# Thread: Poisson process: quotient of two gamma r.v.'s

1. ## Poisson process: quotient of two gamma r.v.'s

Let {N(t): t≥0} be a Poisson process of rate 1 and let S_r be the time to the rth point. Calculate the density function of X = S_2/S_6 using the Jacobian transformation method.

To use the Jacobian method, I need to define another random variable Y that is a function of S_2 and S_6.
X=S_2/S_6
Y=???
I know that S_2~gamma(2,1) and S_6~gamma(6,1), but I am having some trouble defining the other random variable Y, and the problem is that S_2 and S_6 are not independent.
The Jacobian method expresses the joint density of X and Y in terms of S_2 and S_6, so I must first obtain the joint density of S_2 and S_6. But now it seems like that I have no way of obtaining the joint density of S_2 and S_6 because they are NOT independent random variables (S_2<S_6 always).
So how can we solve this problem using the Jacobian method?

Any help is much appreciated!

2. Usually Y is either S_2 or S_6,
S_6 might be easier
You must find that first, and yes they are dependent.

3. Originally Posted by kingwinner
the problem is that S_2 and S_6 are not independent.
But $S_2$ and $S_6-S_2$ are independent, and the second one has same distribution as $S_4$, so let's call it $S'_4$, so that $\frac{S_2}{S_6}=\frac{S_2}{S_2+S'_4}$ and then you can do the usual method (like MathEagle said, Y may be chosen to be $S_2$ or $S'_4$).

4. I was going to add that S6-S2 is indep of S2 since the time periods are separate, but I wanted kingswinner to figure that out.

5. Originally Posted by matheagle
I was going to add that S6-S2 is indep of S2 since the time periods are separate, but I wanted kingswinner to figure that out.
Oh, sorry to spoil your pedagogy... I feel like we're quits now, since the same happened on an other thread in reverse (that was a couple of weeks ago)

Best wishes,
Laurent.

6. Originally Posted by Laurent
But $S_2$ and $S_6-S_2$ are independent, and the second one has same distribution as $S_4$, so let's call it $S'_4$, so that $\frac{S_2}{S_6}=\frac{S_2}{S_2+S'_4}$ and then you can do the usual method (like MathEagle said, Y may be chosen to be $S_2$ or $S'_4$).
But I have some more questions:

1) If I take Y to be (S2 + S4') which is the denominator, is that also OK? Will I get the correct answer?

2) I don't quite follow your argument that (S6 - S2) would have gamma(4,1) distribution.
What I have learned is the following...
Theorem: if X1~gamma(r1,lambda) and X2~gamma(r2,lambda), with X1 and X2 INDEPENDENT, then the SUM (X1+X2)~gamma(r1+r2,lambda).

But I am not too sure about a SUBTRACTION of gamma r.v.'s, particuarly when S6 and S2 are DEPENDENT?? Is the theorem above applicable?

Thank you very much!

7. Originally Posted by kingwinner
2) I don't quite follow your argument that (S6 - S2) would have gamma(4,1) distribution.
You should try to picture what S6-S2 is.

(and no, you can't "substract", whatever that would mean)

8. 1) Actually, I am puzzled about this for a long time. The Jacobian method always requires TWO random variables. But we are almost always only given one, and somehow we have to "invent" another random variable, but a lot of times I have no idea how...
Is there any general rule or hint about how this second random variable can be constructed?
X=S2/(S2+S4')
Y=???

How do we know what to pick for Y, and how can we know whether it is going to work or not?
Can we take Y=S4' ? How about Y=S2+S4'? Are there any other possible choice of Y that would work?

2) I can picture S6-S2 as the time between the 2nd and 6th point, but I don't get why S6-S2 is necessarily gamma distributed.
Suppose X1~gamma(6,1), X2~gamma(2,1), and X1 and X2 are NOT independent. Does this imply that X1-X2 ~gamma(6-2,1)=gamma(4,1) ? If so, how can we prove it? I haven't seen this theorem before...

Thanks for any help!