Question

Let λ > 0 be a constant. Consider the function f(x) = (λ/2)e^(−λ|x|) , for − [infinity] < x < [infinity] . (a) Is f(x) a pdf? Why? (b) If your answer is "yes" for part (a), find the corresponding µ and σ^2 (as simple functions of λ). If your answer is "no" for part (a), find the integral (from -[infinity] to [infinity]) of xf(x)dx and the integral(from -[infinity] to [infinity]) of x^2f(x)dx (as simple functions of λ).

Answer 1

a. In order for $f$ to be a PDF, $f$ needs to be non-negative (it is) and the integral over its support must evaluate to 1. $f$ is symmetric about $x=0$, i.e. even, so

$\displaystyle\int_{-\infty}^\infty\frac\lambda2e^{-\lambda|x|}\,\mathrm dx=\lambda\int_0^\infty e^{-\lambda x}\,\mathrm dx=-e^{-\lambda x}\bigg|_0^\infty=1$

and so $f$ is indeed a PDF.

b. Let $X$ be a random variable with $f$ as its PDF. Then

$\mu=E[X]=\displaystyle\int_{-\infty}^\infty xf(x)\,\mathrm dx=\frac\lambda2\int_{-\infty}^\infty xe^{-\lambda|x|}\,\mathrm dx$

The integrand is odd, so the integral vanishes and the mean is $\boxed{\mu=0}$.

The variance of $X$ is

$\sigma^2=E[(X-E[X])^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f(x)\,\mathrm dx$

This integrand is even, so

$E[X^2]=\displaystyle2\int_0^\infty x^2f(x)\,\mathrm dx=\lambda\int_0^\infty x^2e^{-\lambda x}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=e^{-\lambda x}\,\mathrm dx\implies v=-\dfrac{e^{-\lambda x}}\lambda$

so that

$\displaystyle E[X^2]=\lambda\left(-\frac{x^2e^{-\lambda x}}\lambda\bigg|_0^\infty+2\int_0^\infty\frac{xe^{-\lambda x}}\lambda\,\mathrm dx\right)=2\int_0^\infty xe^{-\lambda x}\,\mathrm dx$

Integrate by parts again, this time with

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=e^{-\lambda x}\,\mathrm dx\implies v=-\dfrac{e^{-\lambda x}}\lambda$

$\displaystyle E[X^2]=2\left(-\frac{xe^{-\lambda x}}\lambda\bigg|_0^\infty+\int_0^\infty\frac{e^{-\lambda x}}\lambda\,\mathrm dx\right)=\frac2\lambda\int_0^\infty e^{-\lambda x}\,\mathrm dx=-\frac2{\lambda^2}e^{-\lambda x}\bigg|_0^\infty=\frac2{\lambda^2}$

and so the variance is $\boxed{\sigma^2=\frac2{\lambda^2}}$.