Define $\mu_{N}$ to be the sample mean for $N$ i.i.d. random variables that each have mean $\mu$ and variance $\sigma^2$: $$ \mu_{N} := \frac{1}{N} \sum_{i=1}^N X_i. $$
Now consider the sample mean if we removed one of these random variables. That is consider, $\mu_{N-1}$ where w.l.o.g. we remove the last random variable from the sum: $$ \mu_{N-1} := \frac{1}{N-1} \sum_{i=1}^{N-1} X_i. $$
I am interested in describing how close $\mu_{N-1}$ is to $\mu_{N}$ in an appropriate sense..it is this 'sense' that I am unsure of as I haven't been working with probability that long.
Subtracting $\mu_{N-1}$ from $\mu_{N}$ we have \begin{align} \mu_{N} - \mu_{N-1} &= \frac{1}{N} \sum_{i=1}^N X_i - \frac{1}{N-1} \sum_{i=1}^{N-1} X_i \\ & = \frac{1}{N}X_N + \frac{N-1}{N}\frac{1}{N-1} \sum_{i=1}^{N-1} X_i - \frac{1}{N-1} \sum_{i=1}^{N-1} X_i \\ & = \frac{1}{N}X_N + \bigg(\frac{N-1}{N}-1\bigg)\frac{1}{N-1} \sum_{i=1}^{N-1} X_i. \end{align}
So as $N\to \infty$, heurstically it 'looks like' both terms go to zero. But as the $X_i$ are random variables I imagine we need to set some conditions on them for this to hold. For example, we could be unlucky and draw $X_N$'s that are sufficiently large such that $\frac{1}{N} X_N$ does not go to zero as $N \to \infty$.
I feel like the variance $\sigma^2$ could have an effect on the convergence but I'm not sure.
So my questions are
- In the finite sample case can we say something more informative than what I have already derived above regarding how close $\mu_N$ is to $\mu_{N-1}$?
- What conditions are needed on the $X_i$ to ensure $\mu_{N} - \mu_{N-1}$ converges to zero as $N \to \infty$?
- What is an appropriate sense of convergence for $\mu_{N} - \mu_{N-1}$? I am thinking maybe we can say it converges with 'high probability'. Most importantly, is it possible to find an exact rate of convergence?
You need to read about the different concepts of convergence of random variables.
In this case, calling $Z_n = \frac{1}{N} \sum_{i=1}^N X_i - \frac{1}{N-1} \sum_{i=1}^{N-1} X_i$, you want to know if and (in what sense ) $Z_n$ converges to zero. Notice that $Z_n$ is a random variable.
Given that $X_i$ are iid with finite mean and variance, you can deduce that $E[Z_n]=0$ and $\sigma_{Z_n}^2 = \frac{1}{n(n+1)}\sigma_{X}^2$.
This says that $\sigma_{Z_n}^2 \to 0$ (and quite quick).
Then you can apply Chebyshev's_inequality : for any $\epsilon >0$
$$P(|Z_n| > \epsilon) \le \frac{1}{\epsilon}\sigma_{Z_n}^2=\frac{1}{\epsilon \, n(n+1)}\sigma_{X}^2 < \frac{\sigma_{X}^2}{\epsilon \, n^2} $$
Then $P(|Z_n| > \epsilon) \to 0$ as $n\to \infty$
The above amounts to prove convergence in probability from convergence in $L^2$-norm. These are quite strong measures of convergence for random variables, though not so strong as almost sure convergence.