# E[X] and E[X^2] for function of Weibull RV

• May 26th 2010, 10:10 AM
dsmullan
E[X] and E[X^2] for function of Weibull RV
I'm tryng to solve the following integrals:

1) (x/t)*(e^(-x/t))*(1+e^(-x/t))^-2

2) ((x^2)/t)*(e^(-x/t))*(1+e^(-x/t))^-2

Both on negative infinity to positive infinity.

You can see that they are both expectations of functions of a weibull random variable with theta of t and beta of 1.

1) Y = X(1+e^(-X/t))^-2
2) Y = (X^2)(1+e^(-X/t))^-2

I'm not sure if you need to use that or if I just can't solve the integrals properly. Any help?
• May 26th 2010, 12:04 PM
Anonymous1
Quote:

Originally Posted by dsmullan
I'm tryng to solve the following integrals:

1) (x/t)*(e^(-x/t))*(1+e^(-x/t))^-2

2) ((x^2)/t)*(e^(-x/t))*(1+e^(-x/t))^-2

Both on negative infinity to positive infinity.

You can see that they are both expectations of functions of a weibull random variable with theta of t and beta of 1.

1) Y = X(1+e^(-X/t))^-2
2) Y = (X^2)(1+e^(-X/t))^-2

I'm not sure if you need to use that or if I just can't solve the integrals properly. Any help?

See here:

Wolfram Mathematica Online Integrator
• May 26th 2010, 12:15 PM
dsmullan
Quote:

Originally Posted by Anonymous1

Yes, I've done that but when doing part (2), it integrates to a polylog function. I doubt that a prof would expect us to manually integrate to that so I'm wondering if I missed anything.
• May 26th 2010, 01:57 PM
Anonymous1
There may be some slick manipulation you can use.
You may not need it though.
Look at the definition of polylog.
Specifically, examine its behavior when you plug in $\displaystyle \pm \infty.$