Help with finding the expectation

[cse63146]

Homework Statement

Let X1,...,Xn denote a random sample from a N(\mu , \sigma) distribution. Let Y = \Sigma \frac{(X_i - \overline{X})^2}{n}

Homework Equations

The Attempt at a Solution

How would I find E(Y)?

Any help would be greately appreciated.

 

[statdad]

You need to be sure you can justify steps and perform the omitted work

<br /> (X_i - \bar X)^2 = X_1^2 - 2X_1 \bar X + {\bar X}^2<br />

When you compute E[(X_i - \bar X)^2]

<br /> E[X_i^2]<br />

should be easily found, and doesn't depend on i.

<br /> 2E[X_i \bar X] = \frac 2 n E[X_i (X_1 + X_2 + \dots + X_n)]<br />

should be easily determined (and will not depend on i). Also,

<br /> E[{\bar X}^2]<br />

should be easy to find (since you know the distribution of the sample mean).

Work these out individually, then combine them.

 

[cse63146]

X^2_i is a chi-square distribution with n degrees of freedom (since there are n X_i) and it's expectation would be n

\overline{X} -&gt; N(\mu, \frac{\sigma^2}{n}) is a noncentral chisquare distribution \frac{X^2_i}{\sigma^2 /n} and it's expectation is n*sigma^2

Kinda stuck on this one: 2E[X_i \bar X] = \frac 2 n E[X_i (X_1 + X_2 + \dots + X_n)]

I was just wondering, would I be able to use the following property of the chi square distribution:

 

[statdad]

X^2_i is a chi-square distribution with n degrees of freedom (since there are n X_i) and it's expectation would be n

It would be a non-central Chi-square - but you don't need that. For any random variable what do you know about an expression for E[X^2] in terms of the first two moments?

\overline{X} -&gt; N(\mu, \frac{\sigma^2}{n}) is a noncentral chisquare distribution \frac{X^2_i}{\sigma^2 /n} and it's expectation is n*sigma^2

I would make a comment similar to the first one here: you know the distribution of the sample mean, what do you know about the expectation of its square in terms of the first two moments?

Kinda stuck on this one: 2E[X_i \bar X] = \frac 2 n E[X_i (X_1 + X_2 + \dots + X_n)]

Write it as
<br /> \frac 2 n \left( E[X_i^2] + \sum_{j \ne i} E[X_i X_j]\right)<br />
and remember that for i \ne j the Xs are independent.

I was just wondering, would I be able to use the following property of the chi square distribution:

You could - IF you have already obtained that result elsewhere in your class.

 

[cse63146]

Is this what you wre talking about:

E((X_i - \overline{X})^2)= \sigma^2 = E(X)^2 - E(X^2)

not sure how that helps

 

[statdad]

Actually, just the second part:
If
<br /> \sigma^2=E[X^2] - \left(E[X]\right)^2<br />

what does E[X^2] itself equal?

 

[cse63146]

E(X^2_i) = \sigma^2 + E(X_i)^2 = \sigma^2 + (n \mu)^2

Would the expectation for Xbar² be equal to the expectation of of X_i?

 

[statdad]

Why do you have

<br /> E(X_i)^2 = (n\mu)^2<br />

Should the n really be there?

For E[\bar X^2], remember that the sample mean is normally distributed with mean \mu and variance \frac{\sigma^2} n.

 

[cse63146]

Isn't it because \Sigma E[X_i] = E[X_1] + ... + E[X_n] = n \mu and each E[X] = mu, so there are n of them?

 

[statdad]

I asked because you didn't have the sum. My point is that for any random variable that has a variance,

<br /> E[X^2] = \sigma^2_X + \mu^2_x<br />

that is - the expectation of the square equals the variance plus the square of the mean.