5
$\begingroup$

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 $L^1$.

Does anyone have any advice?

$\endgroup$
  • $\begingroup$ 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 at 9:25
1
$\begingroup$

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

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

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 $$\int \min_{n\in \mathbb{N}} f_n = \min_{n\in \mathbb{N}} \int f_n$$

$\endgroup$
  • $\begingroup$ 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 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 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 at 22:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.