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Let X have this pdf(probability density function). Find the CDF(cumulative density function) of X.

We are given a probability density function (PDF) \( f_X(x) \), and we are tasked with finding the cumulative distribution function (CDF) \( F_X(x) \).

The PDF is piecewise defined as:

\[
f_X(x) = \begin{cases} 
\frac{1}{4} & \text{for } 0 < x < 1 \\
\frac{3}{8} & \text{for } 3 < x < 5 \\
0 & \text{otherwise}
\end{cases}
\]

### Step-by-step procedure to find the CDF \( F_X(x) \)

1. **CDF Definition**: The cumulative distribution function \( F_X(x) \) is the integral of the PDF up to \( x \), i.e.,

\[
F_X(x) = \int_{-\infty}^{x} f_X(t) \, dt
\]

2. **Different regions**: Since the PDF is piecewise, we will compute the CDF piecewise, considering the intervals given in the problem.

### Case 1: \( x \leq 0 \)
For any \( x \leq 0 \), the probability is 0 since the PDF is 0 for \( x < 0 \).

\[
F_X(x) = 0 \quad \text{for } x \leq 0
\]

### Case 2: \( 0 < x \leq 1 \)
In this region, the PDF is \( \frac{1}{4} \). We integrate the PDF from 0 to \( x \):

\[
F_X(x) = \int_{0}^{x} \frac{1}{4} \, dt = \frac{1}{4} x \quad \text{for } 0 < x \leq 1
\]

### Case 3: \( 1 < x \leq 3 \)
For \( x \) in this region, the PDF is 0, so the CDF remains constant and equals the value at \( x = 1 \). At \( x = 1 \):

\[
F_X(1) = \frac{1}{4}(1) = \frac{1}{4}
\]

Thus, for \( 1 < x \leq 3 \):

\[
F_X(x) = \frac{1}{4}
\]

### Case 4: \( 3 < x \leq 5 \)
In this interval, the PDF is \( \frac{3}{8} \). The CDF will add the integral of this portion to the previous value (from \( x = 3 \)):

\[
F_X(x) = \frac{1}{4} + \int_{3}^{x} \frac{3}{8} \, dt = \frac{1}{4} + \frac{3}{8}(x - 3)
\]

Simplifying:

\[
F_X(x) = \frac{1}{4} + \frac{3}{8}(x - 3) = \frac{1}{4} + \frac{3}{8}x - \frac{9}{8}
\]
\[
F_X(x) = \frac{3}{8}x - \frac{7}{8} \quad \text{for } 3 < x \leq 5
\]

### Case 5: \( x > 5 \)
For any \( x > 5 \), the PDF is 0, so the CDF remains constant and equals the value at \( x = 5 \). At \( x = 5 \):

\[
F_X(5) = \frac{3}{8}(5) - \frac{7}{8} = \frac{15}{8} - \frac{7}{8} = 1
\]

Thus, for \( x > 5 \):

\[
F_X(x) = 1
\]

### Final CDF:

\[
F_X(x) = \begin{cases} 
0 & \text{for } x \leq 0 \\
\frac{1}{4}x & \text{for } 0 < x \leq 1 \\
\frac{1}{4} & \text{for } 1 < x \leq 3 \\
\frac{3}{8}x - \frac{7}{8} & \text{for } 3 < x \leq 5 \\
1 & \text{for } x > 5
\end{cases}
\]

Would you like more details on any of these steps, or do you have any questions?

### Related Questions:
1. How is the CDF related to the PDF in general?
2. What are the properties of a valid probability density function (PDF)?
3. How would the CDF change if the constants in the PDF were different?
4. Can the PDF be negative? Why or why not?
5. How would we compute the expected value \( E[X] \) using the given PDF?

### Tip:
When integrating piecewise functions like this, always check that the CDF approaches 1 as \( x \to \infty \), ensuring that the total probability is 1.

Let X have this pdf (probability density function). Find the CDF (cumulative density function) of X.

Learn how to calculate the cumulative distribution function (CDF) from a given piecewise probability density function (PDF). This step-by-step guide covers key concepts like integration of piecewise functions, handling multiple intervals, and ensuring the CDF totals to 1.

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