📂Probability Distribution

Approximation of Normal Distribution Variance Stabilization from a Binomial Distribution

Example ¹

If $Y = Y_{n}$ follows a binomial distribution $\text{Bin} (n,p)$, $$\arcsin \sqrt{ {{ Y } \over { n }} } \overset{D}{\to} N \left( \arcsin \sqrt{p} , n/4 \right)$$

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• $N \left( \mu , \sigma^{2} \right)$ refers to the normal distribution.
• $\overset{D}{\to}$ refers to convergence in distribution.

Explanation

The binomial distribution $\text{Bin} (n, p )$ converges to the normal distribution $N \left( np, np(1-p) \right)$ when $n \to \infty$, so there’s nothing particularly miraculous about the normal distribution itself. However, through such a transformation, it’s possible to obtain a limit distribution with a variance constant, regardless of the parameter $p$. This is more of a mathematical trick than a practical application, interesting because it employs the differentiation technique of arcsine, which seemed like it wouldn’t be seen after freshman year in college.

Let’s assume the variance of $u ( Y/n )$ with $u$ applied to $Y/n$ is independent of $p$. For sufficiently large $n$, since $Y/n \approx p$, then by expanding Taylor’s series, $$u \left( {{ Y } \over { n }} \right) \approx u (p) + \left( {{ Y } \over { n }} - p \right) u ' (p)$$

Taking the expectation on both sides yields the mean of $u (Y/n)$ as $u (p)$, and the variance, by its property, is $\Var (aX + b) = a^{2} \Var (X)$, $$\left[ u ' (p) \right]^{2} {{ p ( 1 - p ) } \over { n }}$$

For this variance to become independent of $p$, square of $u ' (p)$ in the numerator must be canceled out by $p ( 1 - p)$. Hence, setting such that $$u ‘(p) = {{ du(p) } \over { dp }} = {{ c } \over { \sqrt{ p ( 1 - p )} }}$$ erases $p$ from the variance of $u(p)$. The solution to this differential equation can be directly found using the differentiation of the arcsine function.

Differentiation of inverse trigonometric functions: $$\left( \arcsin x \right)' = {{ 1 } \over { \sqrt{1-x^{2}} }}$$

The solution to the differential equation looks a bit different in the denominator, but if you check, it exactly matches due to the differentiation of $\sqrt{p}$. $$u (p) = 2c \arcsin \sqrt{p}$$ Since we can set and solve the differential equation in the same manner regardless of what $c$ is, it doesn’t matter what it is, but for aesthetics, setting it as $c = 1/2$ gives the appearance introduced in the example.