Looking for approach to prove $\bar{x}$ and $s$ are independent for t-statistic

Given t statistic

$t = \frac{\bar{x} - \mu}{s/\sqrt(n)}$

Could anyone mention the approach to prove $\bar(x)$ and $s$ are independent...you do not need to prove it...just tell the methods.

As I knew that there are several methods to find $\bar(x)$ and $s$ are independent

One approach that I have is to to simplify the process , let $n=2$ and under the assumption of normality.

Then $s^2$ is proportional to $(x_1-x_2)^2$ and $\bar(x)$ to $x_1+x_2$. Now, $x_1-x_2$ and $x_1+x_2$ are jointly normal, so you only have to check that they are uncorrelated to show that they are independent. Finally, if $y,z$ are independent then so are $g(y),h(z)$ for any functions $g,h$.

Another approach is that

Let $X = {X_1, X_2, ..., X_n}$ ~ $N_n$, $X_i $are independent.

Let $Y = {Y_1, Y_2, ..., Y_n}$ ~ $N_n$, $Y_i $are independent.

$\bar{Y} = \sum_{i = 1}^n Y_i , s^2 = \sum_{i = 1}^n (Y_i - \bar{Y})^2/ (n-1)$

The general approach is to show that $\bar{Y}$ can be written as a function of $X_1$ and $s^2$ can be written as a function of ${X_2, X_3,..., X_n}$

Since $X_i$ are independent, $\bar{Y}$ and $s^2$are also independent.

My question whether these two approaches the same. If not, do you think they work? Also, can you think of another approach?

THANK YOU

Re: Looking for approach to prove $\bar{x}$ and $s$ are independent for t-statistic

Hey askhwhelp.

The easiest way to do this is if you have some function of two random variables (no matter what it is), if the probability P(X=x,Y=y) = P(X=x)P(Y=y) then you have independence. If the PDF is separable, then you have independence.

Also if you can argue in an algebraic or other way that P(X|Y) = P(X) and P(Y|X) = P(Y) then you get the same end result. In other words, if knowing Y doesn't tell you anything about X and vice-versa, then you have done the exact same thing.

Another way of thinking about this is to show that for any function Y = f(X), no function exists to relate the two variables together.

All statements are all equivalent.

Re: Looking for approach to prove $\bar{x}$ and $s$ are independent for t-statistic

Quote:

Originally Posted by

**chiro** Hey askhwhelp. The easiest way to do this is if you have some function of two random variables (no matter what it is), if the probability P(X=x,Y=y) = P(X=x)P(Y=y) then you have independence. If the PDF is separable, then you have independence. Also if you can argue in an algebraic or other way that P(X|Y) = P(X) and P(Y|X) = P(Y) then you get the same end result. In other words, if knowing Y doesn't tell you anything about X and vice-versa, then you have done the exact same thing. Another way of thinking about this is to show that for any function Y = f(X), no function exists to relate the two variables together. All statements are all equivalent.

You are proposing three method, right?

Re: Looking for approach to prove $\bar{x}$ and $s$ are independent for t-statistic

Basically if X's are I.I.D and Y's are I.I.D then everything is independent to each other by definition (but possibly not identically distributed).

Using that should prove the result.