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

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

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.

[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 $i = 1,2,\dots m$.