# Let $X,Y$ be two independent random variables with exponential distribution and parameter $\lambda > 0$.

Let $$X,Y$$ be two independent random variables with exponential distribution with parameter $$\lambda > 0$$. Let $$S = X + Y$$ and $$T = \frac{X}{S}$$. I want to find the joint density function of $$(S,T)$$ . I want then to calculate the marginals and say whether or not $$S$$ and $$T$$ are independent. I start by finding the density function of $$S$$ using the convolution:

$$f_S(s) = \int_{-\infty}^{+\infty}f_X(s-t)f_Y(t)dt$$ $$= \int_{0}^{s} \lambda^2 e^{-\lambda s} dt = \lambda^2 e^{-\lambda s}$$

Then I tried to calculate the density function of $$T$$ but I am stuck here:

$$F_T(t) = \mathbb{P}(T \leq t) = \mathbb{P}\left(\frac{X}{X+Y} \leq t\right).$$ Is this the right method of solving this? Should I find the joint density first? (The problem is that I do not know how to to that)

• Hint: $\frac{X}{X+Y} \leq t \Leftrightarrow X \leq tX+tY \Leftrightarrow (1-t)X \leq tY$ (note that we can assume $t \in [0,1]$) – Stockfish Nov 5 '18 at 9:46
• I see. So it is right to find the two density function first? Why can we assume t in that interval? – qcc101 Nov 5 '18 at 9:47
• Which values can be attained by T? – Stockfish Nov 5 '18 at 9:48
• @qcc101 because $0\leq X\leq X+Y$ a.s. – drhab Nov 5 '18 at 9:49
• There is a mistake/typo in your calculation of $f_S(s)$. It must end with $\cdots=\lambda^2se^{-\lambda s}$. Btw, IMHO it is handsome to concentrate on $\lambda=1$ at first hand. Afterwards the solution can be translated easily to the general case. – drhab Nov 5 '18 at 9:58

Assuming you mean $$\lambda$$ is the rate parameter here (i.e. Exponential with mean $$1/\lambda$$).

First of all, recheck your density of $$S$$. The correct density as mentioned in comments is $$f_S(s)=\lambda^2se^{-\lambda s}\mathbf1_{s>0}$$

It is easy to verify that the density of $$T$$ is $$f_T(t)=\mathbf1_{0

You can find the joint distribution function of $$(S,T)$$ as follows:

For $$s>0$$ and $$0,

\begin{align} P(S\le s,T\le t)&=P\left(X+Y\le s,\frac{X}{X+Y}\le t\right) \\&=\iint_D f_{X,Y}(x,y)\,dx\,dy\quad\quad,\text{ where }D=\{(x,y):x+y\le s,x/(x+y)\le t\} \\&=\lambda^2\iint_D e^{-\lambda(x+y)}\mathbf1_{x,y>0}\,dx\,dy \end{align}

Change variables $$(x,y)\to(u,v)$$ such that $$u=x+y\quad,\quad v=\frac{x}{x+y}$$

This implies $$x=uv\quad,\quad y=u(1-v)$$

Clearly, $$x,y>0\implies u>0\,,\,0

And $$dx\,dy=u\,du\,dv$$

So again for $$s>0\,,\,0,

\begin{align} P(S\le s,T\le t)&=\lambda^2\iint_R ue^{-\lambda u}\mathbf1_{u>0,0

This proves the independence of $$S$$ and $$T$$, with $$S$$ a Gamma variable and $$T$$ a $$U(0,1)$$ variable.

And from the joint distribution function, it is readily seen (without differentiating) that the joint density of $$(S,T)$$ is $$f_{S,T}(s,t)=\lambda^2 se^{-\lambda s}\mathbf1_{s>0,0

Note that the change of variables isn't really necessary if you are comfortable with the first form of the double integral.

The joined density is defined as

$$f_{S,T}(s,t) \\= \int _0^{\infty }\int _0^{\infty }\lambda ^2 e^{-\lambda (x+y)} \delta (s-x-y) \delta \left(t-\frac{x}{x+y}\right)dydx\tag{1a}$$

Here $$\delta(x)$$ is the Dirac delta function.

Changing variables $$x\to u v, y\to u(1-v)$$ gives $$u\in (0,\infty), v\in (0,1)$$, the modulus of the Jacobian $$u$$, so that the integral becomes

$$f_{S,T}(s,t) \\=\int _0^{\infty }\int _0^1\lambda ^2 u e^{-\lambda u} \delta (s-u) \delta (t-v)dvdu\\= \lambda ^2 s e^{-s \lambda } \left\{ \begin{array} {ll} 1 & 0\lt t \lt 1 \\ 0 & \, \textrm{otherwise} \end{array} \right.\tag{1b}$$

On the other hand we have for the separate density functions

$$f_{S}(s)\\= \int _0^{\infty }\int _0^{\infty }\lambda ^2 e^{-\lambda (x+y)} \delta (s-x-y)dydx = \lambda ^2 s \; e^{-\lambda s}\tag{2a}$$

$$f_{T}(t) \\= \int _0^{\infty }\int _0^{\infty }\lambda ^2 e^{-\lambda (x+y)} \delta \left(t-\frac{x}{x+y}\right)dydx \\ = \left\{ \begin{array} {ll} 1 & 0\lt t \lt 1 \\ 0 & \, \textrm{otherwise} \end{array} \right.\tag{2b}$$

And, finally, from $$(1)$$ and $$(2)$$ we have

$$f_{S,T}(s,t) = f_{S}(s) f_{T}(t)\tag{3}$$

which shows the independence of $$s$$ and $$t$$.