## Change of variable

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

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

2. Relevant equations

3. The attempt at a solution

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

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

Thanks.

Recognitions:
Homework Help
 Quote by autobot.d 1. The problem statement, all variables and given/known data Let x,y be iid and $$x, y \sim U(0,1)$$ (uniform on the open set (0,1)) and let $$z = xy^2.$$ Find the density of z. 2. Relevant equations 3. The attempt at a solution $P(z \leq w) = P(xy^2 \leq w) = P(- \sqrt{\frac{w}{x}} \leq y \leq \sqrt{\frac{w}{x}}) = \int^{ \sqrt{\frac{w}{x}}}_{ -\sqrt{\frac{w}{x}}} dy = 2 \sqrt{\frac{w}{x}}$ Is this right. Seems like I am missing something, not sure. Thanks.
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'.

 Quote by autobot.d 1. The problem statement, all variables and given/known data Let x,y be iid and $$x, y \sim U(0,1)$$ (uniform on the open set (0,1)) and let $$z = xy^2.$$ Find the density of z. 2. Relevant equations 3. The attempt at a solution $P(z \leq w) = P(xy^2 \leq w) = P(- \sqrt{\frac{w}{x}} \leq y \leq \sqrt{\frac{w}{x}}) = \int^{ \sqrt{\frac{w}{x}}}_{ -\sqrt{\frac{w}{x}}} dy = 2 \sqrt{\frac{w}{x}}$ Is this right. Seems like I am missing something, not sure. Thanks.
$P(z \leq w)$ should be a function of w only, so something's wrong.

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

## Change of variable

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.

Recognitions:
Homework Help
 Quote by 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.
Look at the two cases x > w and x < w.
 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.