What is the distribution of A given B and B's distribution?

[spitz]

Homework Statement

I need to find the distribution of $A$

Homework Equations

$A|B\sim \text{Poisson}(B)$

$B\sim \text{Exponential}(\mu)$

The Attempt at a Solution

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

Not sure how to calculate this...

 

[sunjin09]

Homework Statement

I need to find the distribution of $A$

Homework Equations

$A|B\sim \text{Poisson}(B)$

$B\sim \text{Exponential}(\mu)$

The Attempt at a Solution

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

Not sure how to calculate this...

Let me guess the A|B is Poisson with parameter B, where B is exponential with parameter μ, so A is discrete and B is continuous, the problem is finding the marginal distribution of the discrete A, the joint pdf of A and B is f(A,B)=Poisson(A;B)*Exp(B;μ), so that the marginal pdf of A is ∫dB f(A,B)=∫dB Poisson(A;B)*Exp(B;μ);
If it ends up becoming evaluating $\int dB e^{-B} B^k$ for some k, start with k=0, and try integration by parts to come up with a recursive relation from k to k+1, and evaluate in closed form if possible.

 

[spitz]

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

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

This is where I am at... not sure what to do now.

 

[sunjin09]

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

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

This is where I am at... not sure what to do now.

Integration by parts to go from k to k+1

 

[Ray Vickson]

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

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

This is where I am at... not sure what to do now.

Change variables in the integral; you ought to be able to get a Gamma function; seehttp://en.wikipedia.org/wiki/Gamma_function .

RGV

 

[spitz]

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

$= \frac{\mu}{1+\mu}\left(1-\frac{\mu}{1+\mu}\right)^k$

Would this be geometric then? What is the parameter?