Question

Suppose the distribution function of \(X\) is given by $$ F(b)=\left\\{\begin{array}{ll} 0, & b<0 \\ \frac{1}{2}, & 0 \leq b<1 \\ 1, & 1 \leq b<\infty \end{array}\right. $$ What is the probability mass function of \(X ?\)

Short answer

The probability mass function of the random variable X is given by: $$ P(X=k)=\left\\{\begin{array}{ll} \frac{1}{2}, & k=0 \\ \frac{1}{2}, & k=1 \\ 0, & \text{otherwise} \end{array}\right. $$

Step-by-step solution

Identify the support of the random variable

First, let's identify the support of X, which is the set of values where the function takes on nonzero probabilities. From the given CDF, it is evident that the support of X is {0, 1} as the CDF only changes its value at these points.

Determine the probabilities of each value in the support

Since we know the support of X, we can now determine the probability of each value in the support by analyzing the changes in F(b) as b moves from one value to the next. For b = 0:\\ The CDF jumps from 0 to 1/2 as b goes from a value less than 0 to 0. This indicates that there is a probability of 1/2 that X takes on the value of 0. So, in our PMF, we assign P(0) = 1/2. For b = 1:\\ The CDF jumps from 1/2 to 1 as b goes from a value less than 1 to 1. This indicates that there is a probability of 1/2 that X takes on the value of 1. So, in our PMF, we assign P(1) = 1/2. Now that we have the probabilities of each value in the support, we can write the probability mass function of X:

Write the probability mass function of X

The probability mass function of a discrete random variable X is defined as the probability of X taking on each possible value in its support. In this case, we have: $$ P(X=k)=\left\\{\begin{array}{ll} \frac{1}{2}, & k=0 \\ \frac{1}{2}, & k=1 \\ 0, & \text{otherwise} \end{array}\right. $$ So, this is the probability mass function of X, which gives the probability of each value occurring in the support of the random variable X.