Does expectation of max - expectation of min always equal expectation of max - min?

Question

Suppose $X_1,X_2,\dots,X_N$ are iid.

Does $E[X_{max}]-E[X_{min}]$ equal $E[X_{max}-X_{min}]$ for any distribution?

For uniform, is does.

Is there a counterexample?

Answer 1

The expectation operator is linear, hence it commutes with subtraction. For any random variables,

$$E(X-Y)=E(X)-E(Y),$$

provided these expectations exist.

Answer 2

To repeat what I said in the comments, even if $X_{\min}$ and $X_{\max}$ are not independent, this does not affect the sum of their expectations, because the expectation is always linear.

That is, for any distribution and $X_1,\ldots,X_n $ iid distributed w.r.t it, the equation $$E[X_{\max} - X_{\min}] = E[X_\max] - E[X_\min]$$ will hold true, as long as the expectations are defined and finite, of course.

There is a catch, however : the left hand side could be defined without the right hand side being defined. However, the right hand side being well defined implies the LHS being well defined and equal to the RHS.

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EDIT : If $X_1$ is integrable, then so is $|X_{\max}|$ because $|X_{\max}| \leq N |X_1|$ , for example. Similarly for the minimum. Therefore, the $X_i$ being integrable implies the equality at least. Any counterexamples must come from non-integrable $X_i$ then.

Answer 3

Note that $|X_{max}| \le \sum_{i=1}^n |X_i|$ and similarly for $|X_{min}|$, so if $E(X_i)$ exists so do $E(X_{max})$ and $E(X_{min})$, and then $E(X_{max} - X_{min}) = E(X_{max}) - E(X_{min})$.

On the other hand, suppose $E(X_i)$ does not exist and $n> 1$. Note that

$$|X_{max} - X_{min}| \ge |X_n| - (|\min(X_1,\ldots, X_{n-1}| + |\max(X_1,\ldots, X_{n-1})|)$$ The conditional expectation of the right side given $X_1, \ldots, X_{n-1}$ is $+\infty$, therefore so is the conditional expectation of the left. Therefore $E(X_{max}-X_{min})$ does not exist.