Probability density of two random variables using characteristic function

Question

I've been trying to solve the following question :

$X$ and $Y$ are two real random variables with a probability density of :

$$f(x,y) = e^{-y} *\mathscr{1}_{0<x<y}(x,y)$$

where $\mathscr{1}$ is the characteristic function.

1. Verify that $f$ is a probability density.
2. Give the marginal probability $f_1$ of $X$ and $f_2$ of $Y$.
3. Are $X$ and $Y$ independent?
4. Determine $\mathbb P{\left\{\frac{X}{Y}\le z \mbox{ and } Y \le y \right\}}$ for $z \in [0,1]$. Give then the law of $\frac{X}{Y}$.
5. Are the $\frac{X}{Y}$ and $Y$ independent variables?

For the first question I tried checking the condition of normalization :

$\int_{R^2} f(x,y)dxdy = 1$

so I did :

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-y} *\mathscr{1}_{0<x<y}(x,y)dxdy$

$\int_{0}^{\infty}\int_{x}^{\infty} e^{-y} dydx = \int_{0}^{\infty} e^{-x} dx = 1$

But I'm not sure of this manipulation.

For the second question I did :

$$f_{1}(x) = \int_{-\infty}^{\infty} f(x,y) dy = \int_{-\infty}^{\infty} e^{-y} *\mathscr{1}_{0<x<y}(x,y)dy = \int_{x}^{\infty} e^{-y} dy = e^{-x}$$

$$f_{2}(y) = \int_{-\infty}^{\infty} f(x,y) dx = \int_{-\infty}^{\infty} e^{-y} *\mathscr{1}_{0<x<y}(x,y)dx = \int_{0}^{y} e^{-y} dx = ye^{-y}$$

But I'm also not sure.

For the third question I think I should try to check if $f(x,y) = f_1(x)f_2(y)$ but I don't know how to handle out the characteristic function in the equality.

For question 4 I tried the following :

$P{\left\{\frac{X}{Y}\le z \mbox{ and } Y \le y \right\}}$ for $z \in [0,1]$ = $P${${\frac{X}{Y}\le z}$} * $P${${Y\le y}$}

$P${${X{\le}Y}$} $= \int_{-\infty}^{\infty} \int_{-\infty}^{y} f(x,y) dxdy = \int_{-\infty}^{\infty} \int_{-\infty}^{y} e^{-y} *\mathscr{1}_{0<x<y}(x,y)dxdy = \int_{0}^{\infty} \int_{0}^{y} e^{-y}dxdy = \int_{0}^{\infty}e^{-y} \int_{0}^{y} dxdy = \int_{-\infty}^{\infty}ye^{-y}dy = 1$

and

$P${${Y{\le}y}$} = 1

Thus,

the answer is 1*1=1 and the law of the fraction is yexp(-y).

For the rest sincerely I have no clue.

Thank you for any advice !

Answer 1

Your computation for question 1 are correct. A good way for this kind of exercises is to decompose the indicator as $\mathbf{1}_{0<x<y}=\mathbf{1}_{0<x}\mathbf{1}_{x<y}$ and start integrating in $y$.

For the second question, indicators are missing in the results. Using the $\mathbf{1}_{0<x<y}=\mathbf{1}_{0<x}\mathbf{1}_{x<y}\mathbf{1}_{0<y}$ would avoid to miss them.

For the third question, it is indeed a good idea to look whether the joint density is the product of the marginal densities. You will see that the two expressions differ.

For the four question, the probability you have to compute is the probability that $(X,Y)$ belongs to some subset of $\mathbb R^2$. Use the expression of the joint density for that.