Question

Let X be the random variable such that : P(X = –2) = P(X = –1), P(X = 2) = P(X = 1) and P(X > 0) = P(X < 0) = P(X = 0). Obtain the probability mass function of X and its distribution functions

Answer 1

$$P(X<0)=P(X = –2) +P(X = –1)$$

$$P(X>0)=P(X = 2) +P(X = 1)$$

$$P(X<0)+P(X = 0) +P(X>0)=1$$

Then

$$P(X<0)=P(X = 0) =P(X>0)=1/3$$

$$P(X=-2)=P(X = -1)=1/6$$

$$P(X=2)=P(X = 1)=1/6$$

Probability mass function

$$p(x) = \begin{cases} 1/6 &\text{if } x=-2 \\ 1/6 &\text{if } x=-1 \\ 1/3 &\text{if } x=0 \\ 1/6 &\text{if } x=1 \\ 1/6 &\text{if } x=2 \\ \end{cases}$$

The cumulative distribution function

$$F(x) = \begin{cases} 0 & x<-2 \\ 1/6 & -2\le x<-1 \\ 1/3 &-1\le x<0 \\ 2/3 & 0\le x<1 \\ 5/6 & 1\le x<2 \\ 1 & x\ge2 \\ \end{cases}$$