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Let $X_1, ..., X_n \stackrel{iid}{\sim} N(0,1)$. Is the following true: $$\frac{1}{\sqrt{n}}\sum_{i=1}^n\mathbf{1}\left(0 < X_i \leq \frac{1}{\sqrt{n}}\right) \xrightarrow{P} 0$$

I've tried approaching it with Markov's inequality, i.e.

$$P\left(\frac{1}{\sqrt{n}}\sum_{i=1}^n\mathbf{1}\left(0 < X_i \leq \frac{1}{\sqrt{n}}\right) > \epsilon \right) \leq \frac{\sum_{i=1}^nP\left(0 < X_i \leq \frac{1}{\sqrt{n}}\right)}{\epsilon\sqrt{n}} = \frac{\sqrt{n}(\Phi(1/\sqrt{n})-1/2)}{\epsilon} \xrightarrow{n\rightarrow \infty} \infty$$ where $\Phi$ is the normal cdf and the limit calculation by L'hospital's rule and I don't know how else to proceed, or if it's even true.

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2 Answers 2

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Let $Y_{n,i}:= \mathbf{1}\left(0 < X_i \leq \frac{1}{\sqrt{n}}\right) $ and $R_n=n^{-1/2}\sum_{i=1}^n\left(Y_{n,i}-\mathbb E\left[Y_{n,i}\right]\right)$. Then $\mathbb E\left[R_n^2\right]\to 0$ and we are thus reduced to examine the convergence of $\sqrt n\mathbb P\left(0 < X_1 \leq \frac{1}{\sqrt{n}}\right)$. This can be done by writing the probability as an integral involving the density of $X_1$ and doing the substitution $t=u/\sqrt n$.

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I believe this is false, and the convergence would have to be to $1/\sqrt{2 \pi}$ if anything, since this is intuitively an approximation of the derivative of the cdf of a Gaussian with mean 0 and variance 1 at the point 0.

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