GATE 2024 DA Q.57: solution and answer

GATE 2024 · DAQ.57 · 2 marks · NAT

Let $X$ be a random variable exponentially distributed with parameter $λ > 0$ . The probability density function of $X$ is given by:

f_{X} (x) = {λ e^{- λ x}, 0, x \geq 0 otherwise

If $5 E (X) = Var (X)$ , where $E (X)$ and $Var (X)$ indicate the expectation and variance of $X$ , respectively, the value of $λ$ is ______ (rounded off to one decimal place).

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Answer and solution

ANSWER

$λ = 0.2$

Official answer: 0.2Source: Official answer key, GATE 2024 DA (IISc Bengaluru)

For an exponential random variable with density $f_{X} (x) = λ e^{- λ x}$ for $x \geq 0$ , the mean is $E (X) = \int_{0}^{\infty} x λ e^{- λ x} d x = \frac{1}{λ}$ (integration by parts).
The second moment is $E (X^{2}) = \int_{0}^{\infty} x^{2} λ e^{- λ x} d x = \frac{2}{λ ^{2}}$ , so $Var (X) = E (X^{2}) - [E (X)]^{2} = \frac{2}{λ ^{2}} - \frac{1}{λ ^{2}} = \frac{1}{λ ^{2}}$ .
Impose the condition $5 E (X) = Var (X)$ , which gives $\frac{5}{λ} = \frac{1}{λ ^{2}}$ .
Multiply both sides by $λ^{2}$ (allowed since $λ > 0$ ): $5 λ = 1$ , so $λ = \frac{1}{5} = 0.2$ .

KEY IDEA

For an exponential distribution with rate $λ$ , the mean is $1/ λ$ and the variance is $1/ λ^{2}$ , so the condition $5 E (X) = Var (X)$ reduces to $5/ λ = 1/ λ^{2}$ .

WATCH OUT FOR

Using the mean and variance of an exponential written in terms of the scale (mean) parameter, or taking the variance as $1/ λ$ instead of $1/ λ^{2}$ . That gives $5/ λ = 1/ λ$ , which has no solution.

From the GATE 2024 DA paper. Every question of this paper · Download the full paper · All 10 Continuous distributions questions, with notes

GATE 2024 DA Q.57: solution and answer

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