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

Concept: The probability density function (pdf) of an exponential distribution with parameter $\lambda$ (where mean $\mu = \frac{1}{\lambda}$) is given by: \[ f(x) = \begin{cases} \lambda e^{-\lambda x}, & x \ge 0 \\ 0, & x < 0 \end{cases} \] The expression we need to evaluate is $\int_0^\infty x^2 f(x) \, dx$, which by definition represents the second raw moment of the distribution, i.e., $E[X^2]$. We can solve this either by standard integration by parts or by using the Gamma function definition: \[ \Gamma(n) = \int_0^\infty entries^{n-1} e^{-t} \, dt = (n-1)! \]

Step 1: Substitute $f(x)$ into the integral.
\[ I = \int_0^\infty x^2 \left( \lambda e^{-\lambda x} \right) dx = \lambda \int_0^\infty x^2 e^{-\lambda x} \, dx \]

Step 2: Apply substitution to transform into a Gamma integral.
Let $\lambda x = t \implies x = \frac{t}{\lambda} \implies dx = \frac{1}{\lambda} dt$. As $x \to 0$, $t \to 0$, and as $x \to \infty$, $t \to \infty$. Substituting these into the integral: \[ I = \lambda \int_0^\infty \left( \frac{t}{\lambda} \right)^2 e^{-t} \left( \frac{1}{\lambda} \, dt \right) \] \[ = \lambda \cdot \frac{1}{\lambda^3} \int_0^\infty t^2 e^{-t} \, dt = \frac{1}{\lambda^2} \int_0^\infty t^{3-1} e^{-t} \, dt \]

Step 3: Evaluate using the Gamma function.
The integral $\int_0^\infty t^{3-1} e^{-t} \, dt$ is exactly $\Gamma(3)$. Since $\Gamma(3) = 2! = 2 \times 1 = 2$: \[ I = \frac{1}{\lambda^2} \cdot 2 = \frac{2}{\lambda^2} \]