GATE 2022 EE Q.54: solution and answer

GATE 2022 · EEQ.54 · 2 marks · NAT

Let the probability density function of a random variable $x$ be given as

f (x) = a e^{- 2∣ x ∣}

The value of ' $a$ ' is ________.

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Answer and solution

ANSWER

$a = 1$

Official answer: 0.99 to 1.01Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

Normalisation condition. Every PDF must integrate to 1 over the whole real line: $\int_{- \infty}^{\infty} f (x) d x = 1$ . So $a \int_{- \infty}^{\infty} e^{- 2∣ x ∣} d x = 1$ .
Use even symmetry. Since $∣ - x ∣ = ∣ x ∣$ , the function $e^{- 2∣ x ∣}$ is even, so $\int_{- \infty}^{\infty} e^{- 2∣ x ∣} d x = 2 \int_{0}^{\infty} e^{- 2∣ x ∣} d x$ . For $x \geq 0$ , $∣ x ∣ = x$ , so the integrand becomes $e^{- 2 x}$ .
Evaluate the integral. $2 \int_{0}^{\infty} e^{- 2 x} d x = 2 [\frac{e ^{- 2 x}}{- 2}]_{0}^{\infty} = 2 (0 - (- \frac{1}{2})) = 2 \times \frac{1}{2} = 1$
Solve for $a$ . $a \times 1 = 1 \Rightarrow a = 1$ .
Check. This is the Laplace (double-exponential) density $\frac{λ}{2} e^{- λ ∣ x ∣}$ with $λ = 2$ , whose constant is $\frac{λ}{2} = \frac{2}{2} = 1$ . This agrees.

KEY IDEA

A PDF must have total area 1. Because $e^{- 2∣ x ∣}$ is even, the area is twice the area over $x \geq 0$ , which is $2 \times \frac{1}{2} = 1$ . So $a = 1$ . In general, $a e^{- λ ∣ x ∣}$ is a valid PDF only when $a = λ /2$ .

WATCH OUT FOR

Integrating only from $0$ to $\infty$ and forgetting the negative half of the line gives $a /2 = 1$ , so $a = 2$ . That constant belongs to the one-sided exponential $2 e^{- 2 x}$ for $x \geq 0$ , not to this two-sided density. Another slip is dropping the absolute value and integrating $e^{- 2 x}$ over all of $R$ , which diverges.

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GATE 2022 EE Q.54: solution and answer

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