The following partial problem statement is from J. Blitzstein, Pg 216, Problem 32, Introduction to Probability.

"A discrete distribution has the memoryless property if for X a random variable with that distribution, P(X >= j + k | X >= j) = P(X >= k) for all nonnegative integers j; k."

My question is:

Why is P(X >= j + k | X >= j) = P(X >= k) but not equal to "1" always?

I ask this because, since X is given to be non-negative. If X>=j, then X is ALWAYS >= j+k.

Where is my rationale flawed? Please help.

  • 1
    $\begingroup$ If you would be correct then we should have $j\geq j+k$ always. This while $k$ is nonnegative... $\endgroup$
    – drhab
    Jul 20, 2023 at 6:35

1 Answer 1


BTW, would recommend that you try to typeset this in latex in the future.

To see why your logic is flawed, suppose that $j = 2$ and $k = 5$. You're saying that $$\{ X \ge 2\} \Rightarrow \{ X \ge 7\}.$$ This doesn't make sense; for example, with positive probability, $X = 6,$ in which case your implication is not true.


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