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

$$(X_i - \bar X)^2 = X_1^2 - 2X_1 \bar X + {\bar X}^2$$

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

$$E[X_i^2]$$

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

$$2E[X_i \bar X] = \frac 2 n E[X_i (X_1 + X_2 + \dots + X_n)]$$

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

$$E[{\bar X}^2]$$

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

$$\overline{X} -> N(\mu, \frac{\sigma^2}{n})$$ 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?

$$\overline{X} -> N(\mu, \frac{\sigma^2}{n})$$ 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
$$\frac 2 n \left( E[X_i^2] + \sum_{j \ne i} E[X_i X_j]\right)$$
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
$$\sigma^2=E[X^2] - \left(E[X]\right)^2$$

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

$$E(X_i)^2 = (n\mu)^2$$

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,

$$E[X^2] = \sigma^2_X + \mu^2_x$$

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