Let $Y,Z$ be idd. RV's, $\ Z \sim e(1), \ Y \sim R(0,1)$. Find $f_{X,Y}$ for $(X,Y)$.

Question

Let $Y,Z$ be idd. RV's, $\ Z \sim e(1), \ Y \sim R(0,1), \ X = \frac Z Y$.

($R(0,1)$ denote the continuous uniform distribution)

Compute $P(X>1)$:

I have $P(X>1) = 1-P(X \le 1) = 1 - P(Z\le Y) = 1 - \int^{\frac 1 2}_0 \int^y_0 \frac {e^{-z}} 2 \ dz \ dy = 0.426$

(Please tell me if you disagree).

Find $f_{X,Y}$ for $(X,Y)$ using the transformation theorem:

In order to apply a transformation I must know $f_{X,Y}$ ? As I look it up I have $$\int \int_A f(x,y) \ dx \ dy = \int \int_B f(w_1(u,v),w_2(u,v)) |J| \ du \ dv$$

Have I misunderstood something ? How can I find $f_{X,Y}$ using this ?

Answer 1

The joint density of $(Z,Y)$ is $$ f_{Z,Y}(z,y)=\mathrm{e}^{-z}\mathbf{1}_{0<z}\mathbf{1}_{0<y<1} $$ by independence. The vector $(X,Y)$ is obtained by transformation of $(Z,Y)$ by the one-to-one function $g(z,y)=(z/y,y)$. Now $g$ has inverse function $h(x,y)=(xy,y)$ and if we let $h_1(x,y)=xy$ and $h_2(x,y)=y$ denote the coordinate functions, then the Jacobian of $h$ is $$ J(x,y)= \begin{pmatrix} \frac{\partial h_1}{\partial x}(x,y) & \frac{\partial h_1}{\partial y}(x,y) \\ \frac{\partial h_2}{\partial x}(x,y) & \frac{\partial h_2}{\partial y}(x,y) \end{pmatrix} = \begin{pmatrix} y & x \\ 0 & 1 \end{pmatrix} $$ which has determinant $\det J(x,y)=y$. Thus the density of $(X,Y)$ is $$ \begin{align} f_{X,Y}(x,y)&=|\det J(x,y)|f_{Z,Y}(h(x,y))=|y| f_{Z,Y}(xy,y)\\ &=y\mathrm{e}^{-xy}\mathbf{1}_{0<xy}\mathbf{1}_{0<y<1}=y\mathrm{e}^{-xy}\mathbf{1}_{0<y<1}\mathbf{1}_{0<x} \end{align} $$

Answer 2

$(X,Y) = g(Z,Y)$ with $g(z,y) = (z/y,y)$.

So we have $x = z/y$ and $y = y$. Thus we can form the functions $w_1(x,y)=xy=z$ and $w_2(x,y)=y=y$.

The Jacobian is $|J| = | y \cdot 1 - 0 \cdot x | = |y |$.

So we have $f_{(X,Y)}(x,y) = f_{(Z,Y)}(w_1(x,y), w_2(x,y)) |J| = 2e^{-xy}|y|$ which is the same as $2e^{-xy}y$ because of the new domain.