Given the probability distribution function(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

f(x) = \frac{2}{9}(x-1), -1<x<2

[/tex]

find the pdf of [tex] Y = X^2 [/tex]

My Solution:

When x = -1, y = 1 and when x = 2, y = 4, so the range of y is

[tex]

1 \leq y \leq 4

[/tex]

So to find the pdf of Y = X^2, we need to find the cdf of Y. Since Y is nontrivial over the interval [tex] 1 \leq y \leq 4 [/tex],

[tex]

F(y) = P(1 \leq Y \leq y)

=P(1 \leq x^2 \leq y)

[/tex]

I'm at a loss here. I do not know how I should deal with square rooting the x, since [tex] x^2 \leq y [/tex] is rewritten as [tex] -\sqrt{y} \leq x \leq \sqrt{y}[/tex]

Any suggestions?

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# Homework Help: Find the probability distribution

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