Question

If \(X\) is \(N\left(\mu, \sigma^{2}\right)\), show that \(E(|X-\mu|)=\sigma \sqrt{2 / \pi}\).

Short answer

For a normally distributed random variable X with mean μ and variance σ², it is indeed shown that the expected absolute deviation from the mean is \(\sigma\sqrt{2/\pi}\).

Step-by-step solution

Analysis of problem

Recognize that the problem is to prove that for a normally distributed random variable \(X\), with mean \(\mu\) and variance \(\sigma^2\), it holds that the expected absolute deviation from the mean is \(\sigma\sqrt{2/\pi}\). To solve the problem, we use the definition of the expected value and the given properties of the normal distribution.

Breaking down absolute deviation

We must remember that the absolute value |z| of a real number z can be broken down into two cases: z when z >= 0, and -z when z < 0. Thus, we decompose |X-\mu| into: \(X - \mu\) when \(X \geq \mu\) and \(\mu - X\) when \(X < \mu\). This allows us to write the expectation \(E(|X-\mu|)\) as a sum of two integrals. Note that both cases occur with probability 1/2 in a normally distributed random variable.

Evaluate the integrals

Now, E(|X-\mu|) can be broken down into two parts using the standard normal distribution \(Z\) = (X-\(\mu\))/\(\sigma\), which has mean 0 and variance 1. Thus, \(E(|X-\mu|)\) = \(\sigma\) * 2 * \(\int_{0}^{\infty}z * f(z) dz\), where \(f(z)\) is the PDF of the standard normal distribution. Substituting the formula for the PDF, we must evaluate \(\sqrt{2}\) * \(\int_{0}^{\infty}z * e^{-z^2/2} dz\) / \(\sqrt{\pi}\) using integration by substitution, namely \(z^2/2\) = u. The result is \(\sigma\sqrt{2/\pi}\)

Final Result

It is shown through this calculation that \(E(|X-\mu|) = \sigma\sqrt{2/\pi}\). The key step was breaking the absolute deviation into two cases that could be handled using properties of the normal distribution.