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- Oct 4th 2008, 03:05 PM #1

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- Oct 4th 2008, 09:35 PM #2
Not quite. According to your pdf, Pr(Y > 1) = 0 ...... Is that correct ....?

If -1 < x < 2 it's clear that 0 < x^2 < 4 => 0 < y < 4. Therefore f(y) = 0 for y > 4 or y < 0.

So you have to find the pdf for Y when $\displaystyle 1 \leq y < 4$. For such values of y, the cumulative density function is given by $\displaystyle F(y) = \Pr(Y < y) = \Pr(X^2 < y) = \Pr(-1 < X < \sqrt{y})$.

- Oct 9th 2008, 07:02 AM #3

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- Oct 9th 2008, 02:28 PM #4
Yes it is $\displaystyle \Pr( - \sqrt{y} \leq X \leq \sqrt{y})$. But note:

1. $\displaystyle \Pr( - \sqrt{y} \leq X \leq \sqrt{y}) = \Pr( - \sqrt{y} \leq X \leq -1) + \Pr( -1 \leq X \leq \sqrt{y}) $.

2. It's given that -1 < x < 2. So the pdf for X is zero if x < -1. So $\displaystyle \Pr( - \sqrt{y} \leq X \leq -1) = 0$ when $\displaystyle 1 \leq y < 4$.

3. So $\displaystyle \Pr( -1 \leq X \leq \sqrt{y}) $ is the only calculation required when $\displaystyle 1 \leq y < 4$.