Question

If the distribution function of \(F\) is given by $$ F(b)=\left\\{\begin{array}{ll} 0, & b<0 \\ \frac{1}{2}, & 0 \leq b<1 \\ \frac{3}{3}, & 1 \leq b<2 \\ \frac{4}{3}, & 2 \leq b<3 \\ \frac{9}{10}, & 3 \leq b<3.5 \\ 1, & b \geq 3.5 \end{array}\right. $$ calculate the probability mass function of \(X\).

Short answer

The probability mass function of the random variable \(X\) is: $$ p(x)=\left\{\begin{array}{ll} \frac{1}{2}, & x=0 \\ \frac{1}{2}, & x=1 \\ \frac{1}{3}, & x=2 \\ \frac{1}{10}, & x=3 \end{array}\right. $$

Step-by-step solution

Identify intervals and values for the PMF

First, let's identify the intervals and the values where the distribution function F(b) changes: - \(0 \leq b < 1\) - \(1 \leq b < 2\) - \(2 \leq b < 3\) - \(3 \leq b < 3.5\) - \(3.5 \leq b\) Note that these are the intervals when the distribution function has different values.

Calculate the differences of the function F(b) for the intervals

Now, we will calculate the differences in the distribution function for each interval. 1. For interval \(0 \leq b < 1\), \(p(0) = F(0) - F(-0.1) = \frac{1}{2} - 0 = \frac{1}{2}\). 2. For interval \(1 \leq b < 2\), \(p(1) = F(1) - F(0.9) = 1 - \frac{1}{2} = \frac{1}{2}\). 3. For interval \(2 \leq b < 3\), \(p(2) = F(2) - F(1.9) = \frac{4}{3} - 1 = \frac{1}{3}\). 4. For interval \(3 \leq b < 3.5\), \(p(3) = F(3) - F(2.9) = \frac{9}{10} - \frac{4}{3} = \frac{1}{10}\). 5. For interval \(3.5 \leq b\), since \(F(b) = 1\) for \(b \geq 3.5\), there are no more possible values for the random variable after \(b=3.5\).

State the probability mass function for X

The probability mass function of the random variable \(X\) is: $$ p(x)=\left\{\begin{array}{ll} \frac{1}{2}, & x=0 \\ \frac{1}{2}, & x=1 \\ \frac{1}{3}, & x=2 \\ \frac{1}{10}, & x=3 \end{array}\right. $$