If $f(x)$ is a strictly increasing function on the unit interval, what is the distribution of $f(\mathcal{U})$? Prove it.

Question

$\mathcal{U}$ is distributed uniformly on the interval $[0,1]$. If $f(x)$ is a strictly increasing function on the unit interval, what is the distribution of $f(\mathcal{U})$? Prove it.

Well if $f(x)$ is a strictly increasing function on the unit interval, then $f'(x)>0$ on $[0,1]$. But I don't know where to go?

Answer 1

Given: $f:[0,1] \to \mathbf{R}$ strictly increasing and $U$ uniformly distributed over $[0,1]$.

$U$ uniformly distributed over $[0,1]$ means: $\forall_{x \in [0,1]}: P(U \le x) = x$. We say that U's cumulative distribution function is $x \mapsto x$.

Let's now look at the cumulative distribution function of $f(U)$: $\forall_{x \in \mathbf{R}}: P(f(U) \le x) = P(f(U) \le x_{f}) \stackrel{(*)}{=} P(U \le f^{-1}(x_{f})) \stackrel{(**)}{=} f^{-1}(x_{f})$, where $x_{f}=\sup\{y \in f([0,1]) | y \le x\}$, $(*)$ holds for strictly increasing $f$ (which also means that $f$ is invertible), and $(**)$ follows from $U$ being uniformly distributed.

Thus $f(U)$'s cumulative distribution function is $x \mapsto f^{-1}(x_f)$.

Answer 2

Hint: $f(\mathcal U) \le f(x)$ iff $\mathcal U \le x$ for $0 \le x \le 1$.