Does the square of uniform distribution have density function?

Solution 1:

They are not uniform distribution.$$F_{X^2}(x) = \mathbb{P}(X^2 \leq x) = \mathbb{P}(X \in [0,\sqrt{x}]) = \sqrt{x}$$ Hence, $$f_{X^2}(x) = \dfrac{dF_{X^2}(x)}{dx} = \dfrac1{2\sqrt{x}}$$ You can do the same thing for $Y \mapsto Y^2$. $$F_{Y^2}(y) = \mathbb{P}(Y^2 \leq y) = \mathbb{P}(Y \in [-\sqrt{y}, \sqrt{y}]) = \sqrt{y}$$ Hence, $$f_{Y^2}(y) = \dfrac{dF_{Y^2}(y)}{dy} = \dfrac1{2\sqrt{y}}$$

Solution 2:

First, find the CDF of the transformed RV. Let $S = X^2$.

$P(S \leq s) = P(X^2 \leq s) = P(X \leq \sqrt s) = \sqrt s$, for $0 \leq s \leq 1$ and $0$ otherwise. Where the last equality is true since $X$ ~ $\text{Unif}(0,1).$ Note, in general, the support is not the same under the transform.

Next, differentiate to find the PDF. $\frac{d}{dx} \sqrt{s} = \frac{1}{2 \sqrt s}$ for $0 \leq s \leq 1$ and $0$ otherwise.

The PDF of $Y^2$ follows similarly. Also, in general, uniformity is not preserved under non-linear transforms.