5
$\begingroup$

$X,Y$ 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!.

$\endgroup$
  • 1
    $\begingroup$ why did you take such $Y_1,Y_2$? $\endgroup$ – MAN-MADE Sep 11 '17 at 17:21
  • 1
    $\begingroup$ 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 $\endgroup$ – Henry Sep 11 '17 at 17:22
  • $\begingroup$ @MANMAID I found a similar technique used in an example of Rohatgi Probability and statistics book. $\endgroup$ – BAYMAX Sep 11 '17 at 17:24
  • $\begingroup$ 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. $\endgroup$ – MAN-MADE Sep 11 '17 at 17:27
  • $\begingroup$ 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? $\endgroup$ – BAYMAX Sep 11 '17 at 17:30
3
$\begingroup$

\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 $u<0.$

$\endgroup$
  • $\begingroup$ Nice! but how we get $P(X-Y \leq u) = E[P(X-Y \leq u | Y)]$ ? $\endgroup$ – BAYMAX Sep 12 '17 at 1:31
  • $\begingroup$ This is the nicest approach! $\endgroup$ – MAN-MADE Sep 12 '17 at 2:19
  • 1
    $\begingroup$ @BAYMAX The Probability for an event is the Expectation for the indicator random variable for the event. $\mathsf P(X-Y\leq u) ~{=\mathsf E(\mathbf 1_{X-Y\leq u}) \\ = \mathsf E(\mathsf E(\mathbf 1_{X-Y\leq u} \mid Y)) \\ = \mathsf E(\mathsf P(X-Y\leq u\mid Y)) }$ $\endgroup$ – Graham Kemp Sep 12 '17 at 2:24
  • $\begingroup$ @BAYMAX : That is the law of total probability: the prior expected value of the posterior probability equals the prior probability. $\endgroup$ – Michael Hardy Sep 12 '17 at 2:31
  • 1
    $\begingroup$ @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 $\operatorname{E}(h(Y)) = \operatorname{E}( \Pr(X-Y\le u \mid Y)). \qquad$ $\endgroup$ – Michael Hardy Sep 12 '17 at 2:48
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Sep 11 '17 at 20:37
  • $\begingroup$ Nice! here when $Y_{2} = \frac{X+Y}{X-Y}$what is the range of $Y_{2}?$,is it $\Bbb{R} / [-1,1] ?$ $\endgroup$ – BAYMAX Sep 12 '17 at 1:35
  • $\begingroup$ @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. $\endgroup$ – MAN-MADE Sep 12 '17 at 1:56
1
$\begingroup$

$$ P(X-Y<z) = \sum_y P(X-y<z)P(Y=y) = \int_{y \in \mathbb{R}} P(X<y+z)f(y) \, dy $$ 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 $e^{-\lvert z \rvert}/2$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Sep 11 '17 at 19:04
1
$\begingroup$

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 $\displaystyle f_{X-Y}(u) = \frac 1 2 e^{-|u|}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.