# Determine constant c that random variable will have a t distribution?

1. Dec 2, 2011

### Mixer

Determine constant c so that random variable will have a t distribution?

1. The problem statement, all variables and given/known data

Suppose that five random variables x1, x2, x3, x4, x5 are independent and have normal distribution N(0,1). Determine a constant c such that the random variable

c*(x1+x2)/$\sqrt{x_3^2 + x_4^2 + x_5^2}$
will have a t distribution?

2. Relevant equations

3. The attempt at a solution

I'm pretty confuced what to do. My cource material doesn't offer any clues how to approach this problem and my teacher didn't show anything related to this kind problem. Any clues for me? Thanks

Last edited: Dec 2, 2011
2. Dec 2, 2011

### LCKurtz

Re: Determine constant c so that random variable will have a t distribution?

Well, if Z ~ N(0,1) and Y ~ $\chi^2(\nu)$, then you know that $$\frac{Z}{\sqrt{\frac Y \nu}}$$ has a student's T distribution. That has a similarity to your problem doesn't it? What is the distribution of x1 + x2? What about that sum of squares under the square root sign? Can you pick c to make it work?

3. Dec 3, 2011

### Mixer

Re: Determine constant c so that random variable will have a t distribution?

Hi there!

If I'm understanding correctly:

x1 + x2 has distribution N(0,2)

So I need to divide x1 + x2 by two in order to have distribution N(0,1) -->

(x1 + x2) / 2

In order to make the denominator to be in form $${\sqrt{\frac Y \nu}}$$

I have to divide x32 + x42 + x52 by three --->

(x32 + x42 + x52) / 3

So if I pick c = $$\frac{\sqrt{3}}{2}$$

It should work?

4. Dec 3, 2011

### LCKurtz

Yes. That looks good.

5. Dec 3, 2011

### Mixer

Thanks very much for your help!

6. Dec 3, 2011

### Ray Vickson

Re: Determine constant c so that random variable will have a t distribution?

Var(X1+X2) = 2, so to get N(0,1) we need to look at (X1+X2)/sqrt(2), not (X1+X2)/2.

RGV

7. Dec 3, 2011

### LCKurtz

Re: Determine constant c so that random variable will have a t distribution?

Woops! Yes.

8. Dec 3, 2011

### Mixer

Ok, so is the correct answer then

c = √(3/2) ??

I had similar kind of problem earlier:

Determine c such that Y will have chi-squared distribution,

Y = (x1 + x2 + x3)^2 + (x4 + x5 + x6)^2

x1,x2,x3,x4,x5,x6 have N(0,1) distribution.

x1 + x2 + x3 has distribution N(0,3)

So in order to have N(0,1) distribution I have to divide x1 + x2 + x3 by three:

x1 + x2 + x3 / 3

I have to square it in order to have chi-squred distribution. Same for x4 + x5 + x6. So if i pick c = 1/9 it should work?

Is this correct then 1/3 ?

9. Dec 3, 2011

### Ray Vickson

The notation N(0,3) is ambiguous. In some books and papers the "3" is the variance, while in others it is the standard deviation. Your statement X1+X2+X3~N(0,3) is correct if "3" is the variance. The standard deviation is sqrt(3), so (X1+X2+X3)/sqrt(3) ~ N(0,1).