Question

Let the probability density function of a random variable x be given as \( f(x) = ae^{-2|x|} \). The value of 'a' is

• A. \( 0.1 \)
• B. \( 0.5 \)
• C. \( 1 \)
• D. \( 1.5 \)

Hint:

The key property of any probability density function \(f(x)\) is that its integral over all possible values of x must equal 1. For even functions, remember that \( \int_{-L}^{L} g(x) dx = 2 \int_{0}^{L} g(x) dx \).

Answer 1

The Correct Option is C

Step 1: Use the property of a probability density function (PDF). For any valid PDF, the total area under the curve must be equal to 1. Mathematically, this is expressed as: \[ \int_{-\infty}^{\infty} f(x) \,dx = 1 \] Substituting the given function: \[ \int_{-\infty}^{\infty} ae^{-2|x|} \,dx = 1 \]
Step 2: Simplify the integral using symmetry. The function \( e^{-2|x|} \) is an even function because \( |-x| = |x| \). For any even function \( g(x) \), the integral from \(-\infty\) to \(\infty\) can be written as twice the integral from 0 to \(\infty\). \[ a \int_{-\infty}^{\infty} e^{-2|x|} \,dx = 2a \int_{0}^{\infty} e^{-2|x|} \,dx \] For \( x \ge 0 \), \( |x| = x \). So the integral becomes: \[ 2a \int_{0}^{\infty} e^{-2x} \,dx = 1 \]
Step 3: Evaluate the integral. \[ \int_{0}^{\infty} e^{-2x} \,dx = \left[ -\frac{1}{2}e^{-2x} \right]_0^\infty \] Evaluating at the limits: \[ = \left( \lim_{x \to \infty} -\frac{1}{2}e^{-2x} \right) - \left( -\frac{1}{2}e^{-2(0)} \right) = (0) - \left( -\frac{1}{2}e^0 \right) = \frac{1}{2} \]
Step 4: Solve for 'a'. Substitute the value of the integral back into the equation from Step 2: \[ 2a \left( \frac{1}{2} \right) = 1 \] \[ a = 1 \]