# Asymptotic behavior of a uniform mixture distribution

Let $$X = \{x_1= -\alpha, x_2, \ldots, x_n= \alpha\}$$ be a set with $$x_{i+1} = x_i + \beta$$ for some $$\alpha, \beta \in \mathbb{R}$$.

$$Y$$ is a random variable that is sampled from a mixture distribution as: $$Y ~ \sim \sum_{i=1}^n p_i \mathbb{U}[x_i, x_{i+1}]$$

where $$\mathbb{U}[x_i, x_{i+1}]$$ denotes a uniform random variable which is sampled from the interval $$[x_i, x_{i+1}]$$.

Let’s pick a distribution, e.g., Gaussian distribution, and let $$CDF(x)$$ denote the cumulative distribution function value of this distribution at $$x$$.

My question is the following: let us give weights $$p_i = CDF(x_{i+1}) - CDF ( x_{i})$$, e.g., the probability given to variable $$\mathbb{U}[x_i, x_{i+1}]$$ is the density assigned by the Gaussian distribution on interval $$[x_i, x_{i+1}]$$. Obviously, this is valid when we have $$\alpha \rightarrow \infty$$. Does the distribution of Y converge to (also) a Gaussian distribution (more generally the distribution used in the CDF), when $$\alpha \rightarrow \infty$$ and $$\beta \rightarrow 0$$?

My intuition says yes, but I cannot prove it.

This is true assuming you are free to choose $$\alpha, \beta$$ however you wish. Convergence in distribution of a sequence of real-valued random variables means their cdfs $$F_n$$ satisfy $$\lim_{n\rightarrow\infty} F_n(x) = F(x)$$ for each point $$x \in \mathbb{R}$$ at which $$F$$ is continuous. We can show that, for any $$\varepsilon > 0$$, there are $$A$$ and $$B$$ such that for all $$\alpha > A$$, $$\beta < B$$, $$\sup_{x \in \mathbb{R}} |F_{\alpha,\beta}(x) - F(x)| < \varepsilon.$$ This is enough to extract a sequence $$\alpha_n, \beta_n$$.

This turned into quite a lengthy post, so let me just say the idea is simple: you approximate the density with piecewise constant functions, and all that matters is the areas under the curves converge uniformly.

Let then $$\varepsilon > 0$$ be given, and let $$\Phi$$ denote the cdf of a standard Gaussian. There is $$A > 0$$ large enough that $$\Phi(-A) < \varepsilon/4$$, which by symmetry also implies $$\Phi(A) > 1-\varepsilon/4$$. Fix some $$\alpha > A$$. We have just cut off the tails.

Given $$x_i = -\alpha + i\beta$$ with $$n = 2\alpha/\beta \in \mathbb{Z}$$, there are $$n$$ intervals $$I_i = [x_i,x_{i+1})$$ that cover $$[-\alpha, \alpha)$$. Assuming $$p_i = \Phi(x_{i+1}) - \Phi(x_i)$$, the total probability mass allocated is $$1 - 2\Phi(-\alpha)$$; the remaining mass can be assigned anywhere outside of $$[-\alpha,\alpha)$$; say it is assigned to $$x > \alpha$$. I'll ignore any technicalities with the right end-point (it has probability 0).

Define a "locator" map $$\ell : [-\alpha, \alpha) \rightarrow \{0, ..., n-1\}$$ which associates to any $$x$$ the unique index $$i$$ of the left end-point in the interval $$I_i$$ (so in particular $$\ell(x_i) = i)$$. Remembering that the density of the $$i^{th}$$ uniform random variable is $$(1/\beta)1_{I_i}$$, the cdf $$F_{\alpha, \beta}$$ satisfies $$F_{\alpha, \beta}(x) = p_{\ell(x)}\frac{x - x_{\ell(x)}}{\beta} + F_{\alpha,\beta}(x_{\ell(x)}),$$ and note that the approximate cdf agrees with $$\Phi$$ at the discretization points $$x_i$$ up to a shift by $$\Phi(-\alpha)$$: $$F_{\alpha,\beta}(x_i) = \sum_{i'=1, ..., i-1} p_{i'} = \sum_{i' = 1,...,i-1} (\Phi(x_{i'+1}) - \Phi(x_{i'})) = \Phi(x_{i}) - \Phi(-\alpha).$$ Thus, for any $$x \in [-\alpha, \alpha)$$, \begin{align*} F_{\alpha,\beta}(x) - \Phi(x) &= p_{\ell(x)}(x - x_{\ell(x)})/\beta + F_{\alpha,\beta}(x_{\ell(x)}) - \Phi(x) \\ &= p_{\ell(x)}(x - x_{\ell(x)})/\beta + \Phi(x_{\ell(x)}) - \Phi(-\alpha) - [\Phi(x) - \Phi(x_{\ell(x)}) + \Phi(x_{\ell(x)})]\\ &= [p_{\ell(x)}(x - x_{\ell(x)})/\beta - (\Phi(x) - \Phi(x_{\ell(x)}))] - \Phi(-\alpha).\tag{1} \end{align*} The left term in brackets in the last equality above is $$(\Phi(x_{\ell(x)+1}) - \Phi(x_{\ell(x)}))(x - x_{\ell(x)})/\beta - (\Phi(x) - \Phi(x_{\ell(x)})),$$ which, if you squint, is the fundamental theorem of calculus: $$\Phi'(a)(x-a) \approx \frac{\Phi(b) - \Phi(a)}{\beta}(x - a) \approx (\Phi(x) - \Phi(a)).$$ I leave it to the reader to justify using compactness of $$[-\alpha,\alpha]$$ and differentiability of $$\Phi$$ on $$(-\alpha,\alpha)$$ that one can find $$B > 0$$ such that any $$\beta < B$$ makes the term in brackets as small as desired, less than $$\varepsilon/2$$.

Going back to $$(1)$$, we find that for $$\alpha > A$$ and $$\beta < B$$ and $$x \in [-\alpha, \alpha)$$, we get $$|F_{\alpha,\beta}(x) - \Phi(x)| < \varepsilon/2 + \varepsilon/4.$$ For the remaining $$x$$, we've misplaced at most $$2\Phi(-\alpha)$$ mass, which is bounded by $$\varepsilon/2$$. Thus, $$\sup_{x \in \mathbb{R}} |F_{\alpha,\beta}(x) - \Phi(x)| < \varepsilon,$$ which establishes the desired convergence.

• Thanks for this great answer! It is very clear to me except for the part which, if you squint, is the fundamental theorem of calculus:''. Could you please say what you meant there? Thanks! Aug 20, 2020 at 22:11
• I think you are using first order Taylor approximation and upper bounding the error. But I am wondering if I can formalize this step. Also, doesn't your proof still hold if you take $\Phi(-\alpha) < \epsilon/2$ rather than $< \epsilon/4$? Many thanks for your time once again. Aug 21, 2020 at 0:21
• I had a bit more look now. If you are using the Taylor’s inequality, don’t we need that the CDF is twice differentiable? I gave the example of Gaussian, but I will actually use a Laplace distribution whose pdf is not differentiable and does not have an upper bound at $x=0$. So is your method only for twice differentiable CDFs? Aug 21, 2020 at 16:33