# Probability: What are p.d.f.'s of x+y and x/y?

1. Feb 23, 2014

### sanctifier

1. The problem statement, all variables and given/known data

The probability desity function (p.d.f.) of joint distribution of random variables X and Y is given as

$f(x,y) = \begin{cases} e^{-(x + y)} \;\; when \;\; x > 0 \\ 0 \;\; \;\;\;\;\;\;\;\;\;\;otherwise \end{cases}$

Question 1: What are the p.d.f.'s of X + Y and X/Y ?

Question 2: Does the expectation of X/Y exist ?

2. Relevant equations

Nothing special.

3. The attempt at a solution

$\begin{cases} u = x \\ v = x + y \end{cases}$

$\begin{cases} x = u \\ y = v - u \end{cases}$

$Jacobian = \begin{bmatrix} \frac{dx}{du} & \frac{dx}{dv} \\\frac{dy}{du} & \frac{dy}{dv} \end{bmatrix} = \begin{bmatrix}1 & 0 \\{-1} & 1 \end{bmatrix}$

$g(u,v)=f(u,v-u)|Jacobian|= e^{-v}$

$h(x+y)=h(v) = \int_0^v g(u,v)du = e^{-v} u |_{u=0}^{u=v} = ve^{-v}$

$\begin{cases}z = x\\ w = \frac{x}{y} \end{cases}$

$\begin{cases} x = z\\ y = \frac{z}{w} \end{cases}$

$Jacobian2 = \begin{bmatrix} \frac{dx}{dz} & \frac{dx}{dw} \\\frac{dy}{dz} & \frac{dy}{dw} \end{bmatrix} = \begin{bmatrix}1 & 0 \\{ \frac{1}{w} } & {- \frac{z}{ w^{2}} } \end{bmatrix}$

$g2(z,w)=f(z, \frac{z}{w} )|Jacobian2|= e^{-z- \frac{z}{w} } \frac{z}{ w^{2} }$

$h2(w)= \int_0^ \infty g2(z,w)dz = \int_0^ \infty e^{-z- \frac{z}{w} } \frac{z}{ w^{2} } dz = -\int_0^ \infty \frac{z}{ w^{2} } {(1+ \frac{1}{w} )}^{-1} d e^{-z(1+ \frac{1}{w})} = 0 + \int_0^\infty \frac{e^{-z(1+ \frac{1}{w})}}{ w^{2} \frac{1}{w}} dz = - \frac{1}{ {w+1}^{2} } e^{-z(1+ \frac{1}{w})} |_{z=0}^{z= \infty }= \frac{1}{{w+1}^{2}}$

$E( \frac{x}{y} )=E(w)= \int_0^\infty \frac{1}{{w+1}^{2}} dw = -\frac{1}{w+1}|_{w=0}^{w=\infty} = 1$

2. Feb 23, 2014

### Ray Vickson

There is something wrong with the question as stated: your joint density $f(x,y)$ blows up as $y \to -\infty$, and does not have a finite integral over $y \in (-\infty,\infty)$. You restrict x but not y.

You never define for us what is meant by $h(w)$ and $h2(w)$, so we end up having to try to guess---a good way to lose marks on an assignment.

Also, when you write
$$\frac{1}{w+1^2}$$
you are writing
$$\frac{1}{w+1}$$
If you really mean
$$\frac{1}{(w+1)^2}$$
then use parentheses. BTW: that last form is the correct density of $X/Y$ at $w \geq 0$.

Finally, the integral you need for $E(X/Y)$ is not the integral you wrote.

3. Feb 25, 2014

### sanctifier

The p.d.f. should be

$f(x,y) = \begin{cases} e^{-(x + y)} \;\; when \;\; x > 0 \;\; y > 0\\ 0 \;\; \;\;\;\;\;\;\;\;\;\;otherwise \end{cases}$

I forgot to write the part $y > 0$, my mistake.

$h(v)=h(x+y)$ actually is the marginal p.d.f. of $v$

Similarly, $h2(w) =h2( \frac{x}{y} )$ is the marginal p.d.f. of $w$

The derivation of h2(w) should be

$h2(w)= \int_0^ \infty g2(z,w)dz = \int_0^ \infty e^{-z- \frac{z}{w} } \frac{z}{ w^{2} } dz = -\int_0^ \infty \frac{z}{ w^{2} } {(1+ \frac{1}{w} )}^{-1} d e^{-z(1+ \frac{1}{w})} = 0 + \int_0^\infty \frac{e^{-z(1+ \frac{1}{w})}}{ w^{2} \frac{1}{w}} dz = - \frac{1}{ {(w+1)}^{2} } e^{-z(1+ \frac{1}{w})} |_{z=0}^{z= \infty }= \frac{1}{{(w+1)}^{2}}$

Yes, as you said, I lost the parentheses when writing $\frac{1}{{(w+1)}^{2}}$

The last integral is wrong, it should be

$E( \frac{x}{y} )=E(w)=\int_0^ \infty w \frac{1}{{(w+1)}^{2}} dw = \int_0^ \infty (\frac{w+1}{{(w+1)}^{2}} - \frac{1}{{(w+1)}^{2}} )dw= \int_0^ \infty \frac{1}{{(w+1)}^{2}} d(w+1)^2 - \int_0^ \infty \frac{1}{{(w+1)}^{2}} dw = ln(w+1)^2|_{w=0}^{w= \infty } + \frac{1}{w+1}|_{w=0}^{w= \infty } = \infty$

Hence, $E( \frac{x}{y} )$ doesn’t exist.

4. Feb 25, 2014

### Ray Vickson

You are correct now.

5. Feb 25, 2014

### sanctifier

Thank you again, Ray.