0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – papasmurfete Mar 5 at 20:20
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ why does y take values from 0 to 1/2? $\endgroup$ – The Poor Jew Mar 5 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. $\endgroup$ – Graham Kemp Mar 5 at 23:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.