Other way to express $e^{|x|+|y|}$

Question

I have a joint PDF with $e^{|x|+|y|}$. I know I can separate the function in two functions, $e^{|x|}$ and $e^{|y|}$. The limits for $x$ and $y$ are from $-\infty$ to $\infty$. Can I integrate from $0$ to $\infty$ and multiply the integral by $2$: $$ 2 \int_{0}^{\infty} e^x \,dx = \int_{-\infty}^{\infty} e^x \,dx \quad ? $$

The complete problem is I have a pdf $f(x,Y)=(1/4)2 \exp[-1/2 (|x| + |y|)].$ the goal is to find the joint PDF f(W,Z) taking in consideration this $W=XY$ and $Z=Y/X$

If the joint pdf is as you describe, then it can only be over a small region. If you left out a minus sign in the description, then indeed if you are "integrating $x$ out" you can go from $0$ to $\infty$ and double. — André Nicolas, May 07 '13 at 01:56
This is the complete pdf f(x,Y)=(1/4)2 exp[-1/2 (|x| + |y|)]. the goal is to find the joint f(W,Z) taking in consideration this W=XY and Z=Y/X. — Lorain, May 07 '13 at 01:58
$$\begin{align}\int_{-\infty}^\infty e^{-|x|},\mathrm dx &= \int_{-\infty}^0 e^{-|x|},\mathrm dx + \int_0^\infty e^{-|x|},\mathrm dx\&= \int_{-\infty}^0 e^{x},\mathrm dx + \int_0^\infty e^{-x},\mathrm dx\&=-\int_{\infty}^0 e^{-y},\mathrm dy + \int_0^\infty e^{-x},\mathrm dx\&=\int_0^{\infty} e^{-x},\mathrm dx + \int_0^\infty e^{-x},\mathrm dx\&= 2\int_0^\infty e^{-x},\mathrm dx\end{align}$$ Notice that the absolute value signs disappeared when we used the fact that if $x < 0$, then $-|x| = x$ while if $x > 0$, then $-|x|=-x$. — Dilip Sarwate, May 07 '13 at 13:32

score 4 · Answer 1 · answered May 07 '13 at 01:58

4

Use this fact.

$$\int_{\mathbb{R}^2} e^{-(|x| + |y|)}\, dx\, dy = \left(\int_{-\infty}^\infty e^{-|x|}\, dx\right)^ 2 = 4$$

answered May 07 '13 at 01:58

ncmathsadist

50,127

I don't get it. Can you walk me through? what happened with y? – Lorain May 07 '13 at 02:04
This is the complete pdf f(x,Y)=(1/4)2 exp[-1/2 (|x| + |y|)]. the goal is to find the joint f(W,Z) taking in consideration this W=XY and Z=Y/X. – Lorain May 07 '13 at 02:18

score 2 · Answer 2 · edited Apr 13 '17 at 12:20

The question in the title does not seem to match what is being insisted on in the comments. Cleaning up the question, it seems that the joint pdf is $$f_{X,Y}(x,y) = \left(\frac{1}{4}\right)^2 \exp\left(-\frac{|x|+|y|}{2}\right), -\infty < x, y < \infty$$ which is the product of the two LaPlacian pdfs $\frac{1}{4}e^{-|x|/2}$ and $\frac{1}{4}e^{-|y|/2}$ of the independent random variables $X$ and $Y$. The OP desires to find the joint pdf of $W = XY$ and $Z = Y/X$.

This is a standard problem in transformation of random variables and the calculations involve Jacobians etc. However, if these are not allowed, we can proceed from first principles as follows.

Suppose that $\alpha$ and $\beta$ are two positive numbers. Then, $$F_{W,Z}(\alpha, \beta)= P\{W\leq \alpha, Z \leq \beta\}$$ can be found by integrating $f_{X,Y}(x,y)$ over two regions:

the fourth quadrant and the region in the first quadrant below the line $y = \beta x$ and the hyperbola $xy = \alpha$
the second quadrant and the region in the third quadrant above the line $y = \beta x$ and the hyperbola $xy = \alpha$

By the symmetry of the joint pdf, the integrals over the two regions have the same value, the integrals over the quadrants have value $\frac{1}{4}$ each, and the integral over the region in the first quadrant below the line $y = \beta x$ and the hyperbola $xy = \alpha$ is just $$\int_{y=0}^{\sqrt{\alpha\beta}} \int_{x=y/\beta}^{\alpha/ y} \frac{1}{4} e^{-(x+y)/2}\,\mathrm dx\,\mathrm dy.$$ Note the absence of absolute value signs since we know we are in the first quadrant and so $x$ and $y$ are positive. One can evaluate this integral and then take the partial derivative with respect to $\alpha$ and $\beta$ or one can use the method of differentiating under the integral sign (see for example, the comment following this answer) to get $f_{W,Z}(\alpha, \beta)$ without actually computing the integral.

Finally, this gives the joint pdf in the first quadrant. You need to figure out its value in other quadrants too, and in each case, you can make the absolute value signs go away by using known restrictions on $x$ and $y$ to replace $|x|$ by $x$ or $-x$ etc as the case may be.

I can't use the jacobian and I can't use linear transformation. I am trying to use the distribution definition but the absolute value is there and I am trying to see how can I modify it. — Lorain, May 07 '13 at 02:56
I thought in not using the Jacobian becaus the pdf is a non singular one. Can I use the Jacobian here? — Lorain, May 07 '13 at 23:26
You can use the Jacobian as many times as you need to. The key idea is that if for a given $(\alpha, \beta)$ we have that $(x_1,y_1), (x_2,y_2), \ldots$ etc are solutions to $g(x,y) = \alpha, h(x,y)=\beta$, then $$f_{W,Z}(\alpha, \beta)\Delta\alpha\Delta\beta \approx \sum_i f_{X,Y}(x_i,y_i)\Delta x_i\Delta y_i$$ where in the limit, the ratio of areas $\frac{\Delta\alpha\Delta\beta}{\Delta x_i\Delta y_i}$ is the Jacobian determinant whose value divides $f_{X,Y}(x_i,y_i)$ in the mystical magical formula for transformation of random variables. — Dilip Sarwate, May 08 '13 at 01:04

Other way to express $e^{|x|+|y|}$

2 Answers2