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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.

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    $\begingroup$ 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. $\endgroup$ Sep 2, 2021 at 15:41
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    $\begingroup$ As you said, that's a sufficient condition. Not a necessary a sufficient condition. $\endgroup$ Sep 2, 2021 at 15:51

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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 $$\lim_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx=1.$$

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  • $\begingroup$ 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) $\endgroup$
    – goran6
    Sep 6, 2021 at 14:06
  • $\begingroup$ @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 $n$)? $\endgroup$
    – nejimban
    Sep 6, 2021 at 14:18
  • $\begingroup$ I mean just the integral of the standard normal PDF, not the CDF (as you suggested). Thank you $\endgroup$
    – goran6
    Sep 6, 2021 at 14:19
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    $\begingroup$ I see. Then the cited convergence theorems still do not seem to apply. (Although in this case, all integrals equal $1$…) $\endgroup$
    – nejimban
    Sep 6, 2021 at 14:21
  • $\begingroup$ 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. $\endgroup$
    – nejimban
    Sep 6, 2021 at 14:24

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