Determine PDF Constant 'a'

The given probability density function (PDF) is $f(x) = e^{-a|x|}$. For $f(x)$ to be a valid PDF, its integral over its domain must equal 1:

$ \int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{\infty} e^{-a|x|} dx = 1 $

Since $e^{-a|x|}$ is an even function ($f(-x) = f(x)$), the integral can be written as:

$ 2 \int_{0}^{\infty} e^{-ax} dx = 1 $

For the integral to converge, we require $a > 0$. Evaluating the integral:

$ 2 \left[ \frac{e^{-ax}}{-a} \right]_{0}^{\infty} = 1 $

$ 2 \left( 0 - \frac{e^0}{-a} \right) = 1 \implies 2 \left( \frac{1}{a} \right) = 1 $

This yields $ a = 2 $. The PDF is therefore $f(x) = e^{-2|x|}$.

Calculate Expected Value E[x]

The expected value $E[x]$ is given by:

$ E[x] = \int_{-\infty}^{\infty} x f(x) dx = \int_{-\infty}^{\infty} x e^{-2|x|} dx $

The function $g(x) = x e^{-2|x|}$ is an odd function ($g(-x) = -g(x)$). The integral of an odd function over a symmetric interval $(-\infty, \infty)$ is always 0.

$ E[x] = 0 $

Calculate Expected Value E[x^2]

The second moment $E[x^2]$ is calculated as:

$ E[x^2] = \int_{-\infty}^{\infty} x^2 f(x) dx = \int_{-\infty}^{\infty} x^2 e^{-2|x|} dx $

Since $h(x) = x^2 e^{-2|x|}$ is an even function ($h(-x) = h(x)$), we simplify:

$ E[x^2] = 2 \int_{0}^{\infty} x^2 e^{-2x} dx $

This integral can be solved using the standard formula $ \int_0^\infty x^n e^{-kx} dx = \frac{n!}{k^{n+1}} $. Here, $n=2$ and $k=2$.

$ \int_{0}^{\infty} x^2 e^{-2x} dx = \frac{2!}{2^{2+1}} = \frac{2}{2^3} = \frac{2}{8} = \frac{1}{4} $

Therefore, $ E[x^2] = 2 \times \frac{1}{4} = \frac{1}{2} $.

Compute Variance Var(x)

The variance of a random variable $x$ is defined as $ Var(x) = E[x^2] - (E[x])^2 $.

Substituting the computed values:

$ Var(x) = \frac{1}{2} - (0)^2 $

$ Var(x) = \frac{1}{2} = 0.5 $

The variance is 0.5.