I have found the distance between the MAX and MIN of 2 random variables in a standard normal distribution.
$\text{Distance}=\mathbb{E}|X_1 – X_2|$, where $X$ has a mean of $\mu$ and a variance of $\sigma^2$.
$$X_1-X_2 \sim N(0,2σ^2) \sim {\sqrt{2}\sigma Z}\,,$$
where $Z \sim N(0,1)$.
$$\text{Distance}=\sqrt{2}\sigma\mathbb{E}|Z|$$
Then using LOTUS,
\begin{align}
\sqrt{2}\sigma\mathbb{E}(Z) &= \sqrt{2}\sigma2\int_{0}^{\infty}Z\frac{e^{-\frac{Z^2}{2}}}{\sqrt{2\pi}}dZ
\\Distance &=\frac{\sqrt{2}\sigma2}{\sqrt{2\pi}}(1)
\\Distance &=\frac{2\sigma}{\sqrt{\pi}}
\end{align}
However, I have not been able to find the distance between the Max and MIN of 3 random variables $X_1, X_2, X_3$. I believe I have the solution, but I do not know how to mathematically prove it. I believe the answer is $$\text{Distance} = \frac{3\sigma}{\sqrt{\pi}} = \frac{3\mathbb{E(x-\mu)}}{\sqrt{2}}$$
Does anyone have a proof?
Best Answer
I believe what you are asking for is the expected value of the sample range $$R=\max(X_1,X_2,X_3)-\min(X_1,X_2,X_3)=\max_{i,j}|X_i-X_j|\,,$$ where $X_1,X_2,X_3$ are i.i.d $N(0,1)$.
Since $|X_i-X_j|$ is the distance between $X_i$ and $X_j$ for all $i\ne j$, just note that
$$2R=|X_1-X_2|+|X_2-X_3|+|X_1-X_3|$$
And as $|X_1-X_2|,|X_2-X_3|$ and $|X_1-X_3|$ are all identically distributed, you have $$2\,\mathbb E(R)=3\,\mathbb E\left[|X_1-X_2|\right]=3\times\frac{2}{\sqrt\pi}$$
That is, $$\mathbb E(R)=\frac{3}{\sqrt\pi}$$