19
$\begingroup$

My question is whether the distribution on $\Bbb R$ with probability density $$ f(x) := \frac 2 {\sqrt{2\pi}} e^{-\frac{x^2}{2}} - 2 \vert x\vert \int_{\vert x \vert}^\infty \frac 1{\sqrt{2\pi}} e^{-\frac{y^2} 2} \text d y$$ is already appearing in some context or even has a name or is part of a wider class of distributions.

Here it is. I discovered the following. Let $U_n$ be uniformly distributed on $\{1, \ldots , n \}$ and $X_{n,k}$ normal distributed with mean $0$ and variance $k/n$, independent. Then $$X_{n, U_n} \to Z$$ where $Z$ has the density above. Proof: The characteristic function of $X_{n, U_n}$ converges pointwise to $$t \mapsto \frac{1 - e^{- \frac 1 2 t^2 }}{\frac 1 2 t^2 }$$ By Levy's continuity theorem the laws of $X_{n, U_n}$ have a weak limit. By Fourier inversion one can derive the density of $Z$.

$\endgroup$
  • $\begingroup$ Where did you encounter this density? Is it even a density? (Did you check its integral is 1 and it is positive a.e.?) $\endgroup$ – EpsilonDelta Dec 5 '19 at 18:15
  • $\begingroup$ @EpsilonDelta : It is positive and has integral $1$. $\endgroup$ – Eric Towers Dec 5 '19 at 18:20
  • 1
    $\begingroup$ I discovered it as a weak limit of certain random variables. So yes, it is a distribution. I also checked that it integrates to 1. $\endgroup$ – Falrach Dec 5 '19 at 18:20
  • $\begingroup$ Interesting. Adding this context to the question can he useful I think. $\endgroup$ – EpsilonDelta Dec 5 '19 at 18:23
  • 1
    $\begingroup$ I removed the probability-theory tag. Please read the tag description when using a tag. $\endgroup$ – joriki Dec 5 '19 at 21:05
7
$\begingroup$

Unfortunately, this turns out to be less interesting than I thought. After a short comment of a collegue I found out that this convergence phenomenon is just a special case of the following easy statement:

Let $V_n$ and $V$ be random variables with $V_n \to V$ in distribution. Let $K$ be a stochastic kernel with the $C_b$-Feller property (i.e. $ v \mapsto Kf(v) := \int f(x)K(v, \text d x)$ is continuous for every continuous and bounded function $f$.). Given $V_n = v_n$ let $X_n \sim K(v_n, \cdot )$ and given $V = v$ let $X\sim K(v, \cdot)$. Then $X_n \to X$ in distribution. Proof: Let $f$ be continuous and bounded. Since $Kf$ is continous and bounded and $V_n \to V$ in distribution we have $$\Bbb E [f(X_n)] = \Bbb E[Kf(V_n)] \to \Bbb E[Kf(V)] = \Bbb E[f(X)] \tag*{$\blacksquare$}$$

In this case:

$V_n := U_n /n$, $V$ uniform distributed on $(0,1)$. It is easy to check that $V_n \to V$ in distribution.

$K(v, A) := \int_A \frac{1}{\sqrt{2\pi v}} e^{-{\frac{x^2}{2v}}} \text d x$ (has Feller property by dominated convergence theorem)

Therefore $Z\sim X$ with $X \sim\mathcal N (0, v)$ given $V=v$, which can be also seen by computing the characteristic function of $\mathcal N (0, V)$ (one can compute the density easily, too):

$$\Bbb E[e^{it X}] = \int_0^1 \Bbb E [e^{itX} \vert V=v] \text d v = \int_0^1 e^{-\frac 1 2 t^2 v} \text d v = \frac{1-e^{-\frac 1 2 t^2}}{\frac 1 2 t^2}$$

In retrospect this distribution makes intuitivly and perfectly sense of course.

$\endgroup$
  • $\begingroup$ Nice :) Could you write the definition of a stochastic Kernel ? I tried to look here en.wikipedia.org/wiki/Markov_kernel but it looks slightly different ? $\endgroup$ – Thomas Dec 16 '19 at 14:13
  • $\begingroup$ Stochastic Kernel is the same as Markov kernel. Here I need the additional assumption that the kernel is $C_b$-Feller. $\endgroup$ – Falrach Dec 16 '19 at 16:35
  • $\begingroup$ So according to what I understand for every value of $v$ you have a probability measure $K(v,dx)$. Can you explain me better how $K(v,\cdot)$ defines a random variables? I can think of random variables distributed according to that measure but than we cannot talk about convergence so I guess you are thinking of something else? $\endgroup$ – Thomas Dec 16 '19 at 17:59
  • $\begingroup$ Ok you always talk about convergence in distribution so maybe you can just say that X is any random variable distributed according to that measure. Could you please say if this is the correct interpretation? $\endgroup$ – Thomas Dec 16 '19 at 18:08
  • $\begingroup$ Yes this is exactly what it means. $\endgroup$ – Falrach Dec 16 '19 at 18:15

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.