# Probability Theory（5）

## Problem

Today I got a problem about the difference of two normal distribution variables of confidence interval. Obviously it is much harder than the ratio between two normal distribution variables’ variance if you know $\mu_1$ and $\mu_2$.

Question is as follows:

We can learn that $X\sim N(\mu_1,\sigma^2)$,$Y\sim N(\mu_2,\sigma^2)$ and that the expression of average of samples $\overline X$ and expectation $\mu_1$ and the variance of the sample $S_1^2$ follow T-student distribution,and the degree of freedom is substract one from the number of the samples,namely $\frac{\overline X-\mu_1}{S_1/\sqrt{n}}\sim t(n-1)$.Although we could clearly know what the distribution the $\mu_1-\mu_2$ follows,unknown variance is a big problem.

So let’s remember some equation about it.

Firstly,Let’s define $S_w^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}$

and then,$\frac{(\overline X-\overline Y)-(\mu_1-\mu_2)}{S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$ will follow T-Student distribution,the degree of freedom is the sum of $\mu_1$ and $\mu_2$ then subtract two from it.

So we put data into equaltion.

$S_w=\sqrt{\frac{(2-1)1^2+(3-1)2^2}{2+3-2}}=\sqrt{3}$

$\frac{(15-25)-(\mu_1-\mu_2)}{\sqrt{\frac{3}{2}+\frac{3}{3}}}\sim t(3)$

namely the confidence interval of 0.8 is $[t_{0.9}(3),t_{0.1}(3)]$

We get a range that is $\frac{(15-25)-(\mu_1-\mu_2)}{\sqrt{\frac{3}{2}+\frac{3}{3}}} \in [t_{0.9}(3),t_{0.1}(3)]$

simplify the equation as follows:

$(15-25)-(\mu_1-\mu_2)\in [\sqrt{\frac{5}{2}}t_{0.9}(3),\sqrt{\frac{5}{2}}t_{0.1}(3)]$

$-(\mu_1-\mu_2)\in [\sqrt{\frac{5}{2}}t_{0.9}(3)+10,\sqrt{\frac{5}{2}}t_{0.1}(3)+10]$

$(\mu_1-\mu_2)\in [-\sqrt{\frac{5}{2}}t_{0.1}(3)-10,-\sqrt{\frac{5}{2}}t_{0.9}(3)-10]$

lastly we know the T-student distribution is symmetrical by Y-axis,So $t_{0.9}(3)=-t_{0.1}(3)$

we got the answer $[-\sqrt{\frac{5}{2}}t_{0.1}(3)-10,\sqrt{\frac{5}{2}}t_{0.1}(3)-10]$

## $S_w$

Some equation about $S_w$,here is some proof of the equation.

We clearly know that $\frac{X_i}{\sqrt{\sigma^2}}\sim \chi^2(1)$

It is obvious that $\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{\sigma^2} \sim \chi^2(n_1+n_2-2)$

We should prove the unbiasedness of the $S_w^2$,namely $E(S_w^2)=\sigma^2$

We all know that $E(\chi^2(n))=n$

so $E(S_w^2)=E(\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2})$

Let’s simplify the equation to prove the conclusion.

$E(\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2})=\frac{\sigma^2}{n_1+n_2-2}E(\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{\sigma^2})$

$=\frac{\sigma^2}{n_1+n_2-2}E(\chi^2(n_1+n_2-2))=\sigma^2$

## End

When we get the problem of confidence interval,we should firstly find the expression of target variable and find that what the distribution it follows.

Then simplify the equation to get the confidence interval.