$X$ a random variable with distribution $F_X(x)$

$$Y=g(X) = \left\{ \begin{array}{lr} X-c & : X > c\\ 0 & : -c < X \le c \\ X+c & : X \le -c \end{array} \right\}$$

My approach

This is a mixed distribution

$F_Y(y=0) = P(-c < X < c) = F_X(c)-F_X(-c)$

My question:

Is this the correct way to write $F_Y(y) $ , $ y \ne 0$. If not, then what is the correct way?

$F_Y(y) = P(Y\le y) = P(Y\le y | X>c) P(X>c) + P(Y\le y | X<c) P(X<c)$

This would give. $F_X(y+c) (1-F_X(c)) + F_X(y-c)F_X(c)$

  • 2
    $\begingroup$ A simple geometric approach works for your previous question, too, and shows that the two questions are essentially the same one. Graph $F_X$. Cut out the vertical band over $(-c,c)$. Sew the two halves of the graph back together. Relabel the axes: you have to subtract $c$ from numbers at the right and add $c$ to numbers at the left of the origin. That's the graph of $F_Y$. You can check that this is not what you have written, so stare at it and write a correct formula. (In the previous question the entire negative half of the graph over the interval $(-\infty,0)$ would be excised.) $\endgroup$
    – whuber
    Jun 25, 2014 at 18:20
  • $\begingroup$ stats.stackexchange.com/questions/248003/…. $\endgroup$ Apr 8, 2018 at 20:11


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