Normal/Chi Squared Distribution Question

[mrkb80]

Homework Statement

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

Homework Equations

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?

 

[Ray Vickson]

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

 

[mrkb80]

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?

 

[mrkb80]

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]

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

 

[mrkb80]

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]

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

 

[mrkb80]

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]

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

 

[mrkb80]

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

 

[Ray Vickson]

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

Yes, you have it now.

RGV

 

[mrkb80]

Great! thanks for the help.