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I'm working on the following problem.

We're given a CDF, $F_y(t)$, and a uniformly distributed random variable $X$ on the interval $[0,1]$. We define $Y = f(X)$ where $f(u) = inf\{t \in \mathbb{R} | F_y(t) \geq u\}$. We want to prove that $Y$ has the desired CDF $F_y(t)$. (Note that $Y$ won't necessarily be unique.)

Our professor gave us the following hint, but I'm not sure how it's helpful.

Hint: First show that the following two sets are equal, $(-\inf, F_y(t)] = \{u \in \mathbb{R}: f(u) \leq t\}$.

What I'm thinking is that our CDF $F_y(t)$ need not be continuous, only necessarily continuous from the right, so we need to proceed by cases. I found a similar question here, but I feel like this one is a bit different. Any hints or advice would be much appreciated.

user31415926
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1 Answers1

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Why the hint is helpful:

\begin{align} P(Y \le t) &= P(f(X) \le t)\\ &= P(X \in \{u \in \mathbb{R} : f(u) \le t\}) \\ &= P(X \in (-\infty, F_y(t)]) & \text{used hint here} \\ &= F_y(t) & \text{$X$ is uniform on $[0,1]$} \end{align}

Proving the hint:

If $u$ satisfies $f(u) \le t$, then using the definition of $f$ and the fact that $F_y$ is monotone nondecreasing implies that $F_y(t) \ge u$.

Conversely, if $u \le F_y(t)$, then using the same two facts (definition of $f$, $F_y$ is monotone nondecreasing) implies $f(u) \le t$.

angryavian
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