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