Question

. let ???? be a discrete random variable that is uniformly distributed over the set of integers in the range [????, ????], where ???? and ???? are integers with ???? < 0 < ????. find the pmf of the random variables max(0,????) and min(0,????).

Answer 1

The PMFs for the random variables max(0, X) and min(0, X) can be calculated using the properties of uniform distributions.

Step-by-step explanation:

The first random variable, max(0, X), takes on the maximum value between 0 and X. Since X is uniformly distributed over the set of integers in the range [a, b], where a < 0 < b, the maximum value will always be either 0 or b. The probability mass function (PMF) of this random variable can be represented as:

P(max(0, X) = 0) = P(X < 0) = (0 - a)/(b - a)

P(max(0, X) = b) = P(X >= 0) = (b - 0)/(b - a)

The second random variable, min(0, X), takes on the minimum value between 0 and X. Again, since X is uniformly distributed, the minimum value will always be either a or 0. The PMF of this random variable can be represented as:

P(min(0, X) = a) = P(X < 0) = (0 - a)/(b - a)

P(min(0, X) = 0) = P(X >= 0) = (b - 0)/(b - a)

Answer 2

Filling in the blanks, I assume you're talking about a random variable
`X` distributed uniformly over the integers
`a\le x\le b`. Let both
`a,b>0` so we can write the support of
`X` as the set

`S = \{-a, -a+1, -a+2, \ldots, -1, 0, 1, \ldots, b-2, b-1, b\}`

Note that
`|S| = a+b+1`, so the PMF of
`X` is

`\mathrm{Pr}(X=x) = \begin{cases}\frac1{a+b+1} & \text{if } x\in S \\ 0 & \text{otherwise}\end{cases}`

Let
`Y=\max\{0,X\}`. Then

`Y = \max\{0,X\} = \begin{cases}0 & \text{if } X\le0 \\ X & \text{if } X>0 \end{cases}`

which tells us

`\displaystyle \mathrm{Pr}(Y=0) = \mathrm{Pr}(X\le0) = \sum_(x=-a)^0 \mathrm{Pr}(X=x) = (a+1)/(a+b+1)`

and

`\displaystyle \mathrm{Pr}(Y\\eq0) = \mathrm{Pr}(X>0) = \sum_(x=1)^b \mathrm{Pr}(X=x) = \frac b{a+b+1}`

Hence the PMF of
`Y` is

`\mathrm{Pr}(Y=y) = \begin{cases}(a+1)/(a+b+1) & \text{if } y=0 \\\\ \frac b{a+b+1} & \text{otherwise}\end{cases}`

Let
`Z=\min\{0,X\}`. The same reasoning applies, but this time

`Z = \min\{0,X\} = \begin{cases} 0 & \text{if } X \ge 0 \\ X & \text{if } X < 0 \end{cases}`

Now

`\displaystyle \mathrm{Pr}(Z=0) = \mathrm{Pr}(X\ge0) = \sum_(x=0)^b \mathrm{Pr}(X=x) = (b+1)/(a+b+1)`

and

`\displaystyle \mathrm{Pr}(Z\\eq0) = \mathrm{Pr}(X<0) = \sum_(x=-a)^(-1) \mathrm{Pr}(X=x) = \frac a{a+b+1}`

so that

`\mathrm{Pr}(Z=z) = \begin{cases}(b+1)/(a+b+1) &\text{if }z=0 \\\\ \frac a{a+b+1} & \text{otherwise}\end{cases}`