Help with a dirichlet function

[sneaky666]

Homework Statement

I am stuck on this problem. Can someone give me a hint on how to start?

Prove f_X1,X2 is a density.

Homework Equations

The Attempt at a Solution

I think i am supposed to integrate twice to show it equals 1 to show its a density function, but I don't know how to do this...

 

[fzero]

To integrate this, you might find the definition of the Beta function,

$$B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt, ~\text{Re}(x)> 0, \text{Re}(y)> 0,$$

useful. The Beta function also satisfies the identity

$$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.$$

 

[sneaky666]

Can you show me the next step, as I am still very lost in this question.

 

[fzero]

You might want to look up what properties a probability density function must have. The integral that you refer to in the first post is one of them. You should try to set that integral up. You will probably need to use a change of variables and perhaps the binomial formula to express the integrals in terms of Beta functions.

 

[sneaky666]

i can get to here

But I don't know how to use the beta function.

 

[sneaky666]

i am thinking of doing change of variable, but in any example i find, there is always two new variables defined, all i can think of is one which is u=x1/(1-x2)...

 

[fzero]

It's honestly not a very easy integral to do, but it can be done if you know certain properties of the incomplete beta function. The change of variables I'd use is u = x1+ x2, then the x2 integration will yield an incomplete beta function. You can find a series expansion for the incomplete beta function that will allow you to do the x1 integration in terms of beta functions.

 

[sneaky666]

aww, this is too complicated for me

the farthest i have got is

the right side looks like the beta identity, and somehow i need to use change of variables on the left side, the inner integral looks like the beta function, but there is another term inside (x1) that won't let me change it. How can I get rid of the X1?

In any case if you know how to do this, can you help me get started on this...

 

[fzero]

In any case if you know how to do this, can you help me get started on this...

I made a suggestion in post #7.

 

[sneaky666]

"The change of variables I'd use is u = x1+ x2, then the x2 integration will yield an incomplete beta function"

I can use
u = x1+ x2

but what does "x2 integration" mean?
does that mean after doing the change of variables with u=x1+x2, the integral with respect to x2 will yeild...

Also the thing i am also not sure is, since this is 2 integrals, the way i leanred to do change of variables is get 2 new variables, i can use u1=x1+x2, but what do i make u2 equal...? If i new that i can get a new equation using the jacobian

 

[fzero]

It means integrate over the part of the integrand that depends on x2. I didn't actually do it that way, I just checked the result on Wolfram Alpha.

 

[sneaky666]

so now i have

i assume this is the incomplete beta function.

I don't know how to find a series expansion for the incomplete beta function that will allow me to do the x1 integration in terms of beta functions.

 

[sneaky666]

i can't figure out know how to integrate the du (inner integral) , How do you integrate it?
If only the -x1 were not there, i can use the identity...

I also may have done the cgange of variables wrong, if i had to use the jacobian, then, it requires 2 new variables, but i only have the u=x1+x2...

 

[fzero]

i assume this is the incomplete beta function.

I don't know how to find a series expansion for the incomplete beta function that will allow me to do the x1 integration in terms of beta functions.

I think all of the formulas that you need can even be found on wikipedia if you don't have access to a text that has them.

That change of variables that I suggested isn't the only thing to attempt. I tried expanding (1-x1-x2) using the binomial formula, but that leads to a sum that I haven't been able to compute.

 

[sneaky666]

then i don't think i can even finish this...

 

[fzero]

There's an indirect way to compute the integral that is based on the way you derive the relationship between the beta function and gamma function. That computation is done on the wikipedia page

http://en.wikipedia.org/wiki/Beta_function#Relationship_between_gamma_function_and_beta_function

In your case you will rewrite $$\Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3)$$ using a coordinate change

$$u = z x_1, ~ v = z x_2, ~ w = z(1-x_1-x_2).$$

You can then factor the triple integral into the expression for another beta function and a double integral that is the integral you're trying to compute.

 

[sneaky666]

i am still very confused here, can you please show me the first couple of steps in latex?

 

[fzero]

i am still very confused here, can you please show me the first couple of steps in latex?

Try reproducing that calculation shown on the wikipedia page, the calculation you have to do is almost exactly the same. I can't do the work for you.

 

[sneaky666]

why do you put a z in front of the terms?

if i do that then i will get x1 = u/z, x2 = v/z and x3 = w/z, but in the identity in the wiki page it only had u, v, and w

and i now have

 

[fzero]

It's extremely confusing to figure out what you're trying to do when you keep writing the identity that you are trying to prove over and over again without actually doing anything. The two lines that you wrote down at the bottom make no sense at all.

Forget about the identity right now. I told you that you should compute

$$\Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3) = \left( \int_0^\infty e^{-u} u^{\alpha_1-1} du \right)\left(\int_0^\infty e^{-v} v^{\alpha_2-1} dv \right) \left( \int_0^\infty e^{-w} w^{\alpha_3-1} dw \right)$$

using the trick on that page, namely a coordinate change

$$u = z x_1, ~ v = z x_2, ~ w = z(1-x_1-x_2).$$
When you do this, you will be able to factor the new expression as a z integral that is a gamma function. The x1,x2 integral will be the integral you are trying to compute.

 

[sneaky666]

I don't understand how to do the coordinate change, so its a change of variable u=zx1 ?

 

[fzero]

I don't understand how to do the coordinate change, so its a change of variable u=zx1 ?

Yes it's a change of variable on u,v,w with the form mentioned. If you can reproduce the calculation for the beta function, you can do this one.

 

[sneaky666]

i now have

do i have to do anything to it before i have do the change of variables?

 

[sneaky666]

now i have this after the coordinate change

but i think there's something wrong with this

are the ranges of the integrals right?
and in the end isn't it supposed to be dx2dx1dz?

 

[sneaky666]

i did something, can you tell me if this is 100% correct, and if something is wrong, can you please tell me

 

[sneaky666]

so is the previous post correct??

 

[fzero]

The calculation is correct, the order in which you're writing things is a bit messed up. Just tossing random formulas around before you've proven them, without writing any words to explain what you're doing isn't a proof. You need to explain that you're trying to show that the pdf is normalized and put the steps down in a logical order.

 

[sneaky666]

yeah ok. glad i finally figured this one out...

 

[sneaky666]

by the way how do i remove the pictures i added in my previous posts?, i can't edit them