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 $Z$.

  • $\begingroup$ Where did you encounter this density? Is it even a density? (Did you check its integral is 1 and it is positive a.e.?) $\endgroup$ Dec 5, 2019 at 18:15
  • $\begingroup$ @EpsilonDelta : It is positive and has integral $1$. $\endgroup$ Dec 5, 2019 at 18:20
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    $\begingroup$ 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, 2019 at 18:20
  • $\begingroup$ Interesting. Adding this context to the question can he useful I think. $\endgroup$ Dec 5, 2019 at 18:23
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    $\begingroup$ I removed the probability-theory tag. Please read the tag description when using a tag. $\endgroup$
    – joriki
    Dec 5, 2019 at 21:05

1 Answer 1


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.

  • $\begingroup$ 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, 2019 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, 2019 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, 2019 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, 2019 at 18:08
  • $\begingroup$ Yes this is exactly what it means. $\endgroup$
    – Falrach
    Dec 16, 2019 at 18:15

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