How to calculate the value of √2 performing any common Probability Distribution test

Question

I want to calculate the value of √2 but using any common probability distribution, preferably based on Bernoulli Trials.

I will perform a test in real and observe the output of the test and the output of this test should lead me to the value of √2 like as we can compute the value of π using Buffon's needle test

eg. I used a fair coin, performed many trials, observed the output, calculate PDF using this output, this output is equal to √2

I am new to Probability and ML so please keep it to the simplest and based on very common distributions.

Answer 1

Let $U_i \stackrel{\text{i.i.d.}}{\sim} \text{Uniform(1, 2)}$. I chose 1 and 2 to be the limits, because we know the square root of 2 is in [1,2]. Then compute $$\hat{x}_n=1 + \frac{1}{n}\sum_{i=1}^n \mathbf{1}(U_i^2 \le 2),$$ where $\mathbf{1}(U^2\le 2)$ is equal to $1$ if $U^2 \le 2$, and $0$ otherwise. I assume we can compute the squares of numbers, even if we don't have a routine for computing square roots. Then I take $\hat{x}$ to be an estimate of $\sqrt{2}$.

Why is $\hat{x}_n$ a good estimate? Well, the expected value of $\hat{x}_n$ is equal to $$1 + \int_1^2 \mathbf{1}(U_1^2 \le 2)\,dx = 1 + \int_1^\sqrt{2}\,dx = \sqrt{2}.$$ In fact, by the strong law of large numbers, $\hat{x}_n$ converges to its expectation, which is $\sqrt{2}$, as $n\to\infty$.

Answer 2

Here's a method based on fair Bernoulli trials (e.g. by tossing a fair coin).

In a sequence of tosses of a fair coin, let $\ b_i=1\ $ if the $\ i^\text{th}\ $ toss is a head, or $\ b_i=0\ $ if it is a tail, $\ B_n = \sum_\limits{i=1}^n \frac{b_i}{2^i}\ $, and $\ B = \sum_\limits{i=1}^\infty \frac{b_i}{2^i}\ $. The random variable $\ B\ $ is uniformly distributed over the interval $\ [0,1]\ $, and so $\ \text{Prob}\left(B<\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}} \ $. Since $\ \text{Prob}\left(B=\frac{1}{\sqrt{2}}\right)=0 \ $, then, with probability $1$, either $\ B> \frac{1}{\sqrt{2}}\ $, or $\ B<\frac{1}{\sqrt{2}}\ $. The first of these events will occur if and only if $\ B_n^2>\frac{1}{2}\ $ for some $\ n\ $, and the second will occur if and only if $\ \left(B_n+\frac{1}{2^n}\right)^2 <\frac{1}{2}\ $ for some $\ n\ $.

Thus, if you keep tossing the coin until either $\ B_n^2>\frac{1}{2}\ $ or $\ \left(B_n+\frac{1}{2^n}\right)^2<\frac{1}{2}\ $ (which will eventually occur for some $\ n\ $ with probability $1$), and put $\ X_1=0\ $ in the first case, or $\ X_1=1\ $ in the second, then $\ X_1=\mathbf{1}_{\left\{B<\frac{1}{\sqrt{2}}\right\}}\ $, and $\ E\left(X_1\right)=\frac{1}{\sqrt{2}}\ $.

Now repeat the process to obtain a sequence, $\ X_1, X_2, \dots\ $ of independent random variables with mean $\ \frac{1}{\sqrt{2}}\ $. By the law of large numbers, $\ \lim_\limits{n\rightarrow\infty}\frac{2}{n}\sum_\limits{i=1}^n X_i=\sqrt{2}\ $.