What is the expected absolute difference between sample and population mean?

Question

Intro

A sample mean is an unbiased estimator of the population mean. In other words, the expected difference between the population mean and the sample mean is zero regardless of the population distribution. In other words $E[\bar x - x_p]=0$, where $\bar x$ and $x_p$ are the sample and population mean, respectively.

Question

Given that the population is normally distributed with variance $\sigma^2$ and knowing the sample size $n$, what is the expected absolute difference between the population mean and the sample mean?

or in mathematical form:

$$E[\space| \bar x - x_p |\space] = \space ?$$

The vertical lines stands for "absolute value"

Answer 1

This is an addendum to @Aksakal's answer. As he points out, we need to find the value of $E|Y|]$ where $Y \sim \mathcal N(0,\sigma^2/n)$. This can be done very straightforwadly via the law of the unconscious statistician, without needing to think of $\chi$ random variables etc. We have \begin{align} E[|Y|] &= \int_{-\infty}^\infty |y|\cdot\frac{1}{(\sigma/\sqrt{n})\sqrt{2\pi}} \exp\left(-\frac{y^2}{2\sigma^2/n}\right)\,\mathrm dy\\ &= 2\int_{0}^\infty y\cdot\frac{1}{(\sigma/\sqrt{n})\sqrt{2\pi}} \exp\left(-\frac{y^2}{2\sigma^2/n}\right)\,\mathrm dy\\ &= \sqrt{\frac{2}{\pi}}\cdot \frac{\sigma}{\sqrt{n}}\int_{0}^\infty \frac{y}{\sigma^2/n}\cdot \exp\left(-\frac{y^2}{2\sigma^2/n}\right)\,\mathrm dy\\ &= \sqrt{\frac{2}{\pi}}\cdot \frac{\sigma}{\sqrt{n}}~\left[-\exp\left(-\frac{y^2}{2\sigma^2/n}\right)\right|_0^\infty\\ &= \sqrt{\frac{2}{n\pi}}\cdot\sigma \end{align}

Answer 2

The sample mean is going to be normal since the underlying distribution is normal. The distribution of a sample mean is $\mathcal{N}(\mu,\sigma^2/n)$.

It's easy to compute the expectation of the absolute deviation then:

$$\bar x-\mu\sim\mathcal{N}(0,\sigma^2/n)$$

All you need is the expectation of absolute value of a normal. A distribution of the absolute value of a normal distribution is called "folded normal". In our case the underlying normal (of the deviation from population) has mean zero, hence it reduces to a $\chi$ distribution with degrees of freedom 1. You can find the formulas anywhere: $$\sigma\sqrt{\frac{2}{n\pi}}$$

Answer 3

Consider a normal random variable $Y$ with mean $\mu$ and variance $\tau^2$, and let $Z=\frac{Y-\mu}{\tau}$ (so $Z$ is standard normal).

$$\:\:E(|Z|)=2\int_0^\infty z\cdot \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} dz$$

$\quad$ Let $u=\frac{z^2}{2}$, so $du=z \,dz$.

$$\qquad=\frac{2}{\sqrt{2\pi}}\int_0^\infty e^{-u} du$$

$$=\sqrt{\frac{2}{\pi}}\qquad\quad$$

Hence $E(|Y-\mu|)=\tau E(|Z|)=\tau\sqrt{\frac{2}{\pi}}$.

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Let $X\sim N(\mu,\sigma^2)$. Let $Y=\bar{X}$. Then $\tau=\sigma/\sqrt{n}$.

Hence $E(|\bar{X}-\mu|)=E(|Y-\mu|)=\tau\sqrt{\frac{2}{\pi}}=\sigma\sqrt{\frac{2}{n\pi}}\quad$ ($\approx 0.8 \frac{\sigma}{\sqrt{n}}$)