What is this well known probability distribution?

[Avatrin]

Homework Statement

I have a probability density function:
f$_{X}$(x) = θk$^{θ}$x$^{-θ-1}$ when x > k and 0 everywhere else
And:
Y = θlog(X/k)

Find the probability density function for Y. What well known density function is this?

Homework Equations

I don't really know where to start...

The Attempt at a Solution

I first tried solving for x and:
X = ke$^{Y/θ}$
Plugging this into f(x):
f$_{Y}$(y) = θk$^{θ}$(ke$^{y/θ}$)$^{-θ-1}$

I also tried going the other way around:
f$_{Y}$(x) = θlog(θk$^{θ-1}$x$^{-θ-1}$)

But, I don't really know what to do and where to start.. ...

 

[vela]

You can't simply rewrite f_X(x) in terms of Y the way you did. It's a little more complicated. The correct relationship is
$$f_X(x)\,dx = f_Y(y)\,dy$$ where I've assumed both dx and dy are positive. The idea here is that interval (x, x+dx) corresponds to the interval (y, y+dy) so the probability that X is in the interval (x, x+dx), which is given by the lefthand side of the equation, should be equal to the probability Y is in the interval (y, y+dy), which is given by the righthand side of the equation. Solving for f_Y(y), you get
$$f_Y(y) = \frac{f_X(x)}{|dy/dx|}.$$ The absolute value is necessary to cover the case where dx>0 corresponds to dy<0. That is, when y decreases as x increases.

 

[Avatrin]

Actually, I had to use the fundamental theorem of calculus:

$\frac{d}{dy}$∫$^{y}_{0}$f(x)dx = f(y)

Therefore;
$\frac{d}{dy}$∫$^{ke^(Y/θ)}_{0}$θk$^{θ}$x$^{-(θ+1)}$dx
= $\frac{d}{dy}$-e$^{-y}$
= e$^{-y}$
And, this is the exponential distribution..

Or... Am I wrong again?

 

[Ray Vickson]

Actually, I had to use the fundamental theorem of calculus:

$\frac{d}{dy}$∫$^{y}_{0}$f(x)dx = f(y)

Therefore;
$\frac{d}{dy}$∫$^{ke^(Y/θ)}_{0}$θk$^{θ}$x$^{-(θ+1)}$dx
= $\frac{d}{dy}$-e$^{-y}$
= e$^{-y}$
And, this is the exponential distribution..

Or... Am I wrong again?

No, you are now correct.

RGV