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I have found the pdf for RV U distribution through the CDF method and am struggling to find the bounds. The transformation was $U = Y^2$ and $Y$'s original bounds were $-1 \leqslant Y \leqslant 1$.

The way I found the bounds for $U$ were the following:

$(-1)^2 \leqslant Y^2 \leqslant 1^2$

$1 \leqslant Y^2 \leqslant 1$

But then I get stuck, what am I doing wrong?

1 Answers 1

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Remember, although $0 b^2>0$.

Break your interval into a disjoint union, and observe how the two halves fold onto the same interval.

You have $(-1\leqslant Y < 0)\wedge(0\leqslant Y\leqslant 1)$ and are transforming via: $U=Y^2$.

The interval transforms to $(1\geqslant U> 0)\vee(0\leqslant U\leqslant 1)$, which is simply: $0\leqslant U\leqslant 1$

  • 0
    and you are logically oring the two inequalities because of having to take the square root correct? – 2017-01-25
  • 0
    No, it's because of the cases considered in the first line. Squaring affects the inequalities differently depending on whether $Y$ is positive or negative. – 2017-01-25