Question

Consider a piecewise constant PDF of the form

fX(x)=c if 1 < x ≤ 3;2 c, if 0≤x≤ 1;0,otherwise

Find the following values.

a)c

b) P(1/2 ≤X≤3/2)=

Answer 1

To find the value of the normalization constant \( c \), we need to ensure that the total probability over the domain of the probability density function (PDF) integrates to 1. The PDF is defined as follows:

\[
f_X(x) =
\begin{cases}
c & \text{if } 1 < x \leq 3 \\
2c & \text{if } 0 \leq x \leq 1 \\
0 & \text{otherwise}
\end{cases}
\]

### a) Finding \( c \):

To find \( c \), we will find the integral of the PDF over its defined intervals and set it equal to 1.

1. **For the interval \( 1 < x \leq 3 \):**

\[
\int_{1}^{3} c \, dx = c \cdot (3 - 1) = 2c
\]

2. **For the interval \( 0 \leq x \leq 1 \):**

\[
\int_{0}^{1} 2c \, dx = 2c \cdot (1 - 0) = 2c
\]

3. **Adding up both integrals:**

\[
\int_{-\infty}^{\infty} f_X(x) \, dx = 2c + 2c = 4c
\]

Setting this equal to 1 for normalization:

\[
4c = 1 \implies c = \frac{1}{4}
\]

### So, \( c = \frac{1}{4} \).

### b) Finding \( P \left( \frac{1}{2} \leq X \leq \frac{3}{2} \right) \):

To find this probability, we can integrate the PDF over the interval \( \left[ \frac{1}{2}, \frac{3}{2} \right] \).

This interval covers two parts:

1. The interval from \( \frac{1}{2} \) to \( 1 \), where \( f_X(x) = 2c = \frac{2}{4} = \frac{1}{2} \).
2. The interval from \( 1 \) to \( \frac{3}{2} \), where \( f_X(x) = c = \frac{1}{4} \).

Calculating each integral:

1. **From \( \frac{1}{2} \) to \( 1 \):**

\[
P\left( \frac{1}{2} \leq X \leq 1 \right) = \int_{\frac{1}{2}}^{1} \frac{1}{2} \, dx = \frac{1}{2} \cdot (1 - \frac{1}{2}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}
\]

2. **From \( 1 \) to \( \frac{3}{2} \):**

\[
P\left( 1 \leq X \leq \frac{3}{2} \right) = \int_{1}^{\frac{3}{2}} \frac{1}{4} \, dx = \frac{1}{4} \cdot \left( \frac{3}{2} - 1 \right) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}
\]

3. **Adding both probabilities:**

\[
P\left( \frac{1}{2} \leq X \leq \frac{3}{2} \right) = P\left( \frac{1}{2} \leq X \leq 1 \right) + P\left( 1 \leq X \leq \frac{3}{2} \right) = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}
\]

### Final Answer:

\[
P\left( \frac{1}{2} \leq X \leq \frac{3}{2} \right) = \frac{3}{8}
\]