- #1

- 15

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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

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- Thread starter neznam
- Start date

- #1

- 15

- 0

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

vela

Staff Emeritus

Science Advisor

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

- #3

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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

- #4

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- #5

Dick

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- #6

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

- #7

Dick

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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.

- #8

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Ok i think i got it

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

Thanks so much

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

Thanks so much

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