Finding the Expectation - Continuous case

Question

For the RV with PDF,

f(x;$\lambda$) = (e^-xx^$\lambda$)/$\lambda$!, x > 0

Find the Expectation.

I tried integration by parts, but it gets complicated. I got some answer like:

-(e^-xx^$\lambda+2$)/$\lambda$!(x-$\lambda$-1).

I am not sure if this is right. Could someone help.

Thanks.

Answer 1

Why not use a computer algebra system to check your work? In this case, your density $f(x)$ is:

_{(source: tri.org.au)}

and $E[X]$ can be found as:

 Expect[x, f]

1 + $\lambda$

where Expect is a function from the mathStatica add-on to Mathematica (I am an author of the former). I am sure you could do this with Maple or other packages too, and then use the package to check your own manual integration, if you need that.

Answer 2

A standard trick is to note that $xf(x;\lambda)=\mathrm e^{-x}x^{\lambda+1}(\lambda!)^{-1}=(\lambda+1)f(x;\lambda+1)$ and that, since $f(\ ;\lambda+1)$ is a PDF, one has $\int\limits_0^\infty f(x;\lambda+1)\mathrm dx=1$.

Thus, $E[X]=\int\limits_0^\infty xf(x;\lambda)\mathrm dx=\int\limits_0^\infty(\lambda+1)f(x;\lambda+1)\mathrm dx=\lambda+1$ with no computation at all.

Exercise:

Use the same trick to compute $E[X^2]$ with no pain. Deduce that $\mathrm{var}(X)=\lambda+1$.