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

Author(s):
Alan Levine (Franklin and Marshall College)

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.

Alan Levine (Franklin and Marshall College), "A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 8 – Solution, Part 3," Convergence (November 2023)