# help with computing density of one variable change.

• Oct 16th 2010, 11:31 AM
Sneaky
help with computing density of one variable change.
Let X~exponential(lambda).Let Y=X^.75. Compute the density of fy of Y.

fY(y) = 4(lambda)(y^3)e^(-lambda*y^4) for y > 0 and equals 0 for y <= 0.

My attempt:
Y=x^0.75
dx = 4dy / x^-0.75

i then get

integral from -inf to +inf of
4(lambda)(y^3)e^(-lambda*y^4) dy

from here i don't understand how to get to the answer
• Oct 16th 2010, 01:46 PM
mr fantastic
Quote:

Originally Posted by Sneaky
Let X~exponential(lambda).Let Y=X^.75. Compute the density of fy of Y.

fY(y) = 4(lambda)(y^3)e^(-lambda*y^4) for y > 0 and equals 0 for y <= 0.

My attempt:
Y=x^0.75
dx = 4dy / x^-0.75

i then get

integral from -inf to +inf of
4(lambda)(y^3)e^(-lambda*y^4) dy

from here i don't understand how to get to the answer

There are many ways of doing this. One simple way is to first get the cdf of Y:

$\displaystyle G(y) = \Pr(Y \leq y) = \Pr(X^{3/4} \leq y) = \Pr(X \leq y^{4/3}) = \int_0^{y^{4/3}} f(x) \, dx$ for $y \geq 0$ and zero otherwise.

Then the pdf is given by $\displaystyle g(y) = \frac{dG}{dy}$. Using the chain rule and the Fundamental Theorem of calculus is a simple way of doing this.