CDF of $\frac{T_1}{T_1+T_2}$ where $T_1,T_2$ are the first two delays in a Poisson process?

Question

$P$ is a Poisson Process with rate $\lambda$. Let $T_1$ be the time of the first event and let $T_2$ be the time of the from the first to the second event. Let $Y = \frac{T_1}{T_1+T_2}$. Find the density of $Y$.

I think I should find the CDF first and then take the derivative, but I do not know how to find $P(Y\leq t)$. Do I need to find the joint density?

Answer 1

\begin{align} & \Pr(Y\le t) = \Pr( T_1 \le t(T_1+T_2)) = \Pr((1-t)T_1\le tT_2) \\[8pt] = {} & \iint\limits_{(u,v)\,:\,(1-t)u\,\le\,tv} e^{-u} e^{-v} \, d(u,v) \\[8pt] = {} & \int_0^\infty \left( \int_0^{tv/(1-t)} e^{-u} e^{-v} \, du \right) \, dv \\[8pt] = {} & \int_0^\infty \big( 1 - e^{-tv/(1-t)} \big) e^{-v} \, dv \\[8pt] = {} & 1 - (1-t) = t \quad \text{provided $0<t<1$}. \\ & \text{So the distribution is uniform on $[0,1]$.} \end{align}

Answer 2

Hints:

• $$P(Y \le y) = P(T_1 \le y(T_1 + T_2)) = P(\frac{1-y}{y}T_1 \le T_2).$$
• Show that $T_1$ and $T_2$ are independent $\text{Exponential}(\lambda)$ random variables using properties of Poisson processes.
• Then, for $y \in (0,1)$, $$P(Y \le y) = \int_0^\infty \int_{\frac{1-y}{y} t_1}^\infty \lambda^2 e^{-\lambda t_1} e^{-\lambda t_2} \, dt_2 \, dt_1 = \int_0^\infty \lambda e^{-\lambda t_1} e^{-\lambda\frac{1-y}{y} t_1} = y$$

Answer 3

HINT For convenience, let $T=T_1$ and $S=T_2$. Note they are iid, so if $f_T(t)$ is the pdf of $T$ then $f_T(s) = f_S(s)$ and $f_{T,S}(t,s) = f_T(t)f_S(s) = f_T(t)f_T(s)$.

Once you know the joint pdf, $$ F_Y(y) = \mathbb{P}[Y \le y] $$ reduces to a regular 2D integral.

Answer 4

nice question! Here is an alternative approach if you have studied transformations of random variables:

Let $X = T_1 + T_2 $ and $Y = \frac{T_1}{T_1 + T_2} $
We have that $(X,Y) = (T_1+T_2 , \frac{T_1}{T_1 + T_2}) := g(T_1,T_2)$
And define $h(X,Y):=g^{-1}(X,Y) = (XY,(1-Y)X)$

The Jacobian of $h$ is the determinant of \begin{pmatrix} x & y \\ -x & 1-y \end{pmatrix} Which is $x$

And hence you can easily find the joint density of $(X,Y)$ and then integrate out $x$ to find the pdf of $Y$