
#1
Dec1112, 02:03 PM

P: 69

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. 



#2
Dec1112, 03:26 PM

HW Helper
Thanks
P: 4,670





#3
Dec1112, 03:26 PM

HW Helper
P: 772

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] 



#4
Dec1112, 04:22 PM

P: 69

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. 



#5
Dec1112, 04:57 PM

HW Helper
Thanks
P: 4,670





#6
Dec1112, 05:06 PM

P: 181

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 squareroot expression.



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