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

Hint:

Compare the given density with the standard normal form 1/(sigma*sqrt(2*pi)) exp(-x^2/(2*sigma^2)); it matches with mean 0 and sigma = 3.

Answer 1

The Correct Option is B

Step 1: Recall the standard form of a normal (Gaussian) density.
A normal random variable with mean \(\mu\) and variance \(\sigma^2\) has probability density function \[ f(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right), \quad x \in (-\infty, +\infty) \] defined over the entire real line, and it is bell-shaped and symmetric about \(\mu\).

Step 2: Compare the given density with this standard form.
The given density is \[ f(x) = \frac{1}{3\sqrt{2\pi}} \exp\left(-\frac{x^2}{18}\right), \quad x \in (-\infty, +\infty) \] Matching the exponent, \(x^2\) in the numerator (instead of \((x-\mu)^2\)) means \(\mu = 0\), and matching denominators, \(2\sigma^2 = 18\), so \(\sigma^2 = 9\) and \(\sigma = 3\). Matching the leading constant, \(\frac{1}{\sigma\sqrt{2\pi}} = \frac{1}{3\sqrt{2\pi}}\) checks out exactly with \(\sigma = 3\).

Step 3: Confirm the domain matches a normal distribution.
The given \(f(x)\) is defined for all real \(x\) from \(-\infty\) to \(+\infty\), just like a normal distribution, which is the only one of the four listed distributions whose support is the entire real line.

Step 4: Rule out the other three distributions using their supports and shapes.
An exponential random variable is only defined for \(x \geq 0\) and its density is \(\lambda e^{-\lambda x}\), which decays monotonically from its peak at \(x=0\), it is never symmetric and never defined for negative \(x\), unlike the given \(f(x)\) which is symmetric and defined for all real \(x\). A Poisson random variable is discrete, taking only nonnegative integer values \(0, 1, 2, \dots\) with a probability mass function, not a continuous density over the real line at all, so it cannot match a continuous \(f(x)\) like this one. A uniform random variable has a constant, flat density over some finite interval \([a,b]\) and zero density outside it, but the given \(f(x)\) involves an exponential term that varies continuously with \(x\) and never actually reaches zero, so it cannot be uniform.

Step 5: Conclusion.
The given density exactly matches the normal distribution formula with mean 0 and standard deviation 3, and its bell-shaped, symmetric, real-line-supported form rules out exponential, Poisson, and uniform.
\[ \boxed{X \text{ is a normal random variable}} \]