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If $X \sim \mathrm{Geometric}(\theta)$, then prove $P(X=n+m \mid X>n)=P(X= m)$

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This is a consequence of the memoryless property of geometric random variables.

$P(X=n+m|X>n) = \frac{P(X=n+m,X>n)}{P(X>n)}$

$= \frac{P(X=n+m)}{P(X>n)} = \frac{(1-\theta)^{m+n-1}\theta}{(1-\theta)^n}= (1-\theta)^{m-1}\theta = P(X=m)$ Note that I have not used the memoryless property here. Rather this proves it holds.