# Density of $X-Y$ where $X,Y$ are independent random variables with common PDF $f(x) = e^{-x}$?

$X,Y$ are independent random variables with common PDF $f(x) = e^{-x}$ then density of $X-Y = \text{?}$

I thought of this let $Y_1 = X + Y$, $Y_2 = \frac{X-Y}{X+Y}$, solving which gives me $X = \frac{Y_1(1 + Y_2)}{2}$, $Y = \frac{Y_1-Y_2}{2}$

then I calculated the Jacobian $J = \begin{bmatrix} \frac{1+y_2}{2} & \frac{y_1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}$ so that $\left|\det(J)\right| = \frac{1+y_1+y_2}{4}$

and the joint density of $Y_1,Y_2$ is the following $W(Y_1,Y_2) = \left|\det(J)\right| e^{-(y_1+y_2)}$ when $y_1,y_2> 0$ and $0$ otherwise.

Next I thought of recovering $X-Y$ as the marginal but I got stuck. I think i messed up in the variables.

Any help is great!.

• why did you take such $Y_1,Y_2$? – MAN-MADE Sep 11 '17 at 17:21
• There are lots of approaches. One is characteristic functions. Another is to note the distribution of the difference should be is symmetric about $0$ and if you look at the right hand half then memorylessness suggests the distribution of $X-Y$ given $X \gt Y$ should have the same exponential distribution as you started with – Henry Sep 11 '17 at 17:22
• @MANMAID I found a similar technique used in an example of Rohatgi Probability and statistics book. – BAYMAX Sep 11 '17 at 17:24
• I guess your $W(Y_1,Y_2)$ is the joint pdf of $Y_1,Y_2$, and if so then it should be $|\det(J)|$ and also I think support of $Y_1,Y_2$ are dependent. – MAN-MADE Sep 11 '17 at 17:27
• thanks! typo there. edited.Also i thought of taking $Y_{1} =X-Y$ and hence I could have obtained the marginal but then what would be the limits of $y_{1},y_{2}$ during the integration? – BAYMAX Sep 11 '17 at 17:30

\begin{align} \underbrace{\text{For } u>0} \text{ we have } f_{X-Y}(u) & = \frac d {du} \Pr(X-Y\le u) \\[10pt] & = \frac d {du} \operatorname{E}(\Pr(X-Y \le u \mid Y)) \\[10pt] & = \frac d {du} \operatorname{E}(\Pr(X \le u+Y\mid Y)) \\[10pt] & = \frac d {du} \operatorname{E}(1-e^{-(u+Y)}) \\[10pt] & = \frac d {du} \int_0^\infty (1 - e^{-(u+y)} ) e^{-y} \, dy \\[10pt] & = \frac d {du} \int_0^\infty (e^{-y} - e^{-u} e^{-2y}) \, dy \\[10pt] & = \frac d {du} \left( 1 - \frac 1 2 {} e^{-u} \right) \\[10pt] & = \frac 1 2 e^{-u}. \end{align} A similar thing applied when $u<0$ gives you $\dfrac 1 2 e^u,$ so you get $\dfrac 1 2 e^{-|u|}.$

But a simpler way to deal with $u<0$ is to say that since the distribution of $X-Y$ is plainly symmetric about $0$ (since $X-Y$ has the same distribution as $Y-X$), if you get $\dfrac 1 2 e^{-u}$ when $u>0,$ you have to get $\dfrac 1 2 e^u$ when $u<0.$

• Nice! but how we get $P(X-Y \leq u) = E[P(X-Y \leq u | Y)]$ ? – BAYMAX Sep 12 '17 at 1:31
• This is the nicest approach! – MAN-MADE Sep 12 '17 at 2:19
• @BAYMAX The Probability for an event is the Expectation for the indicator random variable for the event. $\mathsf P(X-Y\leq u) ~{=\mathsf E(\mathbf 1_{X-Y\leq u}) \\ = \mathsf E(\mathsf E(\mathbf 1_{X-Y\leq u} \mid Y)) \\ = \mathsf E(\mathsf P(X-Y\leq u\mid Y)) }$ – Graham Kemp Sep 12 '17 at 2:24
• @BAYMAX : That is the law of total probability: the prior expected value of the posterior probability equals the prior probability. – Michael Hardy Sep 12 '17 at 2:31
• @BAYMAX : Consider $\Pr(X-Y\le u \mid Y= y),$ the conditional probability of one event given another. It depends on the value of $y.$ As a function of $y,$ say we call it $h(y).$ Then $h(Y)$ is a random variable. Its expected value is $\operatorname{E}(h(Y)) = \operatorname{E}( \Pr(X-Y\le u \mid Y)). \qquad$ – Michael Hardy Sep 12 '17 at 2:48

If I understood you correctly, you have, both $X$ and $Y$ being distributed by an exponential distribution, where $\lambda$ equals one. Now you want to know about the distribution of their difference, namely $Z=X-Y$. Their mass is $$P(z\ge Z)=P(z\ge X-Y)=P(z)$$ which is (for $z\le 0$) $$P(z)=\int^\infty_{0}\int^{\infty}_{x-z}e^{-x}e^{-y}\,dy\,dx,$$ as the area of interest is $y\ge x-z$. Next, we know that the density $$p(z)=\frac{d}{dz}P(z),$$ is the derivative of the mass. Using the Leibnitz rule, this is $$\frac{d}{dz}\int^\infty_{0}\int^\infty_{x-z}e^{-x}e^{-y} \, dy \, dx = \int^\infty_0 \frac{d}{dz}\int^\infty_{x-z}e^{-x}e^{-y}\,dy\,dx$$ $$\int^\infty_{-\infty} e^{-x}e^{-(x-z)} \, dx=\frac{e^z}{2}$$ After repeating the computation of $z\ge 0$, which would entail calculating $$\frac{d}{dz}P(z)=\int^\infty_0 \int^{x+z}_0 e^{-x}e^{-y} \, dy \, dx,$$ we arrive at $$p(z)=\frac{e^{-|z|}}{2}$$

Note that this known as the Laplace distribution.

