If a r. v. X has p. d.f. `f ( x ) = {:(c/x, 1 lt x lt 3"," c lt 0),(0, "otherwise"):}`
Find `c, E(X) and Var (X)`. Also find `F(x)`.

Answer

To solve the problem, we need to find the constant \( c \), the expected value \( E(X) \), the variance \( Var(X) \), and the cumulative distribution function \( F(x) \) for the random variable \( X \) with the given probability density function (pdf):

\[ f(x) = \begin{cases} \frac{c}{x} & \text{for } 1 < x < 3 \\ 0 & \text{otherwise} \end{cases} \]

### Step 1: Find the constant \( c \)

To find \( c \), we use the property that the total probability must equal 1. Therefore, we integrate the pdf over its range:

\[ \int_{1}^{3} f(x) \, dx = 1 \]

Substituting \( f(x) \):

\[ \int_{1}^{3} \frac{c}{x} \, dx = 1 \]

Calculating the integral:

\[ c \int_{1}^{3} \frac{1}{x} \, dx = c [\ln x]_{1}^{3} = c (\ln 3 - \ln 1) = c \ln 3 \]

Since \( \ln 1 = 0 \):

\[ c \ln 3 = 1 \implies c = \frac{1}{\ln 3} \]

### Step 2: Find the expected value \( E(X) \)

The expected value \( E(X) \) is given by:

\[ E(X) = \int_{-\infty}^{\infty} x f(x) \, dx = \int_{1}^{3} x \cdot \frac{c}{x} \, dx \]

Substituting \( c \):

\[ E(X) = \int_{1}^{3} c \, dx = c [x]_{1}^{3} = c (3 - 1) = 2c \]

Now substituting \( c = \frac{1}{\ln 3} \):

\[ E(X) = 2 \cdot \frac{1}{\ln 3} = \frac{2}{\ln 3} \]

### Step 3: Find the variance \( Var(X) \)

Variance is calculated using the formula:

\[ Var(X) = E(X^2) - (E(X))^2 \]

First, we calculate \( E(X^2) \):

\[ E(X^2) = \int_{1}^{3} x^2 \cdot \frac{c}{x} \, dx = \int_{1}^{3} c x \, dx \]

Calculating the integral:

\[ E(X^2) = c \left[ \frac{x^2}{2} \right]_{1}^{3} = c \left( \frac{9}{2} - \frac{1}{2} \right) = c \cdot 4 = 4c \]

Substituting \( c = \frac{1}{\ln 3} \):

\[ E(X^2) = 4 \cdot \frac{1}{\ln 3} = \frac{4}{\ln 3} \]

Now, substituting \( E(X) \):

\[ Var(X) = E(X^2) - (E(X))^2 = \frac{4}{\ln 3} - \left( \frac{2}{\ln 3} \right)^2 \]

Calculating \( (E(X))^2 \):

\[ (E(X))^2 = \frac{4}{(\ln 3)^2} \]

Thus,

\[ Var(X) = \frac{4}{\ln 3} - \frac{4}{(\ln 3)^2} = \frac{4 \ln 3 - 4}{(\ln 3)^2} = \frac{4(\ln 3 - 1)}{(\ln 3)^2} \]

### Step 4: Find the cumulative distribution function \( F(x) \)

The cumulative distribution function \( F(x) \) is defined as:

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

For \( x \leq 1 \):

\[ F(x) = 0 \]

For \( 1 < x < 3 \):

\[ F(x) = \int_{1}^{x} \frac{c}{t} \, dt = c [\ln t]_{1}^{x} = c (\ln x - \ln 1) = c \ln x \]

Substituting \( c = \frac{1}{\ln 3} \):

\[ F(x) = \frac{\ln x}{\ln 3} \]

For \( x \geq 3 \):

\[ F(x) = 1 \]

### Summary of Results

- \( c = \frac{1}{\ln 3} \)
- \( E(X) = \frac{2}{\ln 3} \)
- \( Var(X) = \frac{4(\ln 3 - 1)}{(\ln 3)^2} \)
- \( F(x) = \begin{cases} 0 & \text{for } x \leq 1 \\ \frac{\ln x}{\ln 3} & \text{for } 1 < x < 3 \\ 1 & \text{for } x \geq 3 \end{cases} \)