Limit of the variance of a difference of indicators

Consider a sequence of $n$ iid random variables $Y_i$ having a density with respect to Lebesgue measure at $0$, $g(0)$. Compute the following limit:

$$\lim_{n\rightarrow \infty} \text{Var} \left [\sum_{i=1}^n \int_0^{\eta/\sqrt{n}} \left ( \mathbb 1(Y_i \leq s) - \mathbb 1(Y_i \leq 0) \right) ds\right ]$$

where $\eta$ is some real valued constant.

my attempt

Since the $Y_i$ are iid we can rewrite this as:

$$\lim_{n\rightarrow \infty} n\text{Var} \left [ \int_0^{\eta/\sqrt{n}} \left ( \mathbb 1(Y_i \leq s) - \mathbb 1(Y_i \leq 0) \right) ds\right ]$$

Now, substituting $u = \sqrt n s$:

$$\lim_{n\rightarrow \infty} \text{Var} \left [ \int_0^{\eta} \left ( \mathbb 1(Y_i \leq u/\sqrt n) - \mathbb 1(Y_i \leq 0) \right) du\right ]$$

But since the expression inside the integral goes to $0$ pointwise (since $Y_i$ has a density at $0$), we have that the integral also goes to $0$ (applying dominated convergence). Then, since the variance of a term that goes to a constant goes to $0$, we have that the entire expression evaluates to $0$.

Have I made any incorrect assumptions or manipulations here?

Solution 1:

The following provide a method to calculate the variance what you want. Let \begin{align*} Z_i&=\int_{0}^{\eta/\sqrt{n}}(1(Y_i\le s)-1(Y_i\le0))\,\mathrm{d}s\\ &=\int_{0}^{\infty}1(0< Y_i\le s\le \eta/\sqrt{n})\,\mathrm{d}s\\ &=\Big(\frac{\eta}{\sqrt{n}}-Y_i\Big)^+, \end{align*} then $\{Z_i,1\le i\le n\} $ is a sequence of iid random variables and \begin{align*} \mathbb{E}[Z_i]&=\int_{0}^{\eta/\sqrt{n}}\Big(\frac{\eta}{\sqrt{n}}-Y_i\Big)^+ g(y) \,\mathrm{d}y\\ &=\int_{0}^{\eta/\sqrt{n}}\Big(\frac{\eta}{\sqrt{n}}-Y_i\Big)^+ (g(0)+O(y))\,\mathrm{d}y\\ &=\frac{g(0)\eta^2}{2n}+O(n^{-3/2})\\ \mathbb{E}[Z_i^2]&=\frac{g(0)\eta^3}{3n^{3/2}}+O(n^{-2})\\ \mathsf{Var}[Z_i]&=\mathbb{E}[Z_i^2]-(\mathbb{E}[Z_i])^2\\ &=\frac{g(0)\eta^3}{3n^{3/2}}+O(n^{-2}) \end{align*} Hence, \begin{equation*} \lim_{n\to\infty}\mathsf{Var}\Big[\sum_{i=1}^{n}Z_i\Big]=0. \end{equation*}