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Change of variable

by autobot.d
Tags: variable
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autobot.d
#1
Dec11-12, 02:03 PM
P: 68
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.
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Ray Vickson
#2
Dec11-12, 03:26 PM
Sci Advisor
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P: 4,958
Quote Quote by autobot.d View Post
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.
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
#3
Dec11-12, 03:26 PM
HW Helper
Thanks
P: 946
Quote Quote by autobot.d View Post
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.
[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
#4
Dec11-12, 04:22 PM
P: 68
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.
Ray Vickson
#5
Dec11-12, 04:57 PM
Sci Advisor
HW Helper
Thanks
P: 4,958
Quote Quote by autobot.d View Post
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.
Michael Redei
#6
Dec11-12, 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 square-root expression.


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