Normal/Chi Squared Distribution Question

Homework Statement

X ~ $N(2,9)\\$
Y ~ $\chi_9^2\\$
Z ~ $\dfrac{X-2}{3}\\$
What is $\dfrac{X-2}{\sqrt{Y}}\\$
What is $\dfrac{9Z^2}{Y}\\$

The Attempt at a Solution

I'm not sure about $\dfrac{X-2}{\sqrt{Y}}$, maybe a standard normal, but I'm just guessing.

And for $\dfrac{9Z^2}{Y}$, I'm pretty sure that would be distributed as a $t_9$ since a standard normal over a the square root of a chi squared should be a t distribution?

Last edited:

Ray Vickson
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Are X and Y independent? If so, the ratio is a standard distribution (NOT normal!) that can be found in any decent stats book or by looking on-line. (Google is your friend.) If X and Y are not independent you are out of luck.

RGV

Yeah, I guess that is an important to mention. X and Y are independent.

Well the ratio of 2 chi squared distributions is the F distribution, but I'm not sure what the ratio of 2 normal distributions is?

Can anyone point me the right direction on this one.

I found that the ratio of two standard normal variables is the Cauchy distribution, but I'm not sure how to handle the fact that Y has 9 degrees of freedom.

Ray Vickson
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Can anyone point me the right direction on this one.

I found that the ratio of two standard normal variables is the Cauchy distribution, but I'm not sure how to handle the fact that Y has 9 degrees of freedom.

So, am I to assume you are not willing to look in a stats book, and are not willing to do a Google search?

RGV

No, I'm willing.

I actually spent an hour looking around, but could not find what I needed.

I guess the part I did not understand was that you are allowed to divide the chi squared r.v. by the degrees of freedom to get a r.v. with a chi squared distribution and 1 degree of freedom. That was sort of the trick.

Ray Vickson
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No, I'm willing.

I actually spent an hour looking around, but could not find what I needed.

I guess the part I did not understand was that you are allowed to divide the chi squared r.v. by the degrees of freedom to get a r.v. with a chi squared distribution and 1 degree of freedom. That was sort of the trick.

Who ever gave you the idea that dividing a 9 d.f. chi-squared variable by 9 gives a 1 d.f chi-squared? Nothing could be further from the truth. A chi-squared variable divided by its d.f. is nothing other than a chi-squared variable divided by its d.f. If you plot the probability distribution of chi29(x)/9 and compare it with the plot of chi21(x) you will see that the graphs are very different.

I'm sorry that you cannot seem to locate this material in any source you consulted---I guess you are using poor sources. That material is in every half-way decent stats textbook, because it is so very important in practice.

RGV

Let me clarify....because what I said was wrong, but what I used to solve the problem, I think is correct.

If $Z \sim N(0,1)$ and $Y \sim \chi_\nu^2$ is independent of of $Z$, then $X=\dfrac{Z}{\sqrt{\dfrac{Y}{\nu}}} \sim t_\nu$

I used that to basically solve that this is a $t_1$ distribution

Ray Vickson
Homework Helper
Dearly Missed
Let me clarify....because what I said was wrong, but what I used to solve the problem, I think is correct.

If $Z \sim N(0,1)$ and $Y \sim \chi_\nu^2$ is independent of of $Z$, then $X=\dfrac{Z}{\sqrt{\dfrac{Y}{\nu}}} \sim t_\nu$

I used that to basically solve that this is a $t_1$ distribution

It would only be a t1 distribution if the Chi^squared has 1 d.f., so in your case you are still wrong, but are starting to get close.

RGV

Ah, just saw that...it should be $t_9$

Ray Vickson
Ah, just saw that...it should be $t_9$