# On exponential distributions and dot products

Let

• $$a, b$$ be two variables drawn from an exponential distribution with parameter $$\lambda_1$$.

• $$c, d$$ be two variables drawn from an exponential distribution with parameter $$\lambda_2$$.

I am interested in the probability density function (PDF) of random variable

$$\frac{ac+bd}{c+d}$$

Simulations give the same PDF of that of the random variable $$\frac{a+b}{2}$$, which is the exponential distribution with parameter $$\lambda_1$$.

I've spent a month trying to prove it but without success. Could you please help by proving whether this statement holds true or not?

• OK. apparently, they are very similar but not identical. – Chrysanthi Pas Jul 16 '19 at 12:57
• Ok, apparently, they are not the same :( – Chrysanthi Pas Jul 16 '19 at 12:57
• Do you assume independence? – kjetil b halvorsen Jul 16 '19 at 13:03
• Just posting a question on MO seems to be very enlightening! – Jochen Wengenroth Jul 16 '19 at 13:12
• It’s a good lesson to learn, and something that happens to everyone at least once. When you are spending a lot of time trying to prove something, always remember it may not be true! – Dan Romik Jul 16 '19 at 18:29

I'm assuming you mean $$a, b, c, d$$ to be independent exponential random variables with rate parameters $$\lambda_1, \lambda_1, \lambda_2, \lambda_2$$.

I find that $$(ac+bd)/(c+d)$$ has mean $$\lambda_1^{-1}$$ (the same as $$(a+b)/2$$), but variance $$2 \lambda_1^{-2}/3$$ while $$(a+b)/2$$ has variance $$\lambda_1^{-2}/2$$.

$$\newcommand{\la}{\lambda}$$ Welcome to MathOverflow!

This conjecture is of course false, if the informal term "drawn" you are using means independence. Indeed, by homogeneity, without loss of generality $$\la_2=1$$. Let now $$\la:=\la_1$$, $$(U,V,X,Y):=(a,b,c,d)$$, $$\begin{equation*} S:=\frac{UX+VY}{X+Y},\quad T:=\frac{U+V}2. \end{equation*}$$

We have to show that the pdfs $$f_S$$ and $$f_T$$ of $$S$$ and $$T$$ differ from each other. The random variable (r.v.) $$T$$ has the Gamma distribution with parameters $$2,\la/2$$. So, $$\begin{equation*} f_T(t)=4t\,e^{-2t/\la}/\la^21_{t>0}, \end{equation*}$$ whence $$\begin{equation*} f_T(t)/t\to4/\la^2\quad\text{ as }\quad t\downarrow0. \end{equation*}$$

On the other hand, solving the equation $$s=\frac{ux+vy}{x+y}$$ for $$u$$, to get $$u=\frac{(x+y)s-vy}x$$ and $$\frac{\partial u}{\partial s}=\frac{x+y}x$$, we see that the joint pdf of $$(S,V,X,Y)$$ is given by \begin{align*} f_{S,V,X,Y}(s,v,x,y)&=f_{U,V,X,Y}(\tfrac{(x+y)s-vy}x,v,x,y)\frac{x+y}x \\ &=f_U(\tfrac{(x+y)s-vy}x)f_V(v)f_X(x)f_Y(y)\frac{x+y}x \\ &=\frac1{\la^2}\,\frac{x+y}x\, \exp\Big\{\frac{vy-(x+y)s}{\la x}-\frac v\la-x-y\Big\} 1_{x,y,v>0,\ s>\frac{vy}{x+y}}. \end{align*} Next, for the pdf of $$S$$ we have \begin{align*} f_S(s)&=\iint_{x,y>0} dx\,dy\,\int_0^\infty dv\,f_{S,V,X,Y}(s,v,x,y) \\ &=\frac1{\la^2}\,\iint_{x,y>0} dx\,dy\,\frac{x+y}x\,\exp\Big\{\frac{-(x+y)s}{\la x}-x-y\Big\}\int_0^{(x+y)s/y} dv\,\exp\Big\{\frac{vy}{\la x}-\frac v\la \Big\} \\ &=\frac s{\la^2}\,\iint_{x,y>0} dx\,dy\,e^{-x-y} \frac{(x+y)^2 }{x y}\,r\left(\frac{x+y}{\la y},\frac{x+y}{\la x},s\right), \end{align*} where $$\begin{equation*} r(a,b,s)=\frac{e^{-a s}-e^{-b s}}{(b-a)s} =\frac1{b-a}\,\int_a^b dz\,e^{-z s} \end{equation*}$$ for positive real $$b\ne a$$, so that $$r(a,b,s)$$ is decreasing in $$s$$; as usual, here we let $$\int_a^b:=-\int_b^a$$ if $$b; also, $$r(a,b,s)\uparrow1$$ as $$s\downarrow0$$. So, by the monotone convergence theorem, \begin{align*} f_S(s)/s&\to\frac1{\la^2}\,\iint_{x,y>0} dx\,dy\,e^{-x-y} \frac{(x+y)^2 }{x y} \\ &\ge\frac1{\la^2}\,\iint_{x,y>0} dx\,dy\,e^{-x-y} \frac xy=\infty \end{align*} as $$s\downarrow0$$. Comparing this with (1), we see $$f_S\ne f_T$$, as claimed.

The command of Mathematica 12.2

PDF[TransformedDistribution[(a*c + b*d)/(c + d), {a \[Distributed]
ExponentialDistribution[\[Lambda]1], b \[Distributed] ExponentialDistribution[\[Lambda]1],
c \[Distributed] ExponentialDistribution[\[Lambda]2],d \[Distributed] ExponentialDistribution[\[Lambda]2]}], t]


answers $$\begin{cases} \frac{1}{2} \text{\lambda 1} \left(-\text{Ei}(-t \text{\lambda 1})+2 e^{-2 \text{\lambda 1} t} \text{Ei}(t \text{\lambda 1})+\Gamma (0,t \text{\lambda 1})\right) & t>0 \\ 0 & t<0 \\ \text{Indeterminate} & \text{True} \end{cases}$$

• It should be noticed that the PDF under consideration does not depend on $\lambda2$. – user64494 Jan 8 at 20:48