Moments

**Statistik** · March 17th 2010, 01:28 PM

Hi,

I have a 3-part question from an exam I took and did not solve correctly. I'd love to find out what I should have done... (We already received the exam back, I'm "cleaning up" loose ends.) Thank you!

Assuming Y1, Y2, ..., Yn are iid Exp(1)
a) What are the mean, variance (sigma squared) and third and fourth central moments? (We were given the hint to prove $EY^k = k!$ )
b) Find $E(\bar Y_n)^2$ and $ES_n^4$
c) Find $Ee^{-Y_{(k)}}$

**matheagle** · March 17th 2010, 03:00 PM

Originally Posted by Statistik

Hi,

I have a 3-part question from an exam I took and did not solve correctly. I'd love to find out what I should have done... (We already received the exam back, I'm "cleaning up" loose ends.) Thank you!

Assuming Y1, Y2, ..., Yn are iid Exp(1)
a) What are the mean, variance (sigma squared) and third and fourth central moments? (We were given the hint to prove $EY^k = k!$ )
b) Find $E(\bar Y_n)^2$ and $ES_n^4$
c) Find $Ee^{-Y_{(k)}}$

From the MGF of $(1-t)^{-1}$ you obtain the $k^{th}$ central moment

OR by the definition of the gamma function.......

$E(X^k) = \int_0^{\infty}x^ke^{-x}dx = \int_0^{\infty}x^{(k+1)-1}e^{-x}dx =\Gamma(k+1)=k!$

Then $E((X-1)^3) =E(X^3)-3E(X^2)+3E(X)-1$

$E(\bar Y_n)^2=V(\bar Y_n)+(E(\bar Y_n))^2 ={V(Y_1)\over n}+(E(Y_1))^2$

**Statistik** · March 17th 2010, 03:08 PM

Matheagle,

Thanks. I guess maybe that's why I am stumped - that's what I did and she marked down the 3rd and 4th central moment down as incorrect (I got 6 and 24 for mu3 and mu4, respectively)... ???

**matheagle** · March 17th 2010, 03:12 PM

for the last part I wanted to use MGF of the kth order stat and let t=-1, but I'm not sure i can recognize the MGF of that order stat.

However.......

the density is $\left({n!\over (k-1)!(n-k)!}\right)(1-e^{-x})^{k-1}e^{-x}(e^{-x})^{n-k}$

and using that as a valid density you can manipulate this expectation.
I've never done that with order stats, but it works......

$E(e^{-X_{(k)}})=\int_0^{\infty}e^{-x}\left({n!\over (k-1)!(n-k)!}\right)(1-e^{-x})^{k-1}e^{-x}(e^{-x})^{n-k}dx$

NOW put that into the last term and let m=n+1, so are using a density of the kth order stat from a sample of size n+1.

$=\int_0^{\infty}\left({n!\over (k-1)!(n-k)!}\right)(1-e^{-x})^{k-1}e^{-x}(e^{-x})^{(n+1)-k}dx$

you can pull the constants in and out of the integral and using the fact that a density integrates to one should work.

I get ${n+1-k\over n+1}$

**Statistik** · March 17th 2010, 04:40 PM

matheagle,

Thanks again. Question on the 3rd part ( $E(\bar Y_n)^2$ and $ES_n^4$ : Could I have gotten there using Slutsky? I tried to approach it stating that I knew what Y_nbar converged to in probability and then tried to use Slutsky's Theorem...

Thanks!

**Statistik** · March 19th 2010, 07:37 PM

Hi,

I am still hoping to follow up on this. Maybe I'll post what I did and you could point out where I went wrong?

Take iid sample of random variables that are Exp(1)
b) Find $E(\bar Y_n)^2$

I take that $\bar Y_n \to \mu = 1$ in probability.

I then can use Slutsky with $g(y) = y^2$
to give that $g(\bar Y_n) \to g(\mu)$ in distribution
and thus $(\bar Y_n)^2 \to \mu^2 = 1$ in distribution and probability.

Then $E(\bar Y_n)^2 = \mu^2 = 1$

What am I missing here???

Thanks a ton!

Thread: Moments

LinkBack

Thread Tools

Display

Moments

Slutsky and expected value

Similar Math Help Forum Discussions

Moments

moments estimators

[SOLVED] Moments and MME

Moments

moments

Search Tags