Question

Statistics

Answer 1

For part a, i understand that it has a "chi" squared distribution with n degrees of freedom and that it requires a change of variables in order to prove, but i'm having trouble on that aspect.

Answer 2

No change of variables needed. You're given that \(X_1,...,X_5\) follow a standard normal distribution. It's an established fact that the sum of squared standard normals follow a chi square distribution, i.e.
\[\sum_{i=1}^n X_i\sim \chi^2\text{ with }df=n\]

Answer 3

http://en.wikipedia.org/wiki/Normal_distribution#Combination_of_two_or_more_independent_random_variables

Answer 4

I'm sure your textbook has a proof. I know the book that contains this problem certainly does (Wackerly, Mendenhall, Schaeffer)

Answer 5

And indeed after looking through my book (DeGroot and Schervish) I found that "corollary" hidden in an essay's worth of text. My question now has to do with part B and where to go after expanding the polynomial to
\[\sum_{i=1}^{5}(X _{i}^{2}-2X _{i}Xbar+Xbar ^{2})\]
besides what has already been determined, where the summation from part a is "chi" square

Answer 6

This question actually showed up on a recent exam, and there's a slight trick to it, if you could call it that. Expanding the binomial doesn't seem to help much.

Recall the definition for the sample variance,
\[S^2=\frac{\sum (Y_i-\bar{Y})^2}{n-1}~~\Rightarrow~~\frac{(n-1)S^2}{\sigma^2}=\frac{\sum(Y_i-\bar{Y})^2}{\sigma^2}\]
where the latter follows a chi square distribution with \(n-1\) degrees of freedom.

In the given random variable, \(n=5\), so we have a chi square with 4 degrees of freedom.

Answer 7

This might also be a corrolary in your text.

Answer 8

Allow me to correct myself; expanding the binomial *could* work, but the corollary is a direct result of the expansion, more or less.

Answer 9

Following that same trick that you used, i don't exactly understand how the \[X_{6}^{2}\] like my problem involves will be included. Would it do nothing because it's not a function of i as the summation would need, and would thus be considered as a constant that only shifts the chi distribution?

Answer 10

\(X_6\) has a standard normal distribution, so \(X_6^2\) must have a chi square dist. with 1 deg of freedom.

Adding two distributions, both with chi square, yields another chi square. The degrees of freedom of the sum is the sum of the degrees of freedom. So, the variable for (c) is \(U+X_6^2\), which has \(4+1=5\) degrees of freedom.