How to interpret the mean of geometric distributionWhen I read my textbook, I find one conclusion which states that if the mean of geometric random variable is finite, then with probability 1, we can get the first success within "finite" step.I am confused about the "finite" part. Why there is no chance that we will never get success? Since the mean of geometric random variable is 1/p, doesn't it mean we can always get success within finite experiments if p &gt; 0?

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How to interpret the mean of geometric distributionWhen I read my textbook, I find one conclusion which states that if the mean of geometric random variable is finite, then with probability 1, we can get the first success within &quot;finite&quot; step.I am confused about the &quot;finite&quot; part. Why there is no chance that we will never get success? Since the mean of geometric random variable is 1/p, doesn&#039;t it mean we can always get success within finite experiments if   p  &amp;gt;  0?

Allyson Vance · Accepted Answer

Step 1Probabilities are weights assigned to events, subsets of outcomes. Some events have zero weight.Step 2For example, let&#039;s consider the event A = &quot;no success&quot; = &quot;no success in any finite number of steps&quot;. This event is contained in the event       A    n   = &quot;no success in the first n experiments&quot;. The probability of the latter is   (  1  −  p      )    n  . Therefore by monotonicy of probabilities,   P  (  A  )  ≤  P  (      A    n    )  =  (  1  −  p      )    n   for every n. If   p  &amp;gt;  0, this implies   P  (  A  )  =  0, that is, the weight assigned to &quot;no success&quot; is zero, or, equivalently, the weight assigned to &quot;success in (some) finite time&quot; is 1.

How to interpret the mean of geometric distribution. When I read my textbook, I find one conclusion which states that if the mean of geometric random variable is finite, then with probability 1, we can get the first success within "finite" step.

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