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 $\omega \in \Omega$,

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


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


$$ \min_{\theta \in \mathbb{R}} f_{\theta}$$ 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 $L^1$.

Does anyone have any advice?

  • $\begingroup$ 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. $\endgroup$ – Giuseppe Negro Aug 30 '19 at 9:25

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


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 $$ \int \min_n f_n (x) \, dx= \int \sum_n f_n(x) \, dx = -N.$$

  • $\begingroup$ You can also modify this example so that all $h_n$ are supported in $[0,1]$. $\endgroup$ – Giuseppe Negro Aug 30 '19 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 $$\int \min_{n\in \mathbb{N}} f_n = \min_{n\in \mathbb{N}} \int f_n$$

  • $\begingroup$ 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 $\endgroup$ – Ceeerson Aug 28 '19 at 2:09
  • $\begingroup$ 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. $\endgroup$ – Joe Sjoberg Aug 28 '19 at 14:14
  • $\begingroup$ 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. $\endgroup$ – Ceeerson Aug 28 '19 at 22:27

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