Change of variable

autobot.d

1. The problem statement, all variables and given/known data

Let x,y be iid and [tex]x, y \sim U(0,1) [/tex] (uniform on the open set (0,1)) and let [tex] z = xy^2. [/tex]
Find the density of z.


2. Relevant equations



3. The attempt at a solution

[itex]P(z \leq w) = P(xy^2 \leq w) = P(- \sqrt{\frac{w}{x}} \leq y \leq \sqrt{\frac{w}{x}}) = \int^{ \sqrt{\frac{w}{x}}}_{ -\sqrt{\frac{w}{x}}} dy = 2 \sqrt{\frac{w}{x}}
[/itex]

Is this right. Seems like I am missing something, not sure.

Thanks.

Ray Vickson

No. Y cannot be < 0, so you cannot have the region ## - \sqrt{w/x} \leq y < 0##. Anyway, you need an answer that contains w only, so you still need to get rid of the 'x'.

pasmith

[itex]P(z \leq w)[/itex] should be a function of w only, so something's wrong.

Fix x. Then [itex]P(y^2 \leq w/x) = P(0 \leq y \leq \sqrt{w/x}) = \min(1,\sqrt{w/x})[/itex] since [itex]0 \leq y \leq 1[/itex]. Then
[tex]
P(z \leq w) = P(xy^2 \leq w) = \int_0^1 \min\left(1,\sqrt{\frac wx}\right)\,\mathrm{d}x
= \int_0^w 1\,\mathrm{d}x + \int_w^1 \sqrt{\frac wx}\,\mathrm{d}x
[/tex]

autobot.d

Makes sense it should be a function of w only. I do not understand though how the integral with the minimum is broken up into the two integrals at the end. Any insight?

Thanks for the help.

Ray Vickson

Look at the two cases x > w and x < w.

Michael Redei

The first integral ranges from ##x=0## to ##x=w##, so the minimum is equal to 1. In the second integral, you have ##x\geq w##, and so the minimum is the square-root expression.