Question:medium

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

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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 $Var(X) = 2(1)^2 = 2$.
Updated On: Jun 6, 2026
  • $1$
  • $1/2$
  • $3/2$
  • $2$
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The Correct Option is D

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