# Dependent random variables converging to a density in mean

Let $$X$$ be an absolutely continuous r.v. with density $$f$$ which is continuous on $$(0,\infty)$$. Fix $$x>0$$ and consider some a.s. decreasing sequence $$Y_n$$ bounded by $$X$$ such that $$Y_n\searrow 0$$ a.s. and such that $$\mathbb{P}(x0$$. I stress that all variables are dependent (in a complicated way).

The (initial) question. I would like to know if the sequence $$1_{\{x converges in mean to $$f(x)$$ (i.e., $$\mathbb{E}(1_{\{x) under mild conditions (e.g. $$\mathbb{E}(1/Y_n)<\infty$$ for all $$n$$ or $$X$$ having bounded support).

Special cases. The following examples are not what I am after, but I include them to provide some intuition and perhaps clarity.

1. The independent case. When $$X$$ and $$Y_n$$ are independent for each $$n$$, this is trivial. Indeed, the function $$g_x:y\mapsto\mathbb{P}(x is bounded and continuous on $$[0,\infty)$$ if we set $$g_x(0)=g_x(0+)=f(x)$$. Hence $$\mathbb{E}(1_{\{x by the dominated convergence theorem (as $$g_x(Y_n)\to f(x)$$ a.s.).

2. The quotient can be sandwiched. Assume that some random variables $$0 and $$U_n<1$$ are independent of $$X$$ and satisfy $$L_n\leq Y_n/X\leq U_n$$ for all sufficiently large (deterministic) $$n\in\mathbb{N}$$ and $$U_n/L_n\overset{L^1}{\to} 1$$. In this case, we clearly have $$U_n,L_n\overset{\mathbb{P}}{\to}0$$ and, for sufficiently large $$n$$ $$1_{\{x Moreover, we may bound $$\mathbb{E}\left(\frac{1_{\{x Since convergence in $$L^1$$ implies convergence in distribution, the continuous mapping theorem and the boundedness of $$g_x$$ imply that the last quantity converges in $$L^1$$ to $$g_x(0)=f(x)$$. Similarly, we bound $$\mathbb{E}\left(\frac{1_{\{x which again converges in $$L^1$$ to $$g_x(0)=f(x)$$. By sandwiching, we conclude that $$\mathbb{E}(1_{\{x.

3. Mateusz Kwaśnicki below provided a simple counterexample showing the necessity of some degree of independence between $$Y_n$$ and $$X$$.

The dependence between $$X$$ and $$Y_n$$ (additional assumptions).

1. (Motivated by Special cases(3) above) We may additionally assume that, for each $$n\in\mathbb{N}$$, $$(X,Y_n)$$ has a joint and continuous density.

2. I am particularly interested in the following case. There is a sequence $$\{(\theta_n,\xi_n)\}_{n\in\mathbb{N}}$$ such that (I) $$\theta_n=\prod_{k=1}^n U_k$$ for iid $$U_k\sim U(0,1)$$ (i.e. a stick-breaking process), (II) $$\{\xi_n\}_{n\in\mathbb{N}}$$ are nonnegative and absolutely continuous with continuous densities and $$\sum_{n=1}^\infty \xi_n\in(0,\infty)$$ a.s. and (III) conditional on $$\theta_{n-1}-\theta_n$$, $$\xi_n$$ is independent of $$\{(\theta_{k-1}-\theta_k,\xi_k)\}_{k\neq n}$$. Then we define $$X=\sum_{k=1}^\infty \xi_k$$ and $$Y_n=\sum_{k=n}^\infty \xi_k$$.

Flexibility of the result. It is not necessary that this exact result holds (for me). For instance, it would help if we have sufficient (and mild) conditions for: (I) $$\mathbb{E}(1_{\{x or (II) $$\mathbb{E}[\mathbb{E}(1_{\{x or something similar. If you have such a result, a reference would be much appreciated. Counterexamples are also welcome.

If I am not mistaken, your question has a negative answer. Consider $$Y_n = \begin{cases} \tfrac{1}{n} (X - x) & \text{if X > x,} \\ \tfrac{1}{n} X & \text{if X \leqslant x.} \end{cases}$$ Then $$Y_n$$ is strictly decreasing, $$0 < Y_n < X$$, but $$x < X < x + Y_n$$ never happens: if $$X > x$$, then $$Y_n = \tfrac{1}{n} (X - x)$$, and so $$x + Y_n = x + \tfrac{1}{n} (X - x) = \tfrac{1}{n} X + \tfrac{n - 1}{n} x < \tfrac{1}{n} X + \tfrac{n - 1}{n} X = X .$$ The expectation you are interested in is therefore zero, and it does not converge to the density function of $$X$$ at $$x$$ (unless the density is zero, of course).
• This is a nice and simple counterexample illustrating that a necessary condition is $\mathbb{P}(x<X\leq x+Y_n)>0$. Thanks! I'll edit the statement accordingly. Jul 23 '19 at 22:52
• I do not think that this additional assumption can save you. You can modify $Y_n$ on a tiny set in order to make $x < X < x + Y_n$ possible with very small probability, so that the limit is still zero. Say: $Y_n = \tfrac{1}{n} X$ if $X \leqslant x + n^{-n}$, $Y_n = \tfrac{1}{n} (X - x)$ otherwise; the expectation is of the order $n^{1-n} f(x)$, if I did not make any mistake. Jul 24 '19 at 6:02
• You're absolutely right. One may even add some additional randomness to $Y_n$ (e.g. $Y_n=\frac{U_n}{n}X$ for $U_n\sim U(0,1)$ independent of $X$) and still get a counterexample. It can also be modified to simultaneously fail for countably many such $x$. It seems some 'moderate' degree of independence is necessary. Thanks! Jul 24 '19 at 9:20