Sampling & Confidence

Take a random variable $R$ with an expectation of $\mu$.
Make $n$ trial observations of $R$, and take the average of those observations.
See how close they are to $\mu$.

The observed average of the results will approach the true expectation as the number of trial observations increases.

So, we're doing $n$ observations of $R$. They are mutually independent:
$R_1, \ R_2, \ ..., \ R_n$

They all have the same expectation $\mu$.

Taking the average of those values: $A_n = \frac{R_1 \ + \ R_2 \ + \ ... \ + \ R_n}{n}$
Is this average value close to $\mu$ if $n$ is big?

$\lim_\infty \ \text{Pr}[|A_n - \mu| \leq \delta] = \ ?$

The answer that the Weak Law of Large Numbers gives to that question is $1$:

Another way of saying it is:

So, as the number of trials increases, the probability that the average differs from the expectation more than delta goes to $0$.

As the number of trials increases, the difference between the observed average and the theoretical average gets smaller.

For a clear example to understand the concept of the Law of Large Numbers, The Organic Chemistry Tutor's video is really great.

Let $R_1, \ R_2, \ ..., \ R_n$ be pairwise independent random variables with same mean $\mu$ and variance $\sigma^2$.
Their average $A_n$ is $\frac{R_1 \ + \ R_2 \ + \ ... \ + \ R_n}{n}$.

So, the probability that their average differs from the mean by more than a given tolerance delta is:

It is the Pairwise Independent Sampling Theorem.

Confidence levels refer to the results of estimation procedures for real-world quantities.

So, when some facts are told to have high confidence level, the important thing to remember is that behind this claim, a random experiment lies.

95% confidence level is most common, and as mentioned in the course video, this xkcd comic illustrates the topic:

• previous
• next