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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 $\mathbb{P}(x<X\leq x+Y_n)>0$. I stress that all variables are dependent (in a complicated way).

The (initial) question. I would like to know if the sequence $1_{\{x<X\leq x+Y_n\}}/Y_n$ converges in mean to $f(x)$ (i.e., $\mathbb{E}(1_{\{x<X\leq x+Y_n\}}/Y_n)\to f(x)$) 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 $g_x:y\mapsto\mathbb{P}(x<X<x+y)/y$ is bounded and continuous on $[0,\infty)$ if we set $g_x(0)=g_x(0+)=f(x)$. Hence $$\mathbb{E}(1_{\{x<X\leq x+Y_n\}}/Y_n)= \mathbb{E}(\mathbb{E}(1_{\{x<X\leq x+Y_n\}}/Y_n|Y_n))= \mathbb{E}(g_x(Y_n))\to f(x)$$ 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 $0<L_n$ 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 $n$ $$1_{\{x<X\leq x/(1-L_n)\}}/(U_nX)\leq 1_{\{x<X\leq x+Y_n\}}/Y_n\leq 1_{\{x<X\leq x/(1-U_n)\}}/(L_nX).$$ Moreover, we may bound $$\mathbb{E}\left(\frac{1_{\{x<X\leq x/(1-L_n)\}}}{U_nX}\Big|U_n,L_n\right) \geq \mathbb{E}\left(\frac{1_{\{x<X\leq x/(1-L_n)\}}}{U_nx/(1-L_n)}\Big|U_n,L_n\right) =g_x\left(\frac{xL_n}{1-L_n}\right)\frac{L_n}{U_n}.$$ 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 $$\mathbb{E}\left(\frac{1_{\{x<X\leq x/(1-U_n)\}}}{L_nX}\Big|U_n,L_n\right) \leq \mathbb{E}\left(\frac{1_{\{x<X\leq x/(1-U_n)\}}}{L_nx}\Big|U_n,L_n\right) =g_x\left(\frac{xU_n}{1-U_n}\right)\frac{U_n}{L_n}.$$ which again converges in $L^1$ to $g_x(0)=f(x)$. By sandwiching, we conclude that $\mathbb{E}(1_{\{x<X\leq x+Y_n\}}/Y_n)\to g_x(0)=f(x)$.

  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) $\mathbb{E}(1_{\{x<X\leq x+Y_n\}})/\mathbb{E}(Y_n)\to f(x)$ or (II) $\mathbb{E}[\mathbb{E}(1_{\{x<X\leq x+Y_n\}}|X)/\mathbb{E}(Y_n|X)]\to f(x)$ or something similar. If you have such a result, a reference would be much appreciated. Counterexamples are also welcome.

(Originally asked in math.stackexchange.)

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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).

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  • $\begingroup$ 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. $\endgroup$ – Jorge I. González C. Jul 23 at 22:52
  • $\begingroup$ 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. $\endgroup$ – Mateusz Kwaśnicki Jul 24 at 6:02
  • $\begingroup$ 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! $\endgroup$ – Jorge I. González C. Jul 24 at 9:20
  • $\begingroup$ You're welcome! $\endgroup$ – Mateusz Kwaśnicki Jul 24 at 9:29

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