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A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 8 – Solution, Part 2

Alan Levine (Franklin and Marshal College)


Review statement of Problem 8.


For large values of \(n\), the exact calculation of \(P_\alpha\) requires tiring computations and can hardly represent great interest.

Then the question arises of finding an approximate expression for the probability with the possibility that it is simple and close to exact.

Assuming only \(n\) is large, but \(m\) is not, and considering not the probability of the actual value of the sum \(X_1 + X_2 + \cdots +X_n\) but the probability that this sum lies within given limits, we can return to the general approximate calculations in Chapter III.

To apply these, we should find the mathematical expectation of the first and second powers of the considered variables.

Since the mathematical expectation of any of these variables \(X_1, X_2, \dots X_n\) is equal to \(\frac{1+2 + \cdots + m}{m} = \frac{m+1}{2}\), and the mathematical expectation of their squares is equal to \(\frac{1^2 + 2^2 + \cdots + m^2}{m} = \frac{(m+1)(2m+1)}{6}\), then the difference between the mathematical expectations of their squares and the square of their mathematical expectations19 reduces to \[\frac{(m+1)(2m+1)}{6} - \left(\frac{m+1}{2}\right) ^2 = \frac{m^2 - 1}{12},\] and therefore, the results of the third chapter give, for the probability of the inequalities \[n \frac{m+1}{2} - \tau \sqrt{n \frac{m^2 - 1}{6}} < X_1 + X_2 + \cdots + X_n < n \frac{m+1}{2}  + \tau \sqrt{n \frac{m^2 - 1}{6}} ,\] an approximate expression in the form of the known integral20 \[\frac{2}{\sqrt{\pi}} \int_0^\tau e^{-z^2}\, dz.\]

We will take advantage of the particular example to show other results of this approximate expression of probability, which can be applied in the general case.

And before anything, we note that, in the expansion of any entire function \(F(t)\) in powers of \(t\), the coefficient of \(t^\alpha\) can be represented in the form of the integral \[\frac{1}{2\pi} \int_{-\pi}^{\pi}  F(e^{\phi\sqrt{-1}} )e^{-\alpha\phi \sqrt{-1}}\,d\phi,\] because \(\int_{-\pi}^\pi d\phi = 2\pi\), and, for any integer \(k\) different from 0, we have \(\int_{-\pi}^\pi e^{k\phi \sqrt{-1}} \, d\phi = 0\).21

Therefore,22 \[\begin{align} P_\alpha&= \frac{1}{2\pi} \int_{-\pi}^\pi \frac{ e^{(n-\alpha) \phi \sqrt{-1}}\left(1-e^{m \phi \sqrt{-1}}\right)^n}{m^n\left(1-e^{\phi \sqrt{-1}}\right)^n} \, d\phi \\
    &=\frac{1}{2\pi} \int_{-\pi}^\pi \frac{ e^{ (n \frac{m+1}{2} -\alpha) \phi \sqrt{-1}} \left(e^{\frac{m}{2} \phi \sqrt{-1}} - e^{-\frac{m}{2} \phi \sqrt{-1}}\right)^n}{m^n \left(e^{\frac{1}{2} \phi \sqrt{-1}} - e^{-\frac{1}{2} \phi \sqrt{-1}}\right)^n}\,d\phi \\
    &=\frac{1}{2\pi} \int_{-\pi}^\pi e^{(n \frac{m+1}{2} - \alpha) \phi \sqrt{-1}} \left( \frac{\sin \frac{m}{2} \phi}{m \sin \frac{\phi}{2}}\right)^n\, d\phi \\
    &= \frac{1}{\pi} \int_0^\pi \cos \left(n \frac{m+1}{2} - \alpha\right) \phi\  \left( \frac{\sin \frac{m}{2} \phi}{m \sin \frac{\phi}{2}}\right)^n\, d \phi.


Continue to the third part of Markov's solution of Problem 8.

Skip to Markov's analysis of repeated independent events.


[19] In other words, the variance.

[20] Markov never defines the normal distribution, as we know it today. When he talks about continuous random variables for the first time in Chapter V, his only examples are those which have a constant density function (i.e., what we would now call the “uniform” distribution) and one whose density is defined for all \(x\) and decreases as \(|x|\) increases. He then claims that the density is proportional to \(e^{-x^2}\), although there are many other such possibilities. This gives us a “normal distribution” with mean 0 and variance \(\frac12\).  So, his approximation can be written as \[P\left(\left| \frac{S_n - \mu}{\sqrt{2} \sigma} \right| < \tau \right) = \frac{2}{\sqrt{\pi}} \int_0^\tau e^{-z^2}\, dz.\]

[21] This is a Fourier transform. Markov chose not to use the symbol \(i\) to represent \(\sqrt{-1}\), although that notation had been around for at least a century.

[22] He is using the facts that \(e^\frac{ix}{2} - e^{- \frac{ix}{2}} = 2i \sin\left(\frac{x}{2}\right)\) and \(\cos(x) = \mathrm{Re}\,(e^{ix})\).


Alan Levine (Franklin and Marshal College), "A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 8 – Solution, Part 2," Convergence (November 2023)

A Selection of Problems from A.A. Markov’s Calculus of Probabilities