Question

If $f(x)$ is the probability density function of an exponential distribution and $\frac{1}{\lambda}$ is its mean, then $\int_0^\infty x^2 f(x) \, dx =$}

• A. $\frac{1}{\lambda^2}$
• B. $\frac{3}{\lambda^2}$
• C. $\frac{2}{\lambda^2}$
• D. None of these

Hint:

For an exponential distribution, the general formula for raw moments is $E[X^n] = \frac{n!}{\lambda^n}$. Thus, for $n=1$, $E[X] = \frac{1}{\lambda}$, and for $n=2$, $E[X^2] = \frac{2!}{\lambda^2} = \frac{2}{\lambda^2}$. Remembering this saves substantial calculation time!

Answer 1

The Correct Option is C