Question

The variance of random variable $X$ having density $f_{X}(x)=ce^{-|x|}, -\infty<x<\infty$ is

• A. $1$
• B. $1/2$
• C. $3/2$
• D. $2$

Hint:

For a Laplace distribution $f(x) = \frac{1}{2b}e^{-\frac{|x-\mu|}{b}}$, the variance is always $2b^2$. Here, $\mu=0$ and $b=1$, so $Var(X) = 2(1)^2 = 2$.

Answer 1

The Correct Option is D

The given density function $f_X(x) = ce^{-|x|}$ corresponds to the Laplace (Double Exponential) distribution centered at $0$.

Step 1: \color{redFind the Normalizing Constant c
The total area under the density must be 1:
$\int_{-\infty}^{\infty} ce^{-|x|} dx = 1$
Due to symmetry about 0:
$2c \int_{0}^{\infty} e^{-x} dx = 1$
$2c [-e^{-x}]_{0}^{\infty} = 1 \implies 2c(1) = 1 \implies c = 1/2$.

Step 2: \color{redIdentify Mean and Symmetry
The distribution is symmetric about $x=0$.
Therefore, $E(X) = 0$.
The variance is given by $Var(X) = E(X^2) - [E(X)]^2 = E(X^2)$.

Step 3:\color{red Calculate E(X^2)
$E(X^2) = \int_{-\infty}^{\infty} x^2 f_X(x) dx = \int_{-\infty}^{\infty} x^2 \frac{1}{2} e^{-|x|} dx$.
Using symmetry again:
$E(X^2) = 2 \cdot \frac{1}{2} \int_{0}^{\infty} x^2 e^{-x} dx = \int_{0}^{\infty} x^2 e^{-x} dx$.

Step 4: \color{redEvaluate the Integral using Gamma Function
The integral $\int_{0}^{\infty} x^{n} e^{-ax} dx$ is equal to $\frac{n!}{a^{n+1}}$.
For $n=2$ and $a=1$:
$E(X^2) = \frac{2!}{1^{2+1}} = 2$.
Since $E(X)=0$, $Var(X) = E(X^2) = 2$.