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Homework Help: Urgent Maclaurin Series/Stats Problem

  1. Mar 27, 2010 #1
    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
     
  2. jcsd
  3. Mar 27, 2010 #2

    vela

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    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.
     
  4. Mar 27, 2010 #3
    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
     
  5. Mar 27, 2010 #4
    So my main problem is figuring out the E[X] and as a hint of this problem it is saying to use Maclaurin series.
     
  6. Mar 27, 2010 #5

    Dick

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    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?
     
  7. Mar 27, 2010 #6
    sum of z^k/k! is e^k, so is it sum of z^k/(k-1)! will be e^(k-1)?
     
  8. Mar 27, 2010 #7

    Dick

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    Sum of z^k/k! is e^z. Try that again.
     
  9. Mar 27, 2010 #8
    Ok i think i got it

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

    Thanks so much
     
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