- #1

jimmy1

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So is it true that X/Y follows a beta distribution?

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- Thread starter jimmy1
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- #1

jimmy1

- 61

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So is it true that X/Y follows a beta distribution?

- #2

jimmy1

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Ok, I found the answer (just had a bit of a brain freeze!). It is X/(X+Y), and not X/Y

- #3

ZioX

- 370

- 0

X/Y does follow a beta distribution! (Assuming they have the same second parameter. This is very important). It's called the beta distribution of the second kind with parameters alpha_x and alpha_y. The F distribution is simply b*X/Y where b>0.

I'll show you why X/Y is called the beta distribution of the second kind.

Suppose X~[tex]\Gamma(\alpha_1,\beta)[/tex] and Y~[tex]\Gamma(\alpha_2,\beta)[/tex], X and Y independent. What is the distribution of [tex]U_1=\frac{X}{Y}[/tex]?

Now this is a multivariate transformation, (http://www.ma.ic.ac.uk/~ayoung/m2s1/Multivariatetransformations.PDF see here if you don't know how to do these), so we will use [tex]U_2=Y[/tex] as an auxillary equation.

So, [tex]g_1(x,y)=x/y[/tex] and [tex]g_2(x,y)=y[/tex] where x,y are positive reals (because they come from a gamma distribution) now it should be clear to see that [tex]g_1^{-1}(x,y)=xy[/tex] and [tex]g_2^{-1}(x,y)=y[/tex]. Note how g_1 and g_2 have range (0,+infty).

Therefore, [tex]f_{(U_1,U_2)}(u_1,u_2)=f_{(x,y)}(g_1^{-1}(u_1,u_2),g_2^{-1}(u_1,u_2))|J|[/tex]. As an exercise you can show that [tex]|J|=u_2[/tex]

Since X and Y are independent [tex]f_{(X,Y)}=f_X f_Y[/tex]

Now [tex]f_{(U_1,U_2)}(u_1,u_2)=f_x(u_1u_2)f_y(u_2)u_2=\frac{e^{-\frac{1}{\beta}(1+u_1)u_2}u_1^{\alpha_1-1}u_2^{\alpha_1+\alpha_2-1}}{\beta^{\alpha_1+\alpha_2}\Gamma(\alpha_1)\Gamma(\alpha_2)}[/tex]

(I have done some simplifying)

Now, we don't want the pdf of (U_1,U_2) we want the pdf of U_1, so we integrate over the joint to get the marginal distribution of U_1.

[tex]f_{U_1}(u_1)=\frac{u_1^{\alpha_1-1}}{\beta^{\alpha_1+\alpha_2}\Gamma(\alpha_1)\Gamma(\alpha_2)}\int_0^{+\infty}u_2^{\alpha_1+\alpha_2-1}e^{-\frac{1}{\beta}(1+u_1)u_2}du_2[/tex]

But the integral is just a gamma function (after we change variables). So this means that [tex]\int_0^{+\infty}u_2^{\alpha_1+\alpha_2-1}e^{-\frac{1}{\beta}(1+u_1)u_2}du_2=\frac{\Gamma(\alpha_1+\alpha_2)\beta^{\alpha_1+\alpha_2}}{(1+u_1)^{\alpha_1+\alpha_2}}[/tex].

Plugging this in we get [tex]f_{U_1}(u_1)=\frac{\Gamma(\alpha_1+\alpha_2)u_1^{\alpha_1-1}}{\Gamma(\alpha_1)\Gamma(\alpha_2)(1+u_1)^{\alpha_1+\alpha_2}}=\frac{u_1^{\alpha_1-1}}{\beta(\alpha_1,\alpha_2)(1+u_1)^{\alpha_1+\alpha_2}}[/tex]

So there we go! U_1=X/Y is distributed as that. Now why is this called a beta distribution of the second kind? If you do some transformations you should see that [tex]\beta(\alpha_1,\alpha_2)=\int_0^1x^{\alpha_1}(1-x)^{\alpha_2-1}dx=\int_0^{+\infty}\frac{x^{\alpha_1-1}}{(1+x)^{\alpha_1+\alpha_2}}dx[/tex]

I hope someone finds this interesting ;0

I'll show you why X/Y is called the beta distribution of the second kind.

