# Does $\int \Phi \left( \frac{u}{\xi} \right) f_t(\xi) \mathrm{d} \xi \to \Phi(u)$ imply that $f_t \to \delta_1$?

I'm looking at a family $(f_t)$ of densities of some continuous random variables and know that $$\int_{-\infty}^{\infty} \Phi \left( \frac{u}{\xi} \right) f_t(\xi) \mathrm{d} \xi \xrightarrow{t \to \infty} \Phi(u)$$ for any $u \in \mathbb{R}$, where $\Phi$ denotes the PDF of a standard Gaussian random variable. Does this imply that the sequence $(f_t)$ converges towards a Dirac delta function centered at one, maybe under some additional assumptions?

Writing $$\Phi \left( \frac{u}{\xi} \right) = \frac{1}{\sqrt{2\pi}} \sum_{k=0}^{\infty} \frac{(-(u/\xi)^2/2)^k}{k!}$$ and blatantly integrating term by term, the limit above yields that $$\int_{\infty}^{\infty} \frac{f_t(\xi)}{\xi^{2k}} \mathrm{d}{\xi} \xrightarrow{t \to \infty} 1$$ for any $k \geq 1$ but this doesn't seem to make it easier.

• At the very least: $\Phi$ is even. So there is nothing that prevents $f_t$ to converge to a Dirac delta function centered at $-1$. Or some linear combinations between the two. – Willie Wong Feb 18 '15 at 11:26
• @WillieWong In general you're right, but as the $f_t$ are densities (and thus non-negative), this can't happen here. – herrsimon Feb 18 '15 at 11:31
• What do you mean? Take $a f_t + (1-a) g_t$ for $a \in (0,1)$ and $f_t, g_t$ densities converging to $\delta_1$ and $\delta_{-1}$ respectively. – Willie Wong Feb 18 '15 at 11:33
• @WillieWong You're completely right, please excuse my last nonsense-comment. What I should have written is that the underlying random variables are known to be non-negative. – herrsimon Feb 18 '15 at 11:48

A sketch of an argument (I'll leave it to you to figure out the precise conditions and formulations that makes this argument work):

Write $\Psi(s) = \Phi\circ\exp(s)$ and $g_t(s) = f_t\circ \exp(s)$ for $s\in \mathbb{R}$. The problem now reduces asking about $g_t$ such that $$\lim_{t\to\infty} \int_{-\infty}^{\infty} \Psi(u-s) g_t(s) e^s ~\mathrm{d}s = \Psi(u)$$

We know a few things:

1. $g_t$ is positive, hence also is $g_t(s) e^s$
2. $\Psi$ is positive.
3. $\Psi$ is strictly decreasing
4. $\lim_{s\to -\infty} \Psi(s) = 1$
5. $\Psi$ is concave on $\mathbb{R}_-$, and convex on $\mathbb{R}_+$.

Assumption $e^s g_t(s)\in L^1$. (Equivalently $f_t \in L^1$.), and $s e^s g_t(s) \in L^1$.

Then condition 4 and the equation implies that for sufficiently large $t$, $g_t$ has mass approximately 1.

On the other hand, condition 5 tells you that for sufficiently large $+ u$ translations you have that $$\int_{-\infty}^\infty \Psi(u-s) g_t(s) e^s \mathrm{d}s \geq \Psi(u - m_t)$$ where $m_t$ is the center of mass of $g_t(s) e^s$. The sign is reversed if we look at sufficiently large $+u$ translations. Using that $\Psi$ is strictly decreasing one of the two asymptotic behaviour contradicts $m_t \neq 0$.

With $m_t = 0$ the strict convexity/concavity near the infinities imply that the equation can only be satisfied if the support of $g_t(s) e^s$ is very concentrated near the set $\{0\}$, for $t$ sufficiently large. And a few more lines should give you that $g_t(s) e^s$ converges to the Dirac measure at 0.

Note, however, to get the statement that $f_t$ converges to the dirac support at 1, that one of the conditions that you need to impose would be decay rate on $g_t(s)$ as $s \to -\infty$, which translates also to some statement about behavior of $f_t$ near the origin. This is due to the following fact which poses another obstruction to the result you want in your question.

Let $h_t = 2t \mathbf{1}_{[1/(2t), 1/t]}$ so that $\int h_t = 1$.

Considering now that $$\lim_{t \to \infty} \int_{-\infty}^\infty \Phi(u/\xi) h_t(\xi) \mathrm{d}\xi = 0$$ but that $h_t$ does not converge in $L^1$, and that $h_t$ converges in the sense of distributions to $\delta_0$.

• Thank you very much! I understand everything and can also finish your arguments, except for one step: How exactly does your inequality $\int \Psi(u-s) g_t(s) e^s \mathrm{d}s \geq \Psi(u-m_t)$ follow from convexity/concavity? – herrsimon Feb 18 '15 at 13:18
• For a convex function you have $a f(x) + (1-a) f(y) \geq f( ax + (1-a)y)$. Regard the left hand side as integrating over the measure $a \delta_x + (1-a)\delta_y$. Regard the argument to the right hand side as the center of mass $m$. Now do the usual thing to pass from discrete to continuous by approximations or what not. – Willie Wong Feb 18 '15 at 13:26
• Or you can use the fact that convex functions lie above their linearisations. And for linear functions, convolution against a probability measure is equal to a translation. – Willie Wong Feb 18 '15 at 13:28
• Ok, works like a charm, thank you very much! – herrsimon Feb 18 '15 at 15:14