Suppose we have $X$, a $\operatorname{Uniform}(0, 1)$ random variable which follows with the probability density function $f_X (x)$. Let $Y = \min\{X, 1 − X \}$. It wasn't asked but I want to find the pdf of $Y$.

I think I know how to deal with other transformations but the min here is really bothering me (since $\min(X)=X$ when $X$ is less then $0.5$ maybe I tried to investigate two cases for $Y$ ( $Y$ less then or equal to $0.5$ and greater then $0.5$) but I'm stuck.


There may be more elegant arguments (based on, say, symmetry). Here is one simply doing the bruteforce thing: compute the PDF via the CDF.

Consider the CDF of $Y$, $F$. For $y\in[0,1/2]$, $$\begin{align*} F(y) &= \mathbb{P}\{ Y \leq y\} = \mathbb{P}\{ \min(X,1-X) \leq y\} = \mathbb{P}\{ X \leq y\}\cup \{ 1-X \leq y\} \\ &= \mathbb{P}\{ X \leq y\}\cup \{ X \geq 1-y\} \\ &=\mathbb{P}\{ X \in (0,y]\cup[1-y,1)\} \end{align*}$$ and since $X$ is uniform on $(0,1)$, this gives $$ F(y) = 2y. $$ Therefore, $$ F(y) = \begin{cases} 0 & \text{ if } y < 0\\ 2y & \text{ if } y \in [0,1/2]\\ 1 & \text{ if } y > 1/2\\ \end{cases} $$ Differentiating the CDF, you get the PDF, call it $f$: $$ f(y) = \begin{cases} 0 & \text{ if } y < 0\\ 2 & \text{ if } y \in [0,1/2]\\ 0 & \text{ if } y > 1/2\\ \end{cases} $$ That is, $\boxed{Y\sim\mathrm{Uniform}(0,1/2)}$.

  • $\begingroup$ I don't really get why this : $$\begin{align*} F(y) &= \mathbb{P}\{ Y \leq y\} =\mathbb{P}\{ X \in (0,y]\cup[1-y,1)\} \end{align*}$$ gives us F(y) = 2y. $\endgroup$ – Student number x Mar 3 at 18:01
  • $\begingroup$ @Studentnumberx Which part do you not understand in what you quoted? The last? If so, just note that the measure under the uniform distribution of $(0,y]$ is $y$, and so is the measure of $[1-y,1)$, and the two intervals are disjoint. (Or, if $y=1/2$, their intersection is $\{1/2\}$, which has measure 0) $\endgroup$ – Clement C. Mar 3 at 18:03
  • $\begingroup$ Can density equal to more than 1? isn't the whole integral supposed to be equal to 1? $\endgroup$ – The Poor Jew Mar 6 at 19:11
  • 1
    $\begingroup$ @ThePoorJew Yes, it can. Integrate 2 over [0,1/2], you will get 1. $\endgroup$ – Clement C. Mar 6 at 19:12

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