Question

Let x1, x2, ... , xn be i.i.d. r.v.'s from normal. Given the fact that the r.v. has a chi-square distribution with degress of freedom n. I will attach the file so it is easier to understand

Answer 1

#8a and b

Answer 2

\(S^2\), the sample variance, is defined as
\[S^2=\sum_{i=1}^n\frac{(X_i-\mu)^2}{n-1}\]
With some algebraic manipulation, you can get
\[\begin{align*}S^2&=\sum_{i=1}^n\frac{(X_i-\mu)^2}{n-1}\\
(n-1)S^2&=\sum_{i=1}^n(X_i-\mu)^2\\
\frac{(n-1)S^2}{\sigma^2}&=\sum_{i=1}^n\frac{(X_i-\mu)^2}{\sigma^2}
\end{align*}\]
Hmm, I know the right side is supposed to end up having a chi square distribution with \(n-1\) degrees of freedom, not \(n\)... I must be missing something. Can you look through your text to see what your given definition of \(S^2\) is?

Answer 3

Sure. just give me a second

Answer 4

This is what I found

Answer 5

Ugh, this is proving a lot more confusing than I'd anticipated. Sorry I couldn't be of any more help.

Answer 6

It is okay thank you for trying

Answer 7

I think there might be a typo ... the r.v's should be \(N(\mu,\sigma^2)\), not just \(\sigma\). Otherwise all the results need to have be square rooted on sigma...

a)
\(S^2\) is actually:
\[ S^2=\frac{1}{n-1}\sum_{i=1}^n (X_i-\bar{X})^2\]
So you can do the following:
\[ (n-1)S^2=\sum_{i=1}^n(X_i-\bar{X})^2\\
=\sum_{i=1}^n[(X_i-\mu)+(\mu-\bar{X})]^2\\
=\sum_{i=1}^n\left[(X_i-\mu)^2+2(X_i-\mu)(\mu-\bar{X})+(\mu-\bar{X})^2\right]\\
=\sum_{i=1}^n(X_i-\mu)^2+2\sum_{i=1}^n\left( X_i\mu-X_i\bar{X}-\mu^2+\mu\bar{X} \right) +\sum_{i=1}^n(\mu-\bar{X})^2\\
=\sum_{i=1}^n(X_i-\mu)^2+2\left( n\bar{X}\mu-n\bar{X}\bar{X}-n\mu^2+n\mu\bar{X}\right)+n(\mu-\bar{X})^2\\
=\sum_{i=1}^n(X_i-\mu)^2+2 n\bar{X}\mu-2n\bar{X}^2-2n\mu^2+2n\bar{X}\mu+n\left(\mu^2-2\mu\bar{X}+\bar{X}^2\right)\\
= \sum_{i=1}^n(X_i-\mu)^2+4n\bar{X}\mu-2n\bar{X}^2-2n\mu^2+n\mu^2-2n\bar{X}\mu+n\bar{X}^2\\
=\sum_{i=1}^n(X_i-\mu)^2+-n\bar{X}^2+2n\bar{X}\mu-n\mu^2\\
=\sum_{i=1}^n(X_i-\mu)^2-n(\bar{X}^2-2\bar{X}\mu+\mu^2)\\
=\sum_{i=1}^n(X_i-\mu)^2-n(\bar{X}-\mu)^2\\
=\sum_{i=1}^n(X_i-\mu)^2-\frac{(\bar{X}-\mu)^2}{1/n}\\
=\sum_{i=1}^n(X_i-\mu)^2-\left(\frac{(\bar{X}-\mu)}{1/\sqrt{n}}\right)^2
\]
Then you just divide both sides by \(\sigma^2\)!
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b)You can simply remember this result (which I would as you use this frequently in other stats courses), that it's \(\chi^2(n-1)\)
You can also see this by looking at the first term of the LHS of a) which they tell you is \(\chi^2(n)\) and the 2nd term is a standardized distribution of \(\bar{X}\), and then squared. So it's a N(0,1) rr.v that is squared, meaning it is \(\chi^2(1)\)
So, \(\chi^2(n)-\chi^2(1)=\chi^2(n-1)\)
---------------------------
c) and d) You should Identify this as a \(t(n-1)\) distribution since the numerator is a \(N(0,1)\) random variable, divided by the square root of \(\chi^2(n-1)\) distribution divided by its degrees of freedom, \(n-1). If you do the simplification (just algebra), you end up with the result in d)

Answer 8

The trick of adding and subtracting the same thing comes up a lot in stats too when you have a sum of (something - something)^2 (like in a normal distribution)