Let X Geom(p). For positive integers n, k define
P(X = n + k | X > n) = P(X = n + k) / P(X > n) :
Show that P(X = n + k | X > n) = P(X = k) and then briefly argue, in words, why this is true for geometric random variables.
As geometric distribution is defined as number of trails needed to get the first success. Hence, it does not depends on the past trails whether the next outcome is a success or not, that is, geometric distribution has memoryless property. This is why the above result is true for geometric distribution.
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