1. ## Distribution

Hello, I having doubts on this issue

1) If X~Gamma(a,B) and Y~Beta(γ,B-γ), respectively with B>= 0 and γ>= 0, such that
γ <= B. Which distribution Z= XY?

I tried Jacobian, convolution integral, but I can not get anything

2. ## Re: Distribution

Is this theoretical or do you have data?

3. ## Re: Distribution

This is theoretical, I need to find the distribution of Z=XY.

4. ## Re: Distribution

Both method require similar effort. For Jacobian way you need to first make up a variable, say to pair up with as the original transformation is from . Then you can proceed with the usual Jacobian in this transformation.

5. ## Re: Distribution

I did this BGM, but I can not get somewhere.
Did Z= XY and W = Y, calculated the Jacobian and find the fZ, W (z, w) (joint distribution),
but when I make a marginal, fz (z) distribution get caught.

6. ## Re: Distribution

What I make:
and and solve I have
so and,

now

so

The support of Beta distribution is [0,1] and W=Y~Beta.
Now I do not know how to continue.

7. ## Re: Distribution

Ok just search the question a little bit and this is a well-known result. The problem is that there is a typo in the question:

Let and they are independent. Then

The key fact here is that the sum of the two Beta's parameters need to equal to the shape parameter of the Gamma distribution. Otherwise there will be no nice result.

Continue your integration with some careful simplifications and a change of variable, then it will be done. I have just verified it once.

8. ## Re: Distribution

Originally Posted by BGM
Ok just search the question a little bit and this is a well-known result. The problem is that there is a typo in the question:

Let and they are independent. Then

The key fact here is that the sum of the two Beta's parameters need to equal to the shape parameter of the Gamma distribution. Otherwise there will be no nice result.

Continue your integration with some careful simplifications and a change of variable, then it will be done. I have just verified it once.
This guy tried to change what you said, but still can not solve. Do you have a solution?

9. ## Re: Distribution

Yes as I said I have did it on rough paper completely (and already thrown away). Anyway you are in the correct direction for the convolution step and the remaining task is just a simple integrating process. It is not hard. It will be better to show which part you stuck at.

10. ## Re: Distribution

Ok, now the second part of exercise.
2)Find (conditional distribution)
I can not assume that Z and Y are independent right? What better way to find the conditional distribution.

11. ## Re: Distribution

From the first part of the question you already obtain the joint pdf of in the calculation process. Then by definition you just need to divide it by the marginal pdf of to obtain the conditional density.

Note that it will be also easy to see that

has the same distribution as

as are independent.

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