Transformation of variable

**lindah** · June 21st 2011, 02:29 AM

May I ask if my approach is valid? In attempting this I arrive at:
$f_{X}(x) = \frac{1}{b-a} = \frac{1}{1-0}$

Then finding the cdf of $Y=U^{1/2}$ I have:
$F_{y}(y) = P(Y \leq y) = P(U^{1/2} \leq y) = \int_{0}^{y^2} f_{X}(x)\ dx = y^2$ for $0 \leq y \leq 1$

So my cdf is
$F_{Y}(y) =\begin{cases}0 & y<0 \\ y^2 & 0<y \leq 1\\1 & y>1 \\ \end{cases}<br />$

Then the pdf is the derivative of this
$f_{Y}(y) =\begin{cases} 0 & y<0 \\ 2y & 0<y \leq 1 \\0 & y>1 \\ \end{cases}<br />$

**theodds** · June 21st 2011, 08:08 AM

That's fine.

**lindah** · June 21st 2011, 02:37 PM

Thank you for taking the time to look over it!

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