Find k, if the function f defined by `f(x) = kx, 0 lt x lt 2` =0 , otherwise is the p.d.f of a random variable X.

Answer

To find the value of \( k \) such that the function \( f(x) = kx \) for \( 0 < x < 2 \) is a probability density function (p.d.f) of a random variable \( X \), we need to ensure that the integral of the p.d.f over its entire range equals 1. Here’s how to do it step by step:

### Step 1: Set up the integral of the p.d.f
The p.d.f \( f(x) \) is defined as:
\[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \]
To find \( k \), we need to evaluate the integral of \( f(x) \) from 0 to 2 and set it equal to 1:
\[ \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{2} kx \, dx = 1 \]

### Step 2: Evaluate the integral
Now we compute the integral:
\[ \int_{0}^{2} kx \, dx = k \int_{0}^{2} x \, dx \]
The integral \( \int x \, dx \) is given by:
\[ \int x \, dx = \frac{x^2}{2} \]
Thus, we calculate:
\[ \int_{0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} - 0 = 2 \]
Therefore, we have:
\[ \int_{0}^{2} kx \, dx = k \cdot 2 \]

### Step 3: Set the integral equal to 1
Now we set the result of the integral equal to 1:
\[ k \cdot 2 = 1 \]

### Step 4: Solve for \( k \)
To find \( k \), we divide both sides by 2:
\[ k = \frac{1}{2} \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{\frac{1}{2}} \]

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