1. ## Random variables

$$p_{X,Y}(x,y) = \left\{ \begin{array}{ll} e^{-x} & \mbox{; x>y>0};\\ 0 & \mbox{; otherwise}.\end{array} \right.$$

Find $p_X(x)$ and $p_Y(y)$. Are they independent?

Find also $P(2, $P(Y and $P(2Y.
I'm only having trouble with (formalizing) last two requirements: $P(Y and $P(2Y.

$p_X(x)=...=xe^{-x}\text{ ; }x > 0$
$p_Y(y)=...=e^{-y}\text{ ; }y > 0$
$P(2

Now the last two:
$P(Y is obviously 1, but I don't know how to setup the corresponding definite integral(s).

Similarly for $P(2Y which is reasonably enough $\frac{1}{2}$ (I've peeked at the solutions, but still) ... as would $P(nY, wouldn't it?

2. Originally Posted by courteous
I'm only having trouble with (formalizing) last two requirements: $P(Y and $P(2Y.

$p_X(x)=...=xe^{-x}\text{ ; }x > 0$
$p_Y(y)=...=e^{-y}\text{ ; }y > 0$
$P(2

Now the last two:
$P(Y is obviously 1, but I don't know how to setup the corresponding definite integral(s).

Similarly for $P(2Y which is reasonably enough $\frac{1}{2}$ (I've peeked at the solutions, but still) ... as would $P(nY, wouldn't it?
You should have been taught to draw the required region of integration and then set up the required double integral:

$\displaystyle \Pr(2Y < X) = \Pr(Y < X/2) = \int_{x = 0}^{+\infty} \int_{y = 0}^{y = x/2} f(x, y) \, dy \, dx$.

Yes.

3. Thank you ... $\frac{1}{2}$ indeed.

Further related question:
Let $Z=X+Y$.
What is distribution of $Z$?
I'm told:
"If $Y=g(X,Z)$ holds and $g(x,z)$ is a "well behaved" function, monotonous for each $x$ in $z$ (in $z$?!),
then $p_Z(z)=\int_{-\infty}^{\infty}p_{X,Y}\big(x,g(x,z)\big)\left|\fr ac{\partial g}{\partial z}(x,z)\right|dx$."

How do I find what function $g$ is and then use the given formula? The solution given is $e^{-z/2}-e^{-z} \text{ ; }z>0$.

4. Originally Posted by courteous
Thank you ... $\frac{1}{2}$ indeed.

Further related question:

I'm told:
"If $Y=g(X,Z)$ holds and $g(x,z)$ is a "well behaved" function, monotonous for each $x$ in $z$ (in $z$?!),
then $p_Z(z)=\int_{-\infty}^{\infty}p_{X,Y}\big(x,g(x,z)\big)\left|\fr ac{\partial g}{\partial z}(x,z)\right|dx$."

How do I find what function $g$ is and then use the given formula? The solution given is $e^{-z/2}-e^{-z} \text{ ; }z>0$.
I prefer not to use Z. I'll use U instead.

Use the change of variable theorem:

Make the transformation U = X + Y, V = X. The inverse transformation is X = V, Y = U - V. Then |J| = 1.

Get the joint pdf of U and V: $g(u, v) = e^{-v}$. Integrate wrt to v to get the marginal, which is the required pdf. Note that 0 < y < x => 0 < u - v < v => v < u < 2v and so the region of integration is between the lines v = u/2 and v = u where v > 0:

$g(u) = \int_{v = u/2}^{v = u} e^{-v} \, dv$ etc.

5. Originally Posted by mr fantastic
Then |J| = 1.
I've lost you here ... what is J, a Jacobian matrix?
Originally Posted by mr fantastic
Get the joint pdf of U and V: $g(u, v) = e^{-v}$.
Would you show how did you get joint pdf (I guess it stems from J)?

6. Originally Posted by courteous
I've lost you here ... what is J, a Jacobian matrix?

Would you show how did you get joint pdf (I guess it stems from J)?
Yes.

See for example Walpole, Meyers and Meyers: Probability and Statistics for Engineers and Scientists. It is also explained here: Transformations of Variables (scroll down).