Functionn of two random variables.

**mathsandphysics** · March 2nd 2011, 04:11 PM

Could anyone help me with this question please?

Given X and Y are continuous random variables with the following joint probability density function:

$ f(x,y) = \begin{cases} {\theta}^2 e^{-{\theta} y} & 0 < x < y \\ 0 & \text{otherwise} \end{cases} $

i) Obtain the joint probability density function of X and Y − X.
ii) Show that X and Y − X are independent random variables and write down their (marginal)
probability density functions. Identify these distributions

For part (i)

I let U = X and Y = V - X,

My Jacobian value was 1,
and then I got $f(u,v) = {\theta}^2 e^{-{\theta} , (v-u)} & \text ,u > 0, (v-u) > u$ Is this the correct joint p.d.f for X and Y-X?

Also for the marginal of X, I got $f X(x) = e^{-{\theta x}$ but I'm not quite sure.

**matheagle** · March 2nd 2011, 04:32 PM

don't you want v=y-x?

**matheagle** · March 2nd 2011, 04:33 PM

$f_X(x) = e^{-{\theta x}$ is not a valid density

**matheagle** · March 2nd 2011, 04:36 PM

Let U = X and V = Y - X,

then the Jacobian is 1

and $f(u,v) = {\theta}^2 e^{-{\theta}(v+u)} I(u>0)I(v>0)$

So $f_U(u) = \theta e^{-\theta u} I(u>0)$

and

$f_V(v) = \theta e^{-\theta v}I(v>0)$

**mathsandphysics** · March 2nd 2011, 04:41 PM

Originally Posted by matheagle

Let U = X and V = Y - X,

then the Jacobian is 1

and $f(u,v) = {\theta}^2 e^{-{\theta}(v+u)} I(u>0)I(v>0)$

So $f_U(u) = \theta e^{-\theta u} I(u>0)$

and

$f_V(v) = \theta e^{-\theta v}I(v>0)$

Thanks, I thought I was close to it.

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