Understanding the Probability Density Function (PDF)

The question asks us to find the value of a constant 'a' and a specific probability for a given probability density function (PDF) of a random variable \(X\). The PDF is defined as:

$$f_X(x) = \frac{3}{2}e^{-3x}u(x) + ae^{4x}u(-x)$$

Here, \(u(x)\) represents the unit step function, which is 1 for \(x \ge 0\) and 0 for \(x < 0\). We need to find 'a' and \(P(X \le 0)\).

Determining the Constant 'a'

For any valid PDF, the total integral over the entire domain must equal 1. This means:

$$\int_{-\infty}^{\infty} f_X(x) dx = 1$$

We can split the integral based on the behavior of the unit step functions \(u(x)\) and \(u(-x)\):

• When \( x \ge 0 \): \(u(x) = 1\) and \(u(-x) = 0\). The PDF becomes \( f_X(x) = \frac{3}{2}e^{-3x} \).
• When \( x < 0 \): \(u(x) = 0\) and \(u(-x) = 1\). The PDF becomes \( f_X(x) = ae^{4x} \).

So, the integral condition becomes:

$$\int_{-\infty}^{0} ae^{4x} dx + \int_{0}^{\infty} \frac{3}{2}e^{-3x} dx = 1$$

Let's calculate the first integral (from \(-\infty\) to 0):

$$ \int_{-\infty}^{0} ae^{4x} dx = a \left[ \frac{e^{4x}}{4} \right]_{-\infty}^{0} = a \left( \frac{e^{0}}{4} - \lim_{t \to -\infty} \frac{e^{4t}}{4} \right) = a \left( \frac{1}{4} - 0 \right) = \frac{a}{4} $$

Now, let's calculate the second integral (from 0 to \(\infty\)):

$$ \int_{0}^{\infty} \frac{3}{2}e^{-3x} dx = \frac{3}{2} \left[ \frac{e^{-3x}}{-3} \right]_{0}^{\infty} = \frac{3}{2} \left( \lim_{t \to \infty} \frac{e^{-3t}}{-3} - \frac{e^{0}}{-3} \right) = \frac{3}{2} \left( 0 - (-\frac{1}{3}) \right) = \frac{3}{2} \times \frac{1}{3} = \frac{1}{2} $$

Using the total integral property:

$$ \frac{a}{4} + \frac{1}{2} = 1 $$

Solving for 'a':

$$ \frac{a}{4} = 1 - \frac{1}{2} = \frac{1}{2} $$

$$ a = 4 \times \frac{1}{2} = 2 $$

So, the value of the constant 'a' is 2.

Calculating the Probability \(P(X \le 0)\)

The probability \(P(X \le 0)\) requires integrating the PDF from \(-\infty\) to 0.

$$ P(X \le 0) = \int_{-\infty}^{0} f_X(x) dx $$

From our analysis, for \( x < 0 \), \( f_X(x) = ae^{4x} \). We found that \( a = 2 \). Therefore, we need to calculate:

$$ P(X \le 0) = \int_{-\infty}^{0} 2e^{4x} dx $$

We already calculated this integral part when finding 'a':

$$ P(X \le 0) = 2 \left[ \frac{e^{4x}}{4} \right]_{-\infty}^{0} = 2 \left( \frac{1}{4} - 0 \right) = \frac{2}{4} = \frac{1}{2} $$

Thus, the probability \(P(X \le 0)\) is \(\frac{1}{2}\).

Final Answer

The value of 'a' is 2 and the probability \(P(X \le 0)\) is \(\frac{1}{2}\).