Given a set of $N$ iid samples, $x_k$, of random variable $X$ and the expectation estimator,

$$E[X] \approx \frac{1}{N} \sum_{k=1}^N x_k,$$

how do I find the variance of this estimator (not the variance of $X$) as a function of $N$?

I would only like to know the error in the estimate of $E[X]$ and what I would expect is that more samples (larger $N$) drives the variance down, so that

$$\lim_{N \to \infty} \bigg[\frac{1}{N} \sum_{k=1}^N x_k - E[X]\bigg] = 0.$$

Am I right about this? Perhaps variance is not the correct way to measure the error in the estimate? What I would like to understand is, how many $N$ is sufficient because there is a cost for every extra $x_k$ that is processed.


1 Answer 1


Without more knowledge on $X$, we cannot say a lot. However, if it is IID with mean $\mu$ and variance $\sigma^2$, the sample mean $\hat{\mu} = \frac{1}{N}\left(\sum_{n=0}^{N-1}x_n\right)$ is unbiased, since:

$$E[\hat{\mu}] = E\left(\frac{1}{N}\sum_{n=0}^{N-1}x_n\right) = \frac{1}{N}\sum_{n=0}^{N-1}E\left(x_n\right) = N\mu/N=\mu\,.$$

And its variance goes to zero when $N$ increases:

$$V[\hat{\mu}] = V\left(\frac{1}{N}\sum_{n=0}^{N-1}x_n\right) = \frac{1}{N^2}\sum_{n=0}^{N-1}V\left(x_n\right) = N\sigma^2/N^2=\sigma^2/N\,.$$

Thus, the expectation converges to the actual mean, and the variance of the estimator tends to zero as the number of samples grows. Under these definitions, the sample mean is a consistent estimator. Note that one could try to use other hypotheses: alternative norms, convergence in law, etc.

  • $\begingroup$ Sorry, I corrected them. Indeed, the variance tends to 0 as the number of samples grows $\endgroup$ Commented Jul 5, 2020 at 19:57
  • $\begingroup$ I noticed that $E[x_n] = x_n$ because it is a constant, but you wrote $E[x_n] = \mu$. This seems incorrect. $\endgroup$
    – user827822
    Commented Jul 14, 2020 at 20:34
  • $\begingroup$ We are talking about random variables. What is constant here? $\endgroup$ Commented Jul 14, 2020 at 20:42

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