# Uniform convergence and exchanging limit and integral

I have to perform the following operation $$\lim_{n\to\infty} \int_{-\infty}^{+\infty} \Phi(x+n) \phi(x) dx$$ where $$\Phi$$ and $$\phi$$ are the standard normal CDF and PDF respectively.

From what I understand, a sufficient condition to exchange limit and integral is that the integrand converges uniformly to its limit, which we know to be $$\lim_{n\to\infty} \Phi(x+n) \phi(x) = \phi(x)$$. Here, I don't think we have uniform convergence (but corrections are welcome as my understanding is still lacking), and one way to check it is to see that $$\sup_x |\Phi(x+n)\phi(x)-\phi(x)| \not\rightarrow_{n\to\infty}0$$.

However, solving the integral (using properties of standard normal integrals) yields $$\int_{-\infty}^{+\infty} \Phi(x+n) \phi(x) dx = \Phi(n)$$ and so it turns out that $$\lim_{n\to\infty} \int_{-\infty}^{+\infty} \Phi(x+n) \phi(x) dx = \int_{-\infty}^{+\infty} \lim_{n\to\infty} \Phi(x+n) \phi(x) dx = 1,$$ so we can exchange limit and integral.

What am I missing? I am trying to understand the theory behind, because I will have to deal with more general functions than $$\Phi(x+n)$$, all of which have limits in this sort of additive way.

• There is no simple answer to "when are you allowed to change the order of limit and integration?". All we have are a collection of sufficient conditions. One of these sufficient conditions is the dominated convergence theorem, which will work in this case. Sep 2, 2021 at 15:41
• As you said, that's a sufficient condition. Not a necessary a sufficient condition. Sep 2, 2021 at 15:51

If $$f_n(x)\xrightarrow[n\to\infty]{}f(x)$$ uniformly, then \begin{align*} \left\lvert\int_a^bf_n(x)\,\mathrm dx-\int_a^bf(x)\,\mathrm dx\right\rvert&\le\int_a^b\lvert f_n(x)-f(x)\rvert\,\mathrm dx\\&\le\int_a^b\|f_n-f\|_\infty\,\mathrm dx\\&=(b-a)\|f_n-f\|_\infty\\&\xrightarrow[n\to\infty]{}0.\end{align*} However, this is obviously assuming that $$(a,b)$$ is a finite interval, i.e., both $$a$$ and $$b$$ are finite.
Here you are integrating over $$(-\infty,\infty)$$. Nonetheless, you can use monotone convergence, as $$(\Phi(\cdot+n))_{n\ge0}$$ is non-decreasing (towards the constant function equals to $$1$$).
You can also use local uniform convergence as follows. On the one hand, $$\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\le\int_{-\infty}^\infty\phi(x)\,\mathrm dx=1,$$ so $$\limsup_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\le1.$$ On the other hand, for each fixed $$N>0$$, by the uniform convergence $$\Phi(x+n)\phi(x)\xrightarrow[n\to\infty]{}\phi(x)$$ on $$[-N,N]$$: $$\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\ge\int_{-N}^N\Phi(x+n)\phi(x)\,\mathrm dx\xrightarrow[n\to\infty]{}\int_{-N}^N\phi(x)\,\mathrm dx,$$ which shows that $$\liminf_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\ge\int_{-N}^N\phi(x)\,\mathrm dx.$$ Letting now $$N\to\infty$$ leads to $$\liminf_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\ge\int_{-\infty}^\infty\phi(x)\,\mathrm dx=1.$$ Hence we have $$\lim_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx=1.$$
• Thank you! I suppose that in the case $$\lim_{n\to\infty}\int \phi(x-n)dx$$ instead the change cannot be done because no convergence condition (monotone, dominated, uniform) can be found? (Sorry if this is unrelated, I will post another question in that case. Just trying to improve my understanding of convergence conditions) Sep 6, 2021 at 14:06
• @goran6 If you mean just $\int\Phi(x-n)\,\mathrm dx$ (instead of $\int\Phi(x-n)\phi(x)\,\mathrm dx$), then nothing directly applies (does the integral even converge for each $n$)? Sep 6, 2021 at 14:18
• I see. Then the cited convergence theorems still do not seem to apply. (Although in this case, all integrals equal $1$…) Sep 6, 2021 at 14:21
• If one could make the change of limit and integral, one would get: $$0=\int0\,\mathrm dx=\int\lim_{n\to\infty}\phi(x-n)\,\mathrm dx=\lim_{n\to\infty}\int\phi(x-n)\,\mathrm dx=\lim_{n\to\infty}1=1,$$ which is a contradiction. Sep 6, 2021 at 14:24