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The task is to proof that a discrete distribution is memoryless if and only if it is a geometric distribution.

I haven't wrote this proof myself, but I understand how to prove that a geometric distribution is memoryless:

$\operatorname{P}(X=n|X>m)=\frac{\operatorname{P}(X=n\wedge X>m)}{\operatorname{P}(X>m)}=\frac{\operatorname{P}(X=n)}{\operatorname{P}(X>m)}=\frac{(1-p)^{n-1}p}{(1-p)^m}=(1-p)^{n-m-1}=\operatorname{P}(X=n-m)$

But how to prove the other direction of the implication, that is, how to prove that if a discrete distribution is memoryless then it is a geometric distribution, under assumption that the associated random variable's values can only be natural?

gaazkam
  • 903

1 Answers1

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hint

For any given $k\ge 2,$ we have $$P(X=1)= P(X=k+1\mid X\ge k)\\ P(X=2)=P(X=k+1\mid X\ge k-1).$$ Dividing these equations and using the definition of conditional probability gives $$\frac{P(X\ge k)}{P(X\ge k-1)}=\frac{P(X=2)}{P(X=1)}.$$

Note that the RHS is independent of $k.$

spaceisdarkgreen
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