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Chapter 3: Problem 6

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

Short Answer

Expert verified
The expected value of function \(|X - \mu|\) is \(\sigma \sqrt{2 / \pi}\).

Step by step solution

01

Define the Expected Value

The expected value, or mean, of a continuous random variable is given by the integral \(E[f(X)] = \int f(x)f_{X}(x) dx\), where \(f_{X}(x)\) is the probability density function of \(X\). In this case, \(f(x) = |x - \mu|\) and \(f_{X}(x)\) is the normal density function \(\frac{1}{\sqrt{2\pi\sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)}\).
02

Apply Absolute Value Property

Because of the absolute value, the integral will split into two parts, one for \(x \geq \mu\) and one for \(x < \mu\). Now need to calculate each of them, then sum the two results.
03

Formulate Integral for \(x \geq \mu\)

The integral for \(x \geq \mu\) is \(\int_{\mu}^{\infty} (x-\mu) \cdot \frac{1}{\sqrt{2\pi\sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)} dx\).
04

Formulate Integral for \(x < \mu\)

The integral for \(x < \mu\) is \(\int_{-\infty}^{\mu} (\mu-x) \cdot \frac{1}{\sqrt{2\pi\sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)} dx\).
05

Application of Standard Normal Distribution

Standardize the normal distribution to \(Z = \frac{(x - \mu)}{\sigma}\), for both parts. Then the integrals become \(\int_{0}^{\infty} z \cdot \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz\).
06

Compute the Integral

Computing the integral, you will get the result as \(\sqrt{2/\pi}\).
07

Combine the Results

Since both the integrals for \(x \geq \mu and x < \mu\) are equal, combine them to get the expected value as \(2 \cdot \sigma \cdot \sqrt{2 / \pi}\), which simplifies further to \(\sigma \sqrt{2 / \pi}\).

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