# CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions

Let $X_1$ and $X_2$ be two independent exponential random variables with the PDFs $f_{X_1}(x_1)=\lambda_1 \exp(-\lambda_1 x_1)$ and $f_{X_2}(x_2)=\lambda_2 \exp(-\lambda_2 x_2)$.

Let $X= \frac{X_1X_2}{X_1+X_2+a}$ be a r.v. for which I want to derive the CDF, where $a$ is a positive constant.

In other words, I need to calculate $\mathbb{P}\{X <x \}$.

Any ideas or hints?

Edit (attempt): Using a similar approach to that used here, we have $X_1 \in [0, \infty)$ and $X_2 \in [0, \frac{x(x_1+a)}{x_1-x})$. The latter interval results from: $\frac{x_1x_2}{x_1+x_2+a} < x$, thus $x_2(x_1-x)<x(x_1+a)$ which implies $x_2 < \frac{x(x_1+a)}{x_1-x}$; but I am not sure if this is correct since $x_1-x$ can sometimes be negative (?). Based on the above, we can write: \begin{align} \mathbb{P} \{ \frac{X_1X_2}{X_1+X_2+a} < x \} & = \int_{x_1=0}^\infty \int_{x_2=0}^{ \frac{x(x_1+a)}{x_1-x}} \lambda_1 \exp(-\lambda_1 x_1) \lambda_2 \exp(-\lambda_2 x_2) dx_2 dx_1 \\ &= \int_{x_1=0}^\infty \left(1-\exp(-\lambda_2\frac{x(x_1+a)}{x_1-x}) \right) \lambda_1 \exp(-\lambda_1 x_1) dx_1 \\ & = 1-\lambda_1 \int_{x_1=x}^\infty \exp(-\lambda_2\frac{x(x_1+a)}{x_1-x} ) \exp(-\lambda_1 x_1) dx_1 \end{align} In the last equality: for the integral of $\lambda_1 \exp(-\lambda_1 x_1) dx_1$, I suppose that $x_1 \in [0,\infty)$, whereas for the second term I suppose that $x_1 \ge x$ (since $x_2$ should be positive) (?).

Is my approach correct ?

• If the function has a one to one mapping with the component random variables, ie, there exist only countably finite $X_1,X_2$ for each value of the random variable $X$, then it's just a matter of adding possible values for $X_i's$ Will add calculations as soon as I reach home from work. Dec 19, 2016 at 14:03
• @Giridhur Unfortunately these are continuous random variables and so your method won't work Dec 19, 2016 at 16:54
• Why is that unfortunate? Do continuous rv's not have feelings too? I was looking forwards to the workings ... Dec 19, 2016 at 18:01
• That the upper bound $\frac{x(x_1+a)}{x_1-x}$ of the integral is sometimes negative shows the formula you suggest canot be correct. Instead, consider $$Y=\frac{X_1X_2}{X_1+X_2+a}$$ and, for every $y>0$, use the identity of events $$[Y<y]=[X_1\leqslant y]\cup[X_1>y,X_2<y(X_1+a)(X_1-y)^{-1}]$$ hence $$P(Y<y]=P[X_1\leqslant y]+\int_y^\infty P[X_2<y(x+a)(x-y)^{-1}]f_{X_1}(x)dx$$ Can you finish this?
– Did
Dec 20, 2016 at 11:29
• "I think I am getting the same result in my apporach because in the last equality I change the bounds of integration so that $x_1−x$ stays positive" Indeed this step in your computations is particularly unconvincing. To complete the computation of $P(Y<y)$, I would start from the change of variable $x=y+z$ in the last integral, which yields $$P(Y<y)=1-e^{-\lambda_1y}+\int_0^\infty P(X_2<y+(x+a)/z)\lambda_1e^{-\lambda_1(y+z)}dz$$
– Did
Dec 20, 2016 at 16:26

Thanks to the indications that @Did gave, I was able to derive the CDF of $Y=\frac{X_1X_2}{X_1+X_2+a}$ (note that in the question I use $X$ instead of $Y$) as follows:

