# Thread: probability

1. ## probability

Let $X$ be uniformly distributed over $(0,1)$. Let $Y = X^2$. Find the cdf and pdf for the random variable $Y$, for $0 \leq y \leq 1$.

So $f_{X}(x) = \begin{cases} 1 \ \ \ \ \ \ \ \ 0 < x < 1 \\ 0 \ \ \ \ \ \ \ \ \text{otherwise} \end{cases}$ and $F(x) = \begin{cases} 0 \ \ \ \ \ x < 0 \\ x \ \ \ \ \ 0 < x < 1 \\ 1 \ \ \ \ \ x > 1 \end{cases}$

Then $X = \sqrt{Y}$ and so $f_{Y}(y) = \begin{cases} \frac{1}{2}y^{-1/2} \ \ 0 \leq \sqrt{y} \leq 1 \\ 0 \ \ \ \ \ \ \ \ \text{otherwise} \end{cases}$ and $F(y) = \begin{cases} 0 \ \ \ \ \ \sqrt{y} \leq 0 \\ \sqrt{y} \ \ \ \ \ 0 \leq \sqrt{y} \leq 1 \\ 1 \ \ \ \ \ \sqrt{y} \geq 1 \end{cases}$

Then $P(X \leq 0.70) = F(0.70) = 0.70$ and $P(Y \leq 0.70) = F(0.70) = 0.83$

Is this correct?

2. are the bounds correct?