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A Simple and Probabilistic Proof of the Binomial Theorem Andrew Rosalsky
a
a
Andrew Rosalsky is Professor, Department of Statistics, University of Florida, Gainesville, FL 32611 . The author is grateful to the Associate Editor for offering comments on the initial version and for pointing out the combinatorial proof of the binomial theorem in Ross (1998, pp. 8–9). Published online: 01 Jan 2012.
To cite this article: Andrew Rosalsky (2007) A Simple and Probabilistic Proof of the Binomial Theorem, The American Statistician, 61:2, 161-162, DOI: 10.1198/000313007X188397 To link to this article: http://dx.doi.org/10.1198/000313007X188397
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A Simple and Probabilistic Proof of the Binomial Theorem Andrew Rosalsky preferable to the standard proof since the new proof is far less technical and detail oriented than is the standard proof. This article presents a new and very simple proof of the binomial theorem. Although the binomial theorem in itself is not a probability result, the proof given is based on probabilistic concepts. KEY WORDS: Binomial distribution; Independent and identically distributed random variables; Principle of induction.
2. NEW PROOF OF THE BINOMIAL THEOREM Proof: By a straightforward induction argument, for all n ≥ 1, there exist positive integers C(n,0), . . . ,C(n,n) such that for all a,b ∈ , n
(a + b) =
n
C(n, j )a j bn−j .
(2)
j =0
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1. INTRODUCTION The binomial theorem is an important mathematical result providing the expansion of positive integer powers of sums of two terms. Its statement follows. Binomial theorem. For all integers n ≥ 1 and a,b ∈ , n
(a + b) =
n n j =0
j
a j bn−j .
(1)
If a or b is 0, then 00 is interpreted as 1. The standard proof of the binomial theorem involves a rather tricky argument using the principle of induction; see, for example, Courant and John (1999, pp. 59–60). A simpler proof of the binomial theorem (when a = 1) which also involved an induction argument was obtained by Fulton (1952); it is easy to see that the result for general a follows readily from the a = 1 result. A very nice proof of the binomial theorem based on combinatorial considerations was given by Ross (1998, pp. 8–9). This note presents a new proof of the binomial theorem which we believe is substantially simpler than both the standard proof and that of Fulton (1952) and somewhat simpler than the combinatorial proof in Ross (1998, pp. 8–9). The proof we give is probabilistic in nature although the binomial theorem in itself is not a probability result. The global idea behind our proof is: (i) first to establish by an elementary probability argument involving the binomial distribution the validity of the theorem in the special case 0 < a < 1 and b = 1 − a, and (ii) second to show that the validity of this special case of the theorem yields its validity for all a,b ∈ . Our proof is suitable for inclusion in either an introductory course in mathematical statistics or in a mathematical analysis course which is specifically designed for statistics students. Students having completed a course of the latter type have informed the author that they have found the proof presented below to be
Hence to establish (1) (i.e., to establish the general case described in (ii)), we need to show n C(n, j ) = , j = 0, . . . , n. (3) j To this end, let 0 < a < 1 and let Sn = ni=1 Xi where X1 , . . . ,Xn are independent and identically distributed random variables with P {X1 = 1} = a = 1 − P {X1 = 0}. The sum Sn has the binomial(n,a) distribution which by an elementary probability and counting argument (which does not make use of the binomial theorem) is given by n j P {Sn = j } = a (1 − a)n−j , j = 0, . . . , n j as is well known. Thus, 1=
n
P {Sn = j } =
j =0
n n j a (1 − a)n−j j
(4)
j =0
thereby establishing the theorem in the special case 0 < a < 1 and b = 1 − a described in (i). Next, by (2) 1 = (a + (1 − a))n =
n
C(n, j )a j (1 − a)n−j .
(5)
j =0
It then follows from (4) and (5) that n
C(n, j ) −
j =0
n a j (1 − a)n−j = 0 j
and, consequently, Andrew Rosalsky is Professor, Department of Statistics, University of Florida, Gainesville, FL 32611 (E-mail:
[email protected]). The author is grateful to the Associate Editor for offering comments on the initial version and for pointing out the combinatorial proof of the binomial theorem in Ross (1998, pp. 8–9).
c 2007 American Statistical Association DOI: 10.1198/000313007X188397
n j =0
j n a C(n, j ) − = 0. j 1−a
The American Statistician, May 2007, Vol. 61, No. 2 161
Since 0 < a < 1 is arbitrary, n j =0
n C(n, j ) − x j = 0, 0 < x < ∞. j
[Received July 2006. Revised December 2006.]
(6)
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The assertion (3) now follows immediately since the left-hand side of (6) is the zero polynomial and so all of its coefficients are zero.
162 Teacher’s Corner
REFERENCES Courant, R., and John, F. (1999), Introduction to Calculus and Analysis (Vol. I, reprint of 1989 ed.), Berlin: Wiley. Fulton, C. M. (1952), “A Simple Proof of the Binomial Theorem,” The American Mathematical Monthly, 59, 243–244. Ross, S. M. (1998), A First Course in Probability (5th ed.), Upper Saddle River, NJ: Prentice Hall.
