Let $X$ be a discrete random variable that is uniformly distributed over the set of integers in the range $[a,b],$ where $a$ and $b$ are integers with $a<0<b$. Find the PMF of the random variables $\max\{0,X\}$ and $\min\{0,X\}$
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There are $b-a+1$ integers between $a$ and $b$, inclusive. The number between $a$ and $0$, inclusive, is $1-a$. The remaining $b$ numbers are $\gt 0$.
Let $Y=\max(0,X)$. Then $Y=0$ precisely if $X\le 0$. Thus
$$\Pr(Y=0)=\frac{1-a}{b-a+1}.$$
If $X$ is positive, then $Y=X$. So for any $y$ with $1\le y\le b$ we have $\Pr(Y=y)=\frac{1}{b-a+1}$.
Now it's your turn for the distribution of $Z=\min(0<X)$.
answered Sep 16, 2013 at 3:36
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$\begingroup$ If Z=min(0,X), then Z=0 iff X is greater than or equal to zero, so P(Z=0)=b/(b-a+1). If X is negative, then Z=X. So for any z with a<z<1, we have P(Z=z)=1/(b-a+1) Is that close? $\endgroup$– ElsyeCommented Sep 16, 2013 at 4:00
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$\begingroup$ Close. Little slip on $\Pr(Z=0)$. It is $\frac{b+1}{b-a+1}$, since there are $b$ numbers from $0$ to $b$ inclusive. Also for second part, you want to say for $a\le z\le -1$. $\endgroup$ Commented Sep 16, 2013 at 4:03
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$\begingroup$ How are we ending up with the last part of each of these? (ie P(Y=y)=1/b-a+1. Where are we getting the 1 on top? $\endgroup$– ElsyeCommented Sep 16, 2013 at 13:16
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$\begingroup$ The original $X$ has uniform distribution, with each of the $b-a+1$ points equally likely. So for any integer $x$ between $a$ and $b$, we have $\Pr(X=x)=\frac{1}{b-a+1}$. So in the case of $\max$ and $\min$, for points that don't get collapesed into $0$, probability is unchanged at $\frac{1}{b-a+1}$. $\endgroup$ Commented Sep 16, 2013 at 14:57