Does $\mathbb{E}[\max_{1\leq i,j \leq N}{|Y_i-Y_j|}] = 2\mathbb{E}[\max_{1\leq i,j \leq N} Y_i]$ where $Y_i \sim N(0,1)$?

Question

My lecturer argued that $$\mathbb{E}[\max_{1\leq i,j \leq N}{|Y_i-Y_j|}] = \mathbb{E}[\max_{1\leq i,j \leq N}{(Y_i-Y_j)}] = 2\mathbb{E}[\max_{1\leq i\leq N} Y_i]$$ where $Y_i \sim N(0,1)$ and claimed this was true via symmetry of $Y_i$. However I am unsure if the second equality holds since it appears he is replacing $-Y_j$ by $Y_j$, but surely this would affect the $Y_i$ in the formula?

Any help on seeing why this is true would be much appreciated.

Answer 1

Assuming $Y_1,Y_2,\ldots,Y_N$ are i.i.d $N(0,1)$, so that distribution of $Y_i$ is symmetric about $0$ for each $i$. In other words, $Y_i$ and $-Y_i$ have the same distribution for every $i$: $$Y_i\stackrel{d}{=}-Y_i\quad,\,i=1,2,\ldots,N$$

Therefore by symmetry, $$\max_{1\le i\le N}Y_i\stackrel{d}{=} -\min_{1\le i\le N} Y_i$$

And since $$\max_{1\le i,j\le N}|Y_i-Y_j|=\max_{1\le i\le N}Y_i-\min_{1\le i\le N}Y_i\,,$$

you have

\begin{align} \mathbb E\left[\max_{1\le i,j\le N}|Y_i-Y_j|\right]&=\mathbb E\left[\max_{1\le i\le N}Y_i\right]-\mathbb E\left[\min_{1\le i\le N}Y_i\right] \\&=2\mathbb E\left[\max_{1\le i\le N}Y_i\right] \end{align}