1. ## poisson process

Let {N(t) : t >=0} be a Poisson process of rate 1 and let $T_{1} < T_{2} < ...$ denote the times of the points. Derive the pdf of $Y=T_{2}/T_{4}$

So I've been trying to work with the distribution function. I have:

$P(T_{2}/T_{4} > x) = P(T_{2} > xT_{4}) = P(N(xT_{4}) \in \{0,1\})$

I get stuck there though and can't find a similar example.

2. Originally Posted by Beaky
Let {N(t) : t >=0} be a Poisson process of rate 1 and let $T_{1} < T_{2} < ...$ denote the times of the points. Derive the pdf of $Y=T_{2}/T_{1}$

So I've been trying to work with the distribution function. I have:

$P(T_{2}/T_{4} > x) = P(T_{2} > xT_{4}) = P(N(xT_{4}) \in \{0,1\})$

I get stuck there though and can't find a similar example.
It may help you to prove that $T_n -T_{n-1}$ are i.i.d. exponentials (where T_0=0). Now $Y=\frac{T_2-T_1}{T_1}+1$ where both $T_2-T_1$ and $T_1$ are independent exponentials. Now use the law of total probability and condition one of them.

3. Sorry, typo in original post. It's $Y=T_{2}/T_{4}$. I'll try to use your advice anyways though.

4. I get to $P(\frac{T_{2}}{T_{4}}\le x)=P(\frac{T_{4}-T_{2}}{T_{2}}\ge \frac{1}{x}-1)$

But I get stuck here. I can't figure out how to apply conditioning and the law of total probability since the variables are continuous.

5. Originally Posted by Beaky
I get to $P(\frac{T_{2}}{T_{4}}\le x)=P(\frac{T_{4}-T_{2}}{T_{2}}\ge \frac{1}{x}-1)$

But I get stuck here. I can't figure out how to apply conditioning and the law of total probability since the variables are continuous.
I would suggest getting T_2-T_1 as it is exponential. The law of total probability says (for example)
$
P(X+Y \in A)=\int P(X+y \in A)P(Y \in dy).
$

So suppose that X=T_2-T_1, then X is exponential with parameter one and
$
P(Y \leq y)=\int P(T_4\geq x/y)e^{-x} dx
$

now you should try to figure out the law of T_4.