The problem
Let $X$ be a random variable and let $F_X$ be its distribution function. Suppose $F_X$ admits an inverse function $F_X^{-1}$. Let now $Y=F_X^{-1}(U)$ where $U\hookrightarrow \mathcal{U}([0,1])$. What is the law of $Y$?
My solution
Let $(a,b) \in\mathbb{R}^2, a<b$. We're trying to determine : $\mathbb{P}(a\leq Y \leq b)$.
Let's note that : $a \leq Y\leq b \iff a \leq F_X^{-1}(U) \leq b \iff F_X(a) \leq U \leq F_X(b)$.
Hence,
\begin{align*}
\mathbb{P}(a\leq Y\leq b) &= \mathbb{P}(F_X(a) \leq U \leq F_X(b)) & \\
&= \int_{F_X(a)}^{F_X(b)} \mathbb{1}_{[0,1]}(x)dx & (1)\\
&= \int_{a}^{b} \mathbb{1}_{[0,1]}(F_X^{-1}(t))F_X'(t)dt & (2)
\end{align*}
And we deduce that $Y$ admits $p_Y:t\mapsto\mathbb{1}_{[0,1]}(F_X^{-1}(t))F_X'(t)$ as its density function.
Questions
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Does any of this look right to you?
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It is not said that $F_X$ is of class $C^1$; hence, if it is not, I cannot go from (1) to (2). In this case, would line (1) be enough of an answer to determine the law of $Y$?
Best Answer
Recall that $F(x) \in [0,1]$ for all $x$. So (1) gives $P(a<Y\leq b)=F_X(b)-F_X(a)$. Conclude that $F_X$ is the law of $Y$.