Based on the identity of events \begin{align} \left[ Y <y \right] = \left[ X_1 < y \right] \cup \left[ X_1 \ge y, X_2 < y (X_1+a)(X_1-y)^{-1} \right], \end{align} we get the following \begin{align} \nonumber \mathbb{P} \{ Y < y \} &= \mathbb{P}\{X_1 <y\}+ \int_y^\infty \mathbb{P}\{X_2 < y (x+a) (x-y)^{-1} \} f_{X_1}(x) \, dx \\ \nonumber &= 1-e^{-\lambda_1y} + \lambda_1 \int_y^\infty (1-e^{-\lambda_2 y(x+a)(x-y)^{-1}}) e^{-\lambda_1 x} dx \\ \nonumber &= 1-e^{-\lambda_1y}+e^{-\lambda_1y}- \lambda_1 \int_y^\infty e^{-\lambda_2 y(x+a)(x-y)^{-1}} e^{-\lambda_1 x} dx \\ \nonumber & =_{(i)} 1- \lambda_1 \int_{u=0}^\infty e^{ -\lambda_2 y(u+y+a)u^{-1}} e^{-\lambda_1 (u+y)} du \\ \nonumber &= 1- \lambda_1 e^{-(\lambda_1+\lambda_2)y} \int_{u=0}^\infty e^{-\lambda_2 y (y+a)u^{-1}} e^{-\lambda_1u} du \\ & =_{(ii)} 1- \lambda_1 e^{-(\lambda_1+\lambda_2)y} \, 2 \, \sqrt{ y(y+a) \lambda_2 \lambda_1^{-1} } \, K_1\left(2 \sqrt{ y(y+a) \lambda_2 \lambda_1 } \right), \\ & = 1- e^{-(\lambda_1+\lambda_2)y} \, 2 \, \sqrt{ y(y+a) \lambda_2 \lambda_1 } \, K_1\left(2 \sqrt{ y(y+a) \lambda_2 \lambda_1 } \right) \end{align} in which equality (i) is due to the change of variable $u=x-y$ and equality (ii) follows from [Table of Integrals, Series and Products, 7th edition - equation 3.471.9].

• You are correct. I have edited the answer.
– din
Dec 21, 2016 at 19:00
• Yes. What remains is a success story for this site, it seems... Simply wait a little while before accepting your own answer, to let other people react to it if they wish to.
– Did
Dec 21, 2016 at 19:05

Let $X$ and $Y$ denote two independent exponential random variables and suppose that $V = \frac{XY}{X+Y+a}$ where $a > 0$. What is the CDF of $V$?

First, note that $V > 0$. Let $v$ denote a positive constant, and let us try to determine the complementary CDF $P\{V > v\}$ by integrating the joint density of $X$ and $Y$ over that part of the first quadrantVwhere $V$ exceeds $v$. The set in question, call it $A$, is given by \begin{align} A &= \{(x,y)\colon x > 0, y > 0, \frac{xy}{x+y+a} > v\}\\ &= \{(x,y)\colon x > 0, y > 0, xy > v(x+y+a)\}\\ &= \{(x,y)\colon x > 0, y > 0, xy -vx -vy > av\}\\ &= \{(x,y)\colon x > 0, y > 0, (x-v)(y-v) > v^2 + av\}. \end{align} Now, the graph of the hyperbola $xy = b$ consists of two curves confined to the first and third quadrants respectively and passing through the points $\left(\sqrt{b}, \sqrt{b}\right)$ and $\left(-\sqrt{b}, -\sqrt{b}\right)$ respectively. Therefore, the graph of $(x-v)(y-v) = b$ is just these two curves shifted to the right by $v$ and shifted upwards by $v$, and the two curves now pass through $(\sqrt{b}+v, \sqrt{b}+v)$ and $(-\sqrt{b}+v, -\sqrt{b}+v)$ respectively. Note that the asymptotes of the curves are $x=v, y=v$. Now, when $b$ equals $v^2+av$, $\sqrt{b} > v$ and so the point $(-\sqrt{b}+v, -\sqrt{b}+v)$ is in the third quadrant. Consequently, the lower branch of the hyperbola does not lie in the first quadrant at all. (It does cross the $x$ and $y$ axes into the second and fourth quadrants but that is immaterial in this problem). It follows that we can express $A$ as $$A = \{(x,y)\colon x > v, y > v, (x-v)(y-v) > v^2 + av\}.$$ Hence, \begin{align} 1-F_V(v) &= P((X,Y)\in A)\\ &= \iint_A f_{X,Y}(x,y)\, \mathrm dy \, \mathrm dx\\ &= \int_v^\infty f_X(x) \left[ \int_{y = \frac{v^2+av}{x-v}}^\infty f_Y(y)\,\mathrm dy \right] \, \mathrm dx. \end{align} The inner integral is straightforward to evaluate; the outer one is trickier, needing special functions and tables of integrals to evaluate.

An alternative calculation given in an answer by the OP (with the help of many suggestions from @Did) directly evaluates the CDF $P\{V \leq v\}$ by partitioning the set under consideration into the events $\{X \leq v\}$ and $\left\{X>v, 0 < Y \leq v\frac{x+a}{x-v}\right\}$.