# passing an integral to the inside of a min

I was wondering when I could do something like the following:

$$\int \min\limits_{n\in\mathbb{N}}f_n = \min_{n\in \mathbb{N}} \int f_n$$ when I don't necessarily have a decreasing sequence of functions.

The problem I have is the following:

Imagine we have some really nice random variables $$(X_i)_{i=1}^n \stackrel{iid}{\sim} X$$ in the sense that they are $$L^1$$ and their cdfs are bijective and continuous and perhaps some other properties that maybe I need. Then I want to show that

$$\min_{\theta \in \mathbb{R}} \bigg( \theta + \frac{1}{n(1-\epsilon)} \sum_{i=1}^n \max (-X_i - \theta, 0)\bigg)$$ is an unbiased estimator for

$$\min_{\theta \in \mathbb{R}} \bigg( \theta + \frac{1}{(1-\epsilon)} \mathbb{E}[\max (-X - \theta, 0)] \bigg)$$

I would like to just pass the Expectation from the outside to the inside, but I'm having some trouble showing how to do this. It has been a while since I've taken real analysis.

So far, it seems that for each $$\omega \in \Omega$$,

$$f_{\theta}(\omega) := \theta + \frac{1}{n(1-\epsilon)} \sum_{i=1}^n \max (-X_i(\omega) - \theta, 0)$$ is continuous and convex in $$\theta$$ with negative derivative as $$\theta \rightarrow -\infty$$ and positive derivative as $$\theta \rightarrow \infty$$. So then for each $$\omega \in \Omega$$,

$$\inf_{\theta \in \mathbb{R}} f_{\theta}(\omega) = \min_{\theta \in \mathbb{R}} f_{\theta}(\omega)$$

and

$$\min_{\theta \in \mathbb{R}} f_{\theta}(\omega) = \min_{\theta \in \mathbb{Q}} f_{\theta} (\omega)$$

Hence

$$\min_{\theta \in \mathbb{R}} f_{\theta}$$ is measurable.

All I can think to use would be the Dominated Convergence Theorem, but I'm not sure if I can come up with any sequence that converges to my estimator pointwise which is also bounded in $$L^1$$.

• As you said, the key is dominated convergence. Take your favorite example showing that $\lim \int f_n$ can differ from $\int \lim f_n$, and chances are that it will be a counterexample to $\min \int f_n=\int \min f_n$, too. However, if $f_n$ is dominated by some integrable function, I think that your claim is true. – Giuseppe Negro Aug 30 at 9:25

The proposition is false without additional assumptions on $$f_n$$.

## Counterexample

Let $$h$$ be the hat function on $$[0,1]$$, i.e. the function $$h(x) = \max\left( 1 - |2x-1|,0 \right).$$ Notice that $$h$$ is zero outside of $$[0,1]$$ and $$\int h(x) \,d x=1.$$

Let $$f_n$$ be the negative translate of $$h$$, i.e. $$f_n(x)=-h(x-n)$$ Since those translates have disjoint support (i.e. they do not overlap) $$\min_n f_n = \sum_n f_n.$$ Now if you take the $$\min$$ over $$N$$ of those translates $$\min_n \int f_n(x)\, dx = \min_n (-1) = -1.$$ but $$\int \min_n f_n (x) \, dx= \int \sum_n f_n(x) \, dx = -N.$$

• You can also modify this example so that all $h_n$ are supported in $[0,1]$. – Giuseppe Negro Aug 30 at 9:23

It is true that if you have two functions $$f,g$$ with $$f \le g$$, then $$\int f \le \int g$$. Would this not imply equality if there truly is a $$\min$$ of the functions? However, if you were to replace the minimum with $$\inf$$, I am unsure whether the equality still holds.

Edit: I can expound. If we let $$f_0 = \min f_n$$, then it is true that $$\int \min_{k\in \mathbb{N}} f_k=\int f_0 \le \int f_n$$ for all $$n\in \mathbb{N}$$. So $$\int f_0 = \min_{n\in \mathbb{N}} \int f_n$$ Thus implying that $$\int \min_{n\in \mathbb{N}} f_n = \min_{n\in \mathbb{N}} \int f_n$$

• Well, I'm not quite sure how to show that $\min f_n \in L^1$, but assuming that is true, then I agree that $\int \min f_n \leq \min \int f_n$, but I am having trouble seeing the equality – Ceeerson Aug 28 at 2:09
• I assumed that $f_n \in L^1$ for all $n$. However I may have misunderstood the question in how $\min_{n\in \mathbb{N}} f_n$ was defined. I assumed that $\min_{n\in \mathbb{N}} f_n = f$ for some $f\in \{f_n\}$, where $f$ has the property that $f(x) \leq f_n(x)$ for all $n$. However, maybe it is supposed to be defined pointwise, like this: If $f = \min_{n\in \mathbb{N}} f_n$, then $f(x) = \min_{n\in \mathbb{N}} f_n(x)$. Whatever the intent of the original question, I misinterpreted what was asked. I'm only keeping my answer up so others can see what I initially thought. – Joe Sjoberg Aug 28 at 14:14
• oooh, yeah I see now, yes the min function is defined pointwise, but now that you mentioned your assumption, I am thinking that I can never pass the limit on the inside unless the functions are decreasing or the pointwise min function is in the set of functions considered like you assumed. – Ceeerson Aug 28 at 22:27