Suppose X~[tex]\Gamma(\alpha_1,\beta)[/tex] and Y~[tex]\Gamma(\alpha_2,\beta)[/tex], X and Y independent. What is the distribution of [tex]U_1=\frac{X}{Y}[/tex]?

Now this is a multivariate transformation, (http://www.ma.ic.ac.uk/~ayoung/m2s1/Multivariatetransformations.PDF see here if you don't know how to do these), so we will use [tex]U_2=Y[/tex] as an auxillary equation.

So, [tex]g_1(x,y)=x/y[/tex] and [tex]g_2(x,y)=y[/tex] where x,y are positive reals (because they come from a gamma distribution) now it should be clear to see that [tex]g_1^{-1}(x,y)=xy[/tex] and [tex]g_2^{-1}(x,y)=y[/tex]. Note how g_1 and g_2 have range (0,+infty).

Therefore, [tex]f_{(U_1,U_2)}(u_1,u_2)=f_{(x,y)}(g_1^{-1}(u_1,u_2),g_2^{-1}(u_1,u_2))|J|[/tex]. As an exercise you can show that [tex]|J|=u_2[/tex]

Since X and Y are independent [tex]f_{(X,Y)}=f_X f_Y[/tex]

Now [tex]f_{(U_1,U_2)}(u_1,u_2)=f_x(u_1u_2)f_y(u_2)u_2=\frac{e^{-\frac{1}{\beta}(1+u_1)u_2}u_1^{\alpha_1-1}u_2^{\alpha_1+\alpha_2-1}}{\beta^{\alpha_1+\alpha_2}\Gamma(\alpha_1)\Gamma(\alpha_2)}[/tex]

(I have done some simplifying)

Now, we don't want the pdf of (U_1,U_2) we want the pdf of U_1, so we integrate over the joint to get the marginal distribution of U_1.

[tex]f_{U_1}(u_1)=\frac{u_1^{\alpha_1-1}}{\beta^{\alpha_1+\alpha_2}\Gamma(\alpha_1)\Gamma(\alpha_2)}\int_0^{+\infty}u_2^{\alpha_1+\alpha_2-1}e^{-\frac{1}{\beta}(1+u_1)u_2}du_2[/tex]

But the integral is just a gamma function (after we change variables). So this means that [tex]\int_0^{+\infty}u_2^{\alpha_1+\alpha_2-1}e^{-\frac{1}{\beta}(1+u_1)u_2}du_2=\frac{\Gamma(\alpha_1+\alpha_2)\beta^{\alpha_1+\alpha_2}}{(1+u_1)^{\alpha_1+\alpha_2}}[/tex].

Plugging this in we get [tex]f_{U_1}(u_1)=\frac{\Gamma(\alpha_1+\alpha_2)u_1^{\alpha_1-1}}{\Gamma(\alpha_1)\Gamma(\alpha_2)(1+u_1)^{\alpha_1+\alpha_2}}=\frac{u_1^{\alpha_1-1}}{\beta(\alpha_1,\alpha_2)(1+u_1)^{\alpha_1+\alpha_2}}[/tex]

So there we go! U_1=X/Y is distributed as that. Now why is this called a beta distribution of the second kind? If you do some transformations you should see that [tex]\beta(\alpha_1,\alpha_2)=\int_0^1x^{\alpha_1}(1-x)^{\alpha_2-1}dx=\int_0^{+\infty}\frac{x^{\alpha_1-1}}{(1+x)^{\alpha_1+\alpha_2}}dx[/tex]

I hope someone finds this interesting ;0

Last edited by a moderator:

- #4

alexis_k

- 1

- 0

Hello!

I desperately need a proof of the fact that x/(x+y) has a beta distribution.

I desperately need a proof of the fact that x/(x+y) has a beta distribution.

- #5

SW VandeCarr

- 2,175

- 81

Hello!

I desperately need a proof of the fact that x/(x+y) has a beta distribution.

The mean of the beta distribution is [tex] \mu=\frac{\alpha}{\alpha+\beta}[/tex]. Does this help you?

Edit: Look up the PDF and the MGF of the beta distribution. I assume you know the relationship between the gamma and beta functions. By the way, just saying x/(x+y) doesn't mean much by itself. I'm assuming it's relevant to the ratio of two independent gamma distributions.

Last edited:

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