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- #1

- Thread starter Jason
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- #1

- Feb 12, 2012

- 26

There are three things to know.

First of all, A will have a discrete distribution : it's essentially a Poisson distribution.

Then there are 2 formulas/properties to know.

When we have $\displaystyle P(A=k)$, it's like writing $\displaystyle E[1_{\{A=k\}}]$ where the 1 function is the indicator function.

And finally, $\displaystyle E[E[X|B]]=E[X]$, where X and B are any random variables.

So here, we'll have :

$\displaystyle P(A=k)=E[1_{\{A=k\}}]=E[E[1_{\{A=k\}}|B]]=E[P(A=k|B)]=E\left[e^{-B}\cdot\frac{B^k}{k!}\right]$.

Then you just have to compute this expectation, given that B has an exponential distribution, and you're done. If you encounter any difficulty with this part, please post what you've tried and we'll help

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- #3

- Feb 12, 2012

- 26

So the expectation you're looking for is just $\displaystyle \int_0^\infty \mu e^{-\mu b}\cdot e^{-b}\cdot\frac{b^k}{k!} ~db$, where k is a constant.

If you didn't know this formula, I'll explain it later, I have to go sleep

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- #5

Something like this?

$g(B)=e^{-B}\frac{B^k}{k!}$

$\Rightarrow E(g(B))=\displaystyle\int_{-\infty}^{\infty}g(b)f_B(b)db$

$=\displaystyle\int_{0}^{\infty}e^{-b}\frac{b^k}{k!}\mu e^{-\mu b}db$

$=\frac{\mu}{k!}\displaystyle\int_{0}^{\infty}b^ke^{-b(1+\mu)} db$

$=\frac{\mu}{k!}(0-(-1))=\frac{\mu}{k!}$

erm... ?

$g(B)=e^{-B}\frac{B^k}{k!}$

$\Rightarrow E(g(B))=\displaystyle\int_{-\infty}^{\infty}g(b)f_B(b)db$

$=\displaystyle\int_{0}^{\infty}e^{-b}\frac{b^k}{k!}\mu e^{-\mu b}db$

$=\frac{\mu}{k!}\displaystyle\int_{0}^{\infty}b^ke^{-b(1+\mu)} db$

$=\frac{\mu}{k!}(0-(-1))=\frac{\mu}{k!}$

erm... ?

Last edited:

- Feb 12, 2012

- 26

That's exactly it for the formula !Something like this?

$g(B)=e^{-B}\frac{B^k}{k!}$

$\Rightarrow E(g(B))=\displaystyle\int_{-\infty}^{\infty}g(b)f_B(b)db$

$=\displaystyle\int_{0}^{\infty}e^{-b}\frac{b^k}{k!}\mu e^{-\mu b}db$

$=\frac{\mu}{k!}\displaystyle\int_{0}^{\infty}b^ke^{-b(1+\mu)} db$

$=\frac{\mu}{k!}(0-(-1))=\frac{\mu}{k!}$

erm... ?

But your computation of the integral isn't correct.

Recall that k is a positive integer and that $\displaystyle k!=\Gamma(k+1)=\int_0^\infty e^{-t} t^k ~dt$

Make the proper substitution to get $e^{-b(1+\mu)}$ instead of $e^{-t}$ and it'll be all good ! I think it gives a geometric distribution, but I don't have time doing the whole computation (which shouldn't be too long by the way).

Good luck

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- #7

- Feb 12, 2012

- 26

Well if you want to compute it without mentioning gamma function, it's possible, but you'd have to do successive integrations by parts.Gamma functions?! Haven't done this before...

$=\frac{\mu}{k!}\displaystyle\int_{0}^{\infty}b^ke ^{-(1+\mu)b} db$

$=\frac{u}{k!} \frac {\Gamma (k+1)}{(1+u)^{k+1}}$

$=\frac{u}{(1+u)^{k+1}}$

But this is indeed the solution.

And you'd recognize a geometric distribution because : $\displaystyle \frac{\mu}{(1+\mu)^{k+1}}=\frac{\mu}{1+\mu}\cdot \left(1-\frac{\mu}{1+\mu}\right)^k$

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- #9