CRLB for p: Find E[x] with Maclaurin Series

[neznam]

Random Sampe of size n from distribution with pdf
f(x;p)={(lnp)^x}/px! for x=0,1,...; p>1 and 0 otherwise

Find CRLB for p?

My problem is finding E[x] which is somekind of maclaurin series but can't figure out which one?

Please any suggestions?

Thanks

 

[vela]

I'm not familiar with the abbreviation CRLB. What does it mean?

What expression do you have for E[x]? You need to show more work for us to see where you're getting stuck.

 

[neznam]

E[x] = [1*(lnp)^1]/p*1!+[2*(lnp)^2]/p*2!+[3*(lnp)^3]/p*3!+...

Once I have the expected value E[X] of this distribution I will be able to find the CRLB as well which is defined to be in this case

1/(n*[d/dp ln f(x;p)]^2

Any help is appreciated

 

[neznam]

So my main problem is figuring out the E[X] and as a hint of this problem it is saying to use Maclaurin series.

 

[Dick]

If you factor a 1/p out from your original f then you got z^k/k! (where I'm writing z=ln(p) and k=x to make it look more like a maclaurin series. Do you recognize what function that is? The expectation value is then associated with the maclaurin series k*z^k/k!=z^k/(k-1)!. Can you modify the function you recognized to get the second series?

 

[neznam]

sum of z^k/k! is e^k, so is it sum of z^k/(k-1)! will be e^(k-1)?

 

[Dick]

sum of z^k/k! is e^k, so is it sum of z^k/(k-1)! will be e^(k-1)?

Sum of z^k/k! is e^z. Try that again.

 

[neznam]

Ok i think i got it

E[x]=[(lnp)/p]*e(lnp)=lnp

Thanks so much