A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 8 – Solution, Part 3

Review statement of Problem 8.

Returning to the approximate calculation and letting $n \frac{m+1}{2} - \alpha = \beta = \gamma \sqrt{n \frac{m^2-1}{6}}$, we substitute²³ the exponential function $e^{-n \frac{m^2-1}{24} \phi^2}$ for $\left( \frac{\sin \frac{m}{2} \phi}{m \sin \frac{\phi}{2}}\right)^n$ based on the concept shown in Chapter III, and for the upper limit of the integral, we put $\infty$ in place of $\pi$.

In this way, we get the approximate formula $$P_{n\frac{m+1}{2} - \beta} \approx \frac{1}{\pi} \int_0^\infty \cos\beta\phi \cdot e^{-n\frac{m^2-1}{24} \phi^2}\, d\phi,$$ whose right side is equal to $$\frac{1}{\sqrt{\pi}} \sqrt{\frac{6}{n(m^2-1)}} e^{-\frac{6 \beta^2}{n(m^2-1)}} = \frac{1}{\sqrt{\pi}} \sqrt{\frac{6}{n(m^2-1)}} e^{-\gamma^2}.$$

According to these, the probability of the inequalities $$n \frac{m+1}{2} - \tau \sqrt{n \frac{m^2 - 1}{6}} < X_1 + X_2 + \cdots + X_n < n \frac{m+1}{2}  + \tau \sqrt{n \frac{m^2 - 1}{6}}$$ is represented approximately by the sum of all products $\frac{1}{\sqrt{\pi}} \sqrt{\frac{6}{n(m^2-1)}} e^{-\gamma^2}$ for which $\gamma$ satisfies $-\tau < \gamma < \tau$ and turns the expression $n \frac{m+1}{2} - \gamma \sqrt{n \frac{m^2 - 1}{6}}$ into an integer.

All terms in the sum shown contain a factor $\sqrt{\frac{6}{n(m^2-1)}}$, which is equal to the difference between each two adjacent values of $\gamma$ and will be arbitrarily small for sufficiently large $n$.

Replacing on this basis the sum by the integral, we get for the probability of the inequalities $$n \frac{m+1}{2} - \tau \sqrt{n \frac{m^2 - 1}{6}} < X_1 + X_2 + \cdots + X_n < n \frac{m+1}{2}  + \tau \sqrt{n \frac{m^2 - 1}{6}}$$ the previous approximate expression²⁴ $$\frac{2}{\sqrt{\pi}} \int_0^\tau e^{-\gamma^2}\, d\gamma.$$

This is the end of the solution to this problem.

Continue to Markov's analysis of the binomial distribution.

[23] The Taylor series for $e^{-n \frac{m^2-1}{24} \phi^2}$ and $\left( \frac{\sin \frac{m}{2} \phi}{m \sin \frac{\phi}{2}}\right)^n$ both begin with $1 + \frac{n}{24} (1-m^2) \phi^2$. The third terms of these series differ by $\frac{n}{2880}(1-m^4)\phi^4$. Thus, for sufficiently small $\phi$, the two functions are approximately equal.

[24] In essence, this example shows that the sum of independent, identically-distributed discrete random variables can be approximated by a normal distribution with appropriate mean and standard deviation. In the previous chapter, Markov showed that the distribution of the sample mean can also be approximated by an appropriate normal distribution; i.e., the Central Limit Theorem. He also relaxed the requirement that the variables have the same mean and variance.