$\mathbb{E}[\min (X_{1:n}) + \max(X_{1:n})]/2 = \mathbb{E}[\text{median}(X_{1:n})]$? Say $X$ is continuous random variable, and we have $n$ iid samples, denoted as $X_{1:n}$. Then can we say the following
$$\mathbb{E}[\min (X_{1:n}) + \max(X_{1:n})]/2 = \mathbb{E}[\mathrm{median}(X_{1:n})]$$
I try with the convolution of two and start to take the expectation of it, but this seems exhausting. 
 A: Another way to disprove this equality is to consider the Cauchy distribution, since $\mathbb{E}[\min (X_{1:n})]$ and $\mathbb{E}[\max (X_{1:n})]$ are not defined for this distribution, while $\mathbb{E}[\text{median}(X_{1:n})]$ is defined for $n\ge 3$.
However, the equality holds when the distribution of $X$ is symmetric around a value $\mu$. i.e., when $X-\mu$ and $\mu-X$ have the same distribution, since
$$\mathbb{E}[\min (X_{1:n})]-\mu=\mathbb{E}[\min (X_{1:n}-\mu)]=\mathbb{E}[-\max (X_{1:n}-\mu)]=-\mathbb{E}[-\max (X_{1:n})]+\mu$$
Therefore
$$\mathbb{E}[\min (X_{1:n}) + \max(X_{1:n})]/2 = \mu = \mathbb{E}[\mathrm{median}(X_{1:n})]$$
A: No. Rather than prove otherwise, I wrote some quick code in R to demonstrate this is false.
I generate samples from $X \sim exp(1)$ which has median $\log(2)$. Observe
minX <- rep(NA,5000)
maxX <- rep(NA,5000)
medX <- rep(NA,5000)

for(i in 1:5000){
  X <- rexp(10000,1)
  minX[i] <- min(X)
  maxX[i] <- max(X)
  medX[i] <- median(X)
}

mean(maxX)+mean(minX);
mean(maxX)-mean(minX);
mean(medX); log(2)

> mean(maxX)+mean(minX);
[1] 9.792494
> mean(maxX)-mean(minX);
[1] 9.792292
> mean(medX); log(2)
[1] 0.6931122
[1] 0.6931472

As you can see, the two are clearly not equal. If they were, the WLLN would have ensured that this simulation would reflect this.
Sometimes it's best to just simulate before getting bogged down into deriving something that might be false.
