# Is this distribution already known and has a name?

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 $$Z$$.

• Where did you encounter this density? Is it even a density? (Did you check its integral is 1 and it is positive a.e.?) – QuantumSpace Dec 5 '19 at 18:15
• @EpsilonDelta : It is positive and has integral $1$. – Eric Towers Dec 5 '19 at 18:20
• 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. – Falrach Dec 5 '19 at 18:20
• Interesting. Adding this context to the question can he useful I think. – QuantumSpace Dec 5 '19 at 18:23
• I removed the probability-theory tag. Please read the tag description when using a tag. – joriki Dec 5 '19 at 21:05

## 1 Answer

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.

• 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 ? – Thomas Dec 16 '19 at 14:13
• Stochastic Kernel is the same as Markov kernel. Here I need the additional assumption that the kernel is $C_b$-Feller. – Falrach Dec 16 '19 at 16:35
• 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? – Thomas Dec 16 '19 at 17:59
• 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? – Thomas Dec 16 '19 at 18:08
• Yes this is exactly what it means. – Falrach Dec 16 '19 at 18:15