# Change of variable

1. Dec 11, 2012

### 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.

Last edited: Dec 11, 2012
2. Dec 11, 2012

### 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'.

3. Dec 11, 2012

### pasmith

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

4. Dec 11, 2012

### 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.

5. Dec 11, 2012

### Ray Vickson

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

6. Dec 11, 2012

### 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.