ISRO EC 2023 Q.6: solution and answer

ISRO 2023 · ECQ.6 · 1 mark · MCQ

If a random communication signal has PDF given by $P (X) = 2 a e^{- b ∣ x ∣}$ for all $x$ , $- \infty < x < + \infty$ , then:

See the question in the paper

Answer and solution

ANSWER

$b = 4 a$ , which is option (B).

A valid probability density must integrate to 1 over its whole range: $\int_{- \infty}^{\infty} P (x) d x = 1$ .
Since $e^{- b ∣ x ∣}$ is an even function of $x$ , $\int_{- \infty}^{\infty} 2 a e^{- b ∣ x ∣} d x = 2 \cdot 2 a \int_{0}^{\infty} e^{- b x} d x$ .
For $b > 0$ (needed for the integral to converge), $\int_{0}^{\infty} e^{- b x} d x = \frac{1}{b}$ , so the total area is $\frac{4 a}{b}$ .
Setting $\frac{4 a}{b} = 1$ gives $b = 4 a$ .
Check the other options: $b = 0$ makes the integral diverge, while $b = 2 a$ and $b = a /2$ give total areas $\frac{4 a}{2 a} = 2$ and $\frac{4 a}{a /2} = 8$ , neither equal to 1.

KEY IDEA

A PDF must have total area 1. Normalising the two-sided exponential $2 a e^{- b ∣ x ∣}$ gives $4 a / b = 1$ .

WATCH OUT FOR

Integrating only over $0$ to $\infty$ and forgetting to double it for the symmetric negative half, which gives $b = 2 a$ instead of $b = 4 a$ .