A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 8 – Solution, Part 1

Review statement of Problem 8.

Solution:  Letting \(n= 1,2,3,\dots\) consecutively, we arrive at the conclusion that, for any value of \(n\), the probability of the equation \(X_1 + X_2 + \cdots + X_n = \alpha\), where \(\alpha\) is a given number, can be determined as the coefficient of \(t^\alpha\) in the expansion of the expression \[\left\lbrace\frac{t + t^2 + \cdots + t^m}{m} \right\rbrace ^n\] in powers of the arbitrary number \(t\).¹⁷

On the other hand, we have \[\left\lbrace\frac{t + t^2 + \cdots + t^m}{m} \right\rbrace ^n = \frac{t^n}{m^n} \frac{(1-t^m)^n}{(1-t)^n} = \] \[ = \frac{t^n}{m^n} \left[ 1 - n t^m + \frac{n(n-1)}{1\cdot 2} t^{2m} - \cdots\right] \left[1 + nt + \frac{n(n+1)}{1\cdot 2} t^2 + \frac{n(n+1)(n+2)}{1\cdot 2\cdot 3} t^3 + \cdots \right]. \]

Therefore, denoting the probability of the equation \(X_1 + X_2 + \cdots + X_n = \alpha\) by the symbol \(P_\alpha\), we can establish the formula: \[m^n P_\alpha = \frac{n(n+1) \cdots (\alpha-1)}{1 \cdot 2 \cdot \cdots \cdot (\alpha-n)} - \frac{n}{1} \frac{n(n+1) \cdots (\alpha-m-1)}{1 \cdot 2 \cdot \cdots \cdot (\alpha-n- m)} + \frac{n(n-1)}{1\cdot 2} \frac{n(n+1) \cdots (\alpha-2m-1)}{1 \cdot 2 \cdot \cdots \cdot (\alpha-n-2m)} - \cdots,\] which represents an easy way of calculating \(P_\alpha\) for small values of \(\alpha\).  

It is also not hard to show the equation \(P_{\alpha} = P_{n(m+1) - \alpha}\), which allows us to substitute the number \(\alpha\) for the difference \(n(m+1) - \alpha\) and, in this way, gives the possibility of decreasing \(\alpha\), if \(\alpha > \frac{n(m+1)}{2}\).

Continue to Markov's numerical example for Problem 8.

Skip to the second part of Markov's solution of Problem 8.

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

Skip to Markov's analysis of repeated independent events.

[17] Another instance of the use of probability generating functions. This conclusion is correct because he has assumed that, for each \(n\), \(P(X_n = i) = \frac{1}{m}\), for \(i = 1,2,\dots m\).