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.

  • $\begingroup$ 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. $\endgroup$ – Willie Wong Feb 18 '15 at 11:26
  • $\begingroup$ @WillieWong In general you're right, but as the $f_t$ are densities (and thus non-negative), this can't happen here. $\endgroup$ – herrsimon Feb 18 '15 at 11:31
  • $\begingroup$ 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. $\endgroup$ – Willie Wong Feb 18 '15 at 11:33
  • $\begingroup$ @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. $\endgroup$ – 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):

Since $\Phi$ is even, let us just focus on the problem restricted to $\mathbb{R}_+$.

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 $\delta_0$.

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – herrsimon Feb 18 '15 at 13:18
  • $\begingroup$ 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. $\endgroup$ – Willie Wong Feb 18 '15 at 13:26
  • $\begingroup$ 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. $\endgroup$ – Willie Wong Feb 18 '15 at 13:28
  • $\begingroup$ Ok, works like a charm, thank you very much! $\endgroup$ – herrsimon Feb 18 '15 at 15:14

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.