

A random variable Y follows an exponential distribution with parameter $\theta$. Given $Y = y$, another random variable X follows the Poisson distribution with $\lambda = y$. Show that $P(X = k) = \frac{\theta}{(1 + \theta)^{(k+1)}}$.

I know that $P(X = k) = \frac{e^{-y} y^k}{k!} $ but I don't know how to use the Y random variable to solve the equation.

asked Jul 03 '22 at 23:32

Matheus Sousa

2

You know the conditional $P(X = k\mid Y=y) = \frac{e^{-y} y^k}{k!}$. To get the marginal $P(X = k)$ you need to integrate out $y$ using the distribution of $Y$ – Henry Jul 03 '22 at 23:42
Thanks @Henry for the help. – Matheus Sousa Jul 04 '22 at 00:01
Yes @callculus42, there's a part in this that solves my doubt. Thanks too. – Matheus Sousa Jul 04 '22 at 00:02
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@MatheusSousa Glad to Herr. – callculus42 Jul 04 '22 at 02:52

relating poisson and exponential distributions

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