Maths Olympiad Prep

Track / Stage 6 / 76 of 400 #1076 of 1964

Problem 1076

National olympiad, first round
Algebra Difficulty 6.1 Prove it

Example 2. Prove that

∑k=0∞pk=∑k=0∞ake−ak!=1 \sum_{k=0}^{\infty} p_{k}=\sum_{k=0}^{\infty} \frac{a^{k} e^{-a}}{k!}=1

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution. Taking into account the power series expansion of the function f(x)=exf(x)=e^{x}

ex=1+x1!+x22!+x33!+…+xkk!+… e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots+\frac{x^{k}}{k!}+\ldots

and the resulting equality

ea=1+a1!+a22!+a33!+…+akk!+… e^{a}=1+\frac{a}{1!}+\frac{a^{2}}{2!}+\frac{a^{3}}{3!}+\ldots+\frac{a^{k}}{k!}+\ldots

we obtain

∑k=0∞ake−ak!=e−a∑k=0∞akk!=e−a(1+a1!+a22!+a33!+…+akk!+…)=e−aea=1 \sum_{k=0}^{\infty} \frac{a^{k} e^{-a}}{k!}=e^{-a} \sum_{k=0}^{\infty} \frac{a^{k}}{k!}=e^{-a}\left(1+\frac{a}{1!}+\frac{a^{2}}{2!}+\frac{a^{3}}{3!}+\ldots+\frac{a^{k}}{k!}+\ldots\right)=e^{-a} e^{a}=1

Thus, the series of probabilities of the Poisson distribution converges and its sum is equal to one, i.e., condition (2.1.4) in the definition of the distribution law of a discrete random variable is satisfied.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.