GATE CSE 2026 | Set 2 | Question: 4

Question

The probability density function $f(x)$ of a random variable $X$ which takes real values is
\[
f(x)=\frac{1}{3 \sqrt{2 \pi}} \exp \left(-\frac{x^{2}}{18}\right), \quad x \in(-\infty,+\infty)
\]
Which one of the following statements is correct about the random variable $X$ ?

• A. $X$ is an exponential random variable
• B. $X$ is a normal random variable
• C. $X$ is a Poisson random variable
• D. $X$ is a uniform random variable

Comments

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$μ = 0, σ = 3$

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Support of Discrete random variable is defined as set of all real numbers such that probability that random variable takes that value is positive. That is 

$$\text{Support}(X) = \set{x \in \mathrm{R} \vert P(X = x) > 0}$$. 

Similary, For a continous random variable, support is defined as set of real numbers such that the probability density at that point is positive. That is 

$$\text{Support}(X) = \set{x \in \mathrm{R} \vert f_X(x)> 0}$$. 

We have given a continous random variable $X$, so it erases the option $C$. Then 

• Then support of exponential random variable is $(0, \infty)$, but the support of $X$ is any real number. so option $A$ is not the answer.
• The support of uniform random variable is closed interval $[a, b]$, that is support of uniform is never going to whole real line, so $D$ is not the answer.

So we are left with option $B$. that is the distribution of $X$ follows Normal distribution and indeed $X \sim N(0, 9)$. So option $B$ is the answer.

Answer 1

Poisson is Discrete, eliminates Poisson immediately.
Now check Domain:

• Exponential PDF: f(x) = λe^(-λx) for x ≥ 0 → eliminates Exponential (question says x ∈ (−∞, +∞))
• Uniform: defined on (a, b) → finite interval → eliminates Uniform
• Normal: only distribution that lives on (−∞, +∞)

Answer: Option B

Answer 2

Option-A: PDF for Exponential Distribution is $e^{-\lambda x}\lambda$ for $x \ge 0$.

 

Option-B: PDF for Normal Distribution is $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} 
\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$ where $\mu$ is mean and $\sigma^2$ is variance. 

 

Option-C: PMF for Poisson Distribution is $\frac{\lambda^ke^{-\lambda}}{k!}$ where $k \in \mathbb{N}_{0}$  and $\lambda \in (0,\infty)$

 

Option-D: PDF for Uniform Distribution is $f(x) =
\begin{cases}
\frac{1}{b-a}, & a \le x \le b \\
0, & \text{otherwise}
\end{cases}$

Given PDF follows Normal distribution with mean = 0 and variance = 9

Hence, Option B is correct.