Suppose that the discrete random variable X has cumulative distribution function given by if x < 1 if 1
Suppose that the discrete random variable X has cumulative distribution function given by if x < 1 if 1
A First Course in Probability (10th Edition)
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![### Cumulative Distribution Function of a Discrete Random Variable
Given a discrete random variable \(X\), the cumulative distribution function \(F(x)\) is defined as follows:
\[ F(x) = \begin{cases}
0, & \text{if } x < 1 \\
\frac{1}{3}, & \text{if } 1 \leq x < \frac{4}{3} \\
\frac{1}{2}, & \text{if } \frac{4}{3} \leq x < \frac{3}{2} \\
\frac{3}{4}, & \text{if } \frac{3}{2} \leq x < \frac{9}{5} \\
1, & \text{if } x \geq \frac{9}{5}
\end{cases}
\]
### Task:
Find the possible values and the probability mass function (PMF) of \(X\).
**Solution:**
To find the possible values and the PMF, we need to compute the differences in the cumulative distribution values, as those differences represent the probabilities at specific points.
1. **Calculate the Differences:**
For each interval, the difference in \(F(x)\) indicates the probability mass at the respective value of \(x\):
- For \( x = 1 \): \( P(X = 1) = F\left(\frac{4}{3}\right) - F(1) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \)
- For \( x = \frac{4}{3} \): \( P(X = \frac{4}{3}) = F\left(\frac{3}{2}\right) - F\left(\frac{4}{3}\right) = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \)
- For \( x = \frac{3}{2} \): \( P(X = \frac{3}{2}) = F\left(\frac{9}{5}\right) - F\left(\frac{3}{2}\right) = 1 - \frac{3}{4} = \frac{1}{4} \)
- For \( x \geq \frac{9}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e7c908-505f-4c35-8ac4-e541f2ab69b2%2F4a3cc665-b9a8-4a74-9f99-afdf72dd04cd%2Fexiw1on_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Cumulative Distribution Function of a Discrete Random Variable
Given a discrete random variable \(X\), the cumulative distribution function \(F(x)\) is defined as follows:
\[ F(x) = \begin{cases}
0, & \text{if } x < 1 \\
\frac{1}{3}, & \text{if } 1 \leq x < \frac{4}{3} \\
\frac{1}{2}, & \text{if } \frac{4}{3} \leq x < \frac{3}{2} \\
\frac{3}{4}, & \text{if } \frac{3}{2} \leq x < \frac{9}{5} \\
1, & \text{if } x \geq \frac{9}{5}
\end{cases}
\]
### Task:
Find the possible values and the probability mass function (PMF) of \(X\).
**Solution:**
To find the possible values and the PMF, we need to compute the differences in the cumulative distribution values, as those differences represent the probabilities at specific points.
1. **Calculate the Differences:**
For each interval, the difference in \(F(x)\) indicates the probability mass at the respective value of \(x\):
- For \( x = 1 \): \( P(X = 1) = F\left(\frac{4}{3}\right) - F(1) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \)
- For \( x = \frac{4}{3} \): \( P(X = \frac{4}{3}) = F\left(\frac{3}{2}\right) - F\left(\frac{4}{3}\right) = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \)
- For \( x = \frac{3}{2} \): \( P(X = \frac{3}{2}) = F\left(\frac{9}{5}\right) - F\left(\frac{3}{2}\right) = 1 - \frac{3}{4} = \frac{1}{4} \)
- For \( x \geq \frac{9}{
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