The transformation is $(X,Y)\rightarrow (Y_1,Y_2)$.

$Y_1=X+Y, Y_2=\dfrac{X-Y}{X+Y}$.

Let $y_1=x+y,y_2=\dfrac{x-y}{x+y}$, i.e., $x=\dfrac{y_1(1+y_2)}{2},y=\dfrac{y_1(1-y_2)}{2}$. Now $x>0,y>0$, hence $y_1>0, -1<y_2<1$

$J=\begin{bmatrix}\dfrac{1+y_2}{2}&\dfrac{y_1}{2}\\\dfrac{1-y_2}{2}&\dfrac{-y_1}{2}\end{bmatrix}$. Here, $\det(J)=\dfrac{-y_1}{2}$

Now \begin{align}f_{(Y_1,Y_2)}(y_1,y_2)=|\det(J)|f_{(X,Y)}(x,y)=\dfrac{y_1e^{-y_1}}{2}I(y_1>0,-1<y_2<1)\\=y_1e^{-y_1}I(y_1>0)\cdot\dfrac{1}{2}I(-1<y_2<1)\end{align}

Here $I(\cdot)$ is indicator function.

But I doubt you can recover the pdf of $X-Y$ easily. So, one way to do this analogous to the way you want is taking $Y_1=X-Y, Y_2=\dfrac{X+Y}{X-Y}$.

the reason Rohatgi Probability and statistics used this technique is because of independence of $X+Y,\dfrac{X-Y}{X+Y}$. But that will not work here and eventually the calculation will become very messy.

• I've upvoted this but a much simpler answer is available. I've posted two answers. One of them involves an integral of a function of two variables, but no Jacobians are needed because no changes of variables are done. So my challenge to everyone here: See if you can find a simpler way than that. – Michael Hardy Sep 11 '17 at 20:37
• Nice! here when $Y_{2} = \frac{X+Y}{X-Y}$what is the range of $Y_{2}?$,is it $\Bbb{R} / [-1,1] ?$ – BAYMAX Sep 12 '17 at 1:35
• @BAYMAX the reason Rohatgi Probability and statistics used this technique is because of independence of $X+Y,\dfrac{X-Y}{X+Y}$. That will not work here and eventually the calculation will become very messy. So, I suggest you to see other solutions that are posted here. – MAN-MADE Sep 12 '17 at 1:56

$$P(X-Y<z) = \sum_y P(X-y<z)P(Y=y) = \int_{y \in \mathbb{R}} P(X<y+z)f(y) \, dy$$ by the law of total probability (there's probably a more rigorous way to write that middle expression, but it'll still be that integral). Then this is $$\int_{y+z>0,y>0} (1-e^{-(y+z)})e^{-y} \, dy$$ using the given distributions. This splits into $$\begin{cases} \int_{-z}^{\infty} (e^{-y}-e^{-2y}e^{-z}) \, dy & z<0 \\ \int_{0}^{\infty} (e^{-y}-e^{-2y}e^{-z}) \, dy & z \geq 0 \end{cases} = \begin{cases} \frac{1}{2}e^{z} & z<0 \\ 1-\frac{1}{2}e^{-z} & z\geq 0 \end{cases}.$$ Differentiating then gives the density function as $e^{-\lvert z \rvert}/2$.

• I've up-voted this although I can see that some might object to a discrete sum for a continuous variable. I've also posted an answer with a similar approach but with greater detail. – Michael Hardy Sep 11 '17 at 19:04

I already posted an answer involving no integrals of functions of more than one variable; here's another approach.

\begin{align} \text{First assume } u >0. \text{ Then} \\ \Pr( X-Y > u) & = \int_0^\infty \left( \int_{y+u}^\infty f_{X,Y} (x,y) \, dx \right) \,dy \\[10pt] & = \int_0^\infty \left( \int_{y+u}^\infty e^{-x} e^{-y} \, dx \right) \,dy \\[10pt] & = \int_0^\infty \left( e^{-y} \int_{y+u}^\infty e^{-x} \, dx \right) \,dy \\ & \qquad\text{(This can be done because $e^{-y}$ does not change as $x$ goes from something to $\infty$.)} \\[10pt] & = \int_0^\infty e^{-y} \cdot e^{-(y+ u)} \, dy \\[10pt] & = \frac 1 2 e^{-u}. \end{align} That works if $u>0.$ Then use the fact that $Y-X$ has the same probability distribution as $X-Y$ to conclude that if $u<0$ then $\Pr(X-Y<u) = \frac 1 2 e^{u}.$

Therefore if $u>0$ then $\Pr(X-Y\le u) = 1- \dfrac 1 2 e^{-u}$ and mutatis mutandis if $u<0,$ so we get $\displaystyle f_{X-Y}(u) = \frac 1 2 e^{-|u|}.$