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

[Mixer]

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

Homework Statement

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?

Homework Equations

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

 

[LCKurtz]

Homework Statement

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?

Homework Equations

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

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?

 

[Mixer]

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?

Hi there!

Thanks for reply!

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?

 

[LCKurtz]

Yes. That looks good.

 

[Mixer]

Thanks very much for your help!

 

[Ray Vickson]

Hi there!

Thanks for reply!

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?

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

RGV

 

[LCKurtz]

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

RGV

Woops! Yes.

 

[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 ?

 

[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).