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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?

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  • $\begingroup$ When X<1/2:Y=X with probability P(X<1/2) =1/2 else Y=1-X with probability P(X>1/2) =1/2 $\endgroup$ Mar 5, 2019 at 20:20

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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?

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.

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  • $\begingroup$ why does y take values from 0 to 1/2? $\endgroup$
    – user634512
    Mar 5, 2019 at 23:24
  • $\begingroup$ @ThePoorJew Because $X$ takes values from 0 to 1, and so the minimum of $X,1-X$ will therefore be greater than 0 and at most 1/2. $\endgroup$ Mar 5, 2019 at 23:50

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