# Functions of random variables

1. May 2, 2007

### jimmy1

I have two random variables $$X$$ and $$Y$$. Now the distribution of $$X$$ and $$Y$$, is a bit complicated. Basically they follow Gamma distributions, $$X=\Gamma(k1,\theta)$$ and $$Y=\Gamma(k2,\theta)$$, but $$k1$$ and $$k2$$ are Poisson distributed. But I do have a closed form expression for the distribution of $$X$$ and $$Y$$, and also have an expression for their mean and variances.

Now what I would like to do is find an expression for the mean and variance of $$X/(X+Y)$$.
So I'd like to know, is there a short way of getting the mean and variance of $$X/(X+Y)$$, from the mean and variance of $$X$$ and $$Y$$ and if not, then how would I go about finding the mean and variance of $$X/(X+Y)$$.
Any ideas??

2. May 2, 2007

### EnumaElish

From college prob & stat, I remember quotients of random variables as being notoriously difficult analytically. E.g. their means and variances being expressed with the "squiggly equal sign" rather than the "straight equal sign."

But for practical applications, you can always simulate.

3. May 3, 2007

### EnumaElish

... unless you can show Γ1/(Γ1 + Γ2) ~ F, where F is a known distribution. E.g., if X and Y were i.i.d. Normal with zero means, then X + Y would be Normal with zero mean, and Normal(0,σ12)/Normal(0,σ22) is distributed Cauchy -- which does not happen to have moments, so if this were the case you would have a nonexistence result.

Last edited: May 3, 2007
4. May 3, 2007

### jimmy1

Thanks for that EnumaElish. And yes Γ1/(Γ1 + Γ2) does follow a known distribution, as Beta distribution. So in my case the distribution of $$X/(X+Y)$$ would follow a Beta distribution with parameters k1 and k2, ie $$B(k1,k2)$$.
But the problem with this is that there is no closed form expression of $$B(k1,k2)$$ when k1, and k2 are Poisson distributed, so I'm unable to get a closed form expression for the mean and variance of $$X/(X+Y)$$.
That's why I was wondering if there was any way of getting an expression for the mean and variance of $$X/(X+Y)$$, from the mean and variance of $$X$$ and $$Y$$, or maybe a similar shortcut approach??

5. May 3, 2007

### Jason Swanson

The first parameter in a Gamma distribution must be positive. However, a Poisson random variable has a positive probability of being 0. So this model looks like it is not well-defined.

6. May 5, 2007

### ZioX

X/(X+Y) is a beta distribution with parameters k1 and k2. Therefore, the mean of X/(X+Y) is $\frac{k1}{k1+k2}$ and the variance is $\frac{k1k2}{(k1+k2)^2(k1+k2+1)}$. Now just take the expected values of these. Which isn't the funnest thing in the world.

I think I proved that the distribution of X/(X+Y) is a beta distribution sometime ago on these forums. Or it might have been X/Y is beta (of the 2nd kind).

Last edited: May 5, 2007
7. May 5, 2007

### Jason Swanson

Those expected values are undefined. The probability that k_1 = 0 and k_2 = 0 is positive, in which case both those expressions are 0/0. So it is not even that those expectations are infinite. They simply do not make sense. The OP needs to correct his model, so that it is well-defined.

8. May 13, 2007

### jimmy1

Yes you're right the model is undefined when k1=0 or k2=0, but isn't this just a technicality??
So, am I right in saying that there is nothing wrong with the model, as long as I state that k1>0 and k2>0, and hence making the model well-defined??

9. May 13, 2007

### Jason Swanson

But if k1 > 0 and k2 > 0, then they are not Poisson distributed. So what is their distribution?

10. May 13, 2007

### jimmy1

Well technically k1 and k2 are binomially distributed, but I was thinking of using the Poisson distribution to make things a bit easier.
But anyway from your argument, the problem still exists if I use a binomial dsitribution?
So basically you are saying that I cannot use the binomial or Poisson distribution as the distribution for the two parameters in a Beta distribution?
But can't I just say something like k1 and k2 are 'approximations' to binomial/poisson distributions where k1>0 and k2>0??

11. May 13, 2007

### Jason Swanson

You are right that you cannot use Binomial either. You must use a distribution with no mass at 0. In my opinion, it is not enough to say that it is an approximation to a Poisson or a Binomial. You must tell us what that approximation is. There may be many ways to approximate these, and your results may be different, depending on how you do it. Is this a model of some real-world phenomenon? If so, what is it? The context of the model may help you determine which distribution to use.