Let $X$ be random variable such that $\begin{align} F_X(x) = 1- e^{-x} \end{align}$ if $x \ge 0$ and $F_X(x)=0$ in other case. Find distribution function $Y= \min(1,X)$, $Z=\max(1,X)$.

If I have to find $\max(X,Y)$ or $\min(X,Y)$ ($X,Y$ - random variable) I don't have any problem. But in this case I have number - what should I do?

  • $\begingroup$ If you know how to handle $\text{max}(X,V)$, then why not just think of $V$ as a random variable with mean $1$ and variance $0$? $\endgroup$ – nispio Oct 10 '13 at 22:01

Let $Y=\min(1, X)$. We find $F_Y(y)$.

If $y\le 0$, then $$F_Y(y)=\Pr(Y\le y)=\Pr(\min(1,X)\le y)=\Pr(X\le y)=0.$$

Now let $0\lt y\lt 1$. Then $$F_Y(y)=\Pr(Y\le y)=\Pr(\min(1,X)\le y)=\Pr(X\le y)=1-e^{-y}.$$

If $y\ge 1$, then for sure $\min(1,X)\le 1$, and therefore $F_Y(y)=1$.

Note that $Y$ does not have continuous distribution. The function $F_Y(y)$ is continuous almost everywhere, but there is a sudden jump at $1$. There is a point mass at $1$, which is equal to the probability $e^{-1}$ that $X\ge 1$.

We leave it to you to do max. The analysis is similar to the one for min.

  • $\begingroup$ Could you tell me why in these two first case we have $\Pr(\min(1,X)\le y) = \Pr(X \le y)$ ? I don't understand this equal. $\endgroup$ – Thomas Oct 11 '13 at 6:26
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    $\begingroup$ What is the probability that the smaller of $1$ and $X$ is $\le 0.4$? The smaller of $1$ and $X$ then can't be $1$. So it must be $X$, and $X$ must be $\le 0.4$. $\endgroup$ – André Nicolas Oct 11 '13 at 14:08
  • $\begingroup$ Ok, now i understand. And now I'm doing it for maximum. I suppose that for $z \le0$ and $0 < z < 1$ we have $F_Z(z) = 0$ and for $z \ge 1$ we have $F_Z(z) = 1-e^{-z}$. I'm right? $\endgroup$ – Thomas Oct 11 '13 at 18:28
  • $\begingroup$ Yes, that's right. $\endgroup$ – André Nicolas Oct 11 '13 at 18:36

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