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: Apply the probability density function (PDF) property.For a valid PDF, the total area under the curve equals 1. This is mathematically stated as:\[\int_{-\infty}^{\infty} f(x) \,dx = 1\]Substitute the given function:\[\int_{-\infty}^{\infty} ae^{-2|x|} \,dx = 1\]
Step 2: Simplify the integral using symmetry, noting \( e^{-2|x|} \) is an even function (since \( |-x| = |x| \) ). The integral of an even function \( g(x) \) from \(-\infty\) to \(\infty\) is twice the integral from 0 to \(\infty\):\[a \int_{-\infty}^{\infty} e^{-2|x|} \,dx = 2a \int_{0}^{\infty} e^{-2|x|} \,dx\]Since \( |x| = x \) for \( x \ge 0 \), the integral simplifies to:\[2a \int_{0}^{\infty} e^{-2x} \,dx = 1\]
Step 3: Evaluate the integral:\[\int_{0}^{\infty} e^{-2x} \,dx = \left[ -\frac{1}{2}e^{-2x} \right]_0^\infty\]Evaluate 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: Determine 'a'.Substitute the integral's value from Step 3 into the equation from Step 2:\[2a \left( \frac{1}{2} \right) = 1\]\[a = 1\]