Distribution of sample means and variances

With what you have written, you should be getting an error message; always check your formulas for accuracy and use parentheses liberally.
  • #1
Aria1
21
0

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
 
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  • #2
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 :)
 
  • #3
Aria1 said:

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.
 

FAQ: Distribution of sample means and variances

What is the distribution of sample means?

The distribution of sample means is a theoretical probability distribution that represents all possible means of samples of a certain size that are drawn from a larger population. It is also known as the sampling distribution of the mean.

How is the distribution of sample means different from the population distribution?

The distribution of sample means is different from the population distribution in that it represents the means of samples drawn from the population, while the population distribution represents the entire population. The distribution of sample means is less spread out and more normally distributed compared to the population distribution.

What is the central limit theorem and how does it relate to the distribution of sample means?

The central limit theorem states that as the sample size increases, the distribution of sample means will approach a normal distribution regardless of the shape of the population distribution. This means that the mean of the distribution of sample means will be equal to the mean of the population, and the standard deviation of the distribution of sample means will be equal to the standard deviation of the population divided by the square root of the sample size.

What is the purpose of calculating the standard error of the mean?

The standard error of the mean is a measure of the variability of the sample means from the population mean. It is used to estimate the standard deviation of the population distribution and to calculate confidence intervals for the population mean. It is also used in hypothesis testing to determine the significance of the difference between the sample mean and the population mean.

How does the distribution of sample variances differ from the distribution of sample means?

The distribution of sample variances is a theoretical probability distribution that represents all possible variances of samples of a certain size that are drawn from a larger population. It is different from the distribution of sample means in that it represents the variability of the sample variances, while the distribution of sample means represents the means of the samples. The distribution of sample variances is also not normally distributed and tends to have a positive skew.

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