Let $X \sim \mathrm{Uniform}(0, 1)$ random variable with the probability density function $f_X(x)$ given by
$$
f_X(x) = \begin{cases} 1, &0 < x < 1, \\ 0, &\text{otherwise}\end{cases}
$$
Let $Y$ = $\min\,\{X, 1 − X\}$. Compute the probability density function of $Y$.
So this $Y=\min\,\{X, 1-X\}$ confuses me. I'm not sure how to proceed.
When will $Y=X$ and $Y=1-X$ and how to consider these two cases?
Best Answer
Consider that when the minimum of two variables is less than a value, then at least one from the variables is less than that value. (Also the support for $Y$ will be $(0;1/2]$.) $$\begin{split}\mathsf P(Y\leq y)&=\mathsf P(\min\{X,1-X\}\leq y)~\mathbf 1_{y\in(0;1/2]}\\[1ex]&=\mathsf P(X\leq y\cup 1-X\leq y)~\mathbf 1_{y\in(0;1/2]} \\[1ex]&=\mathsf P(X\leq y\cup X\geq 1-y)~\mathbf 1_{y\in(0;1/2]}\\[1ex] &~~\vdots\end{split}$$
The probability density function will be the unsigned differential of this cumulative distribution function.