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Distribution of sample means and variances

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Homework Statement



Let X1,X2,...,Xn be i.i.d. Normal(μ,σ2) random variables, where the sample size n≥4. For 2≤k≤n-2, we define:

Xbar = (1/n)SUM(Xi) from i=1 to n
Xbark = (1/k)SUMXi) from i=1 to k
Xbarn-k = (1/n-k-1)SUMXi from i=k+1 to n

S2=(1/n-1)SUM(Xi-Xbar)2 from i=1 to n
S2k=(1/k-1)SUM(Xi-Xbark)2 from i=1 to k
S2n-k=(1/n-k-1)SUM(Xi-Xbarn-k)2 from i=k+1 to n

Find distribution of:
a) Xbark + Xbarn-k.

b) ((k-1)S2k + (n-k-1)S2n-k)/σ2

c) S2k/S2n-k

d) Evaluate E(S2 | Xbar = xbar) Explain.


The Attempt at a Solution



**I am not asking for someone to do the proof for me(this is a lot of work), but I would love a verbal explanation for what the distributions of each of those might be/why? THANKS!!

a) I know Xbark is distributed Norm(μ,σ2/k) and Xbarn-k is distributed Norm(u,σ2/n-k), but I don't know how adding the two distributions together impacts the overall distribution.

b) Same idea where I know I am adding a χ2 distribution with parameter k-1 to a χ2 distribution with parameter n-k-1 all over σ2, but do not have a deep enough conceptual understanding to comprehend how adding them together affects it.

c) For this one I got as far as finding a χ2 distribution with parameter k-1 divided by χ2 distribution with parameter n-k-1
 

Answers and Replies

  • #2
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So, I've gotten a little further with this one, and I think part (b) is a chi-square distribution with parameter n and part (c) is an F-distribution with parameters k-1 and n-k-1. I am still not sure just how to do part (a) though, and the concept that is actually giving me trouble is how the constant 1/k affects the distribution. If X had a standard normal distribution, then I think Xbark would yield be distributed Norm(0,k2). However, given the mean and standard deviation as μ and σ2, I am not sure how this might change the answer.
As for part (d), I solved to: (1/(n-1))SUM E(Xi2) -nxbar, but have no idea how to follow it up from there or if I'm going the wrong direction entirely. Any hints would be much appreciated. Thank you :)
 
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Ray Vickson
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Homework Statement



Let X1,X2,...,Xn be i.i.d. Normal(μ,σ2) random variables, where the sample size n≥4. For 2≤k≤n-2, we define:

Xbar = (1/n)SUM(Xi) from i=1 to n
Xbark = (1/k)SUMXi) from i=1 to k
Xbarn-k = (1/n-k-1)SUMXi from i=k+1 to n

S2=(1/n-1)SUM(Xi-Xbar)2 from i=1 to n
S2k=(1/k-1)SUM(Xi-Xbark)2 from i=1 to k
S2n-k=(1/n-k-1)SUM(Xi-Xbarn-k)2 from i=k+1 to n

Find distribution of:
a) Xbark + Xbarn-k.

b) ((k-1)S2k + (n-k-1)S2n-k)/σ2

c) S2k/S2n-k

d) Evaluate E(S2 | Xbar = xbar) Explain.


The Attempt at a Solution



**I am not asking for someone to do the proof for me(this is a lot of work), but I would love a verbal explanation for what the distributions of each of those might be/why? THANKS!!

a) I know Xbark is distributed Norm(μ,σ2/k) and Xbarn-k is distributed Norm(u,σ2/n-k), but I don't know how adding the two distributions together impacts the overall distribution.

b) Same idea where I know I am adding a χ2 distribution with parameter k-1 to a χ2 distribution with parameter n-k-1 all over σ2, but do not have a deep enough conceptual understanding to comprehend how adding them together affects it.

c) For this one I got as far as finding a χ2 distribution with parameter k-1 divided by χ2 distribution with parameter n-k-1
Are you not using a textbook or course notes? Surely many of these items are discussed therein.

However, just in case you do not have a textbook, I will give a couple of hints. You say (correctly) that Xbark and Xbarn-k are normally distributed, and you give their means and variances. Are they dependent or independent? If they were independent, what could you say about their sum? Ditto for the terms (k-1)S2k and (n-k-1)S2n-k.

BTW: your definitions of S2, etc., are incorrect: this is NOT equal to
(1/n-1)SUM(Xi-Xbar)2 from i=1 to n, which is
[tex] \left( \frac{1}{n}-1\right) \sum_{i=1}^n (X_i - \bar{X})^2.[/tex]
The correct formula is
[tex] \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2.[/tex]
which would be written as (1/(n-1)) SUM (Xi-Xbar)2, i=1..n. That is, you need 1/(n-1), not 1/n-1.
 

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