Let m = 1 2 and n = 1 2 be the probabilities of success and failure respectively and let p(k) be a probability mass function such that p ( k ) = m ∑ i = 1 k n i + n ∑ i = 1 k m i .Is this a valid probability mass function?Say we carry out this test infinitely many times, does p ( k ) = 1?We recognise that the probability mass function is a sum of two slightly shifted geometric series. Let's consider one of the two (the other will follow the exact same argument).Say we are summing from 0 to ∞, that is let S 0 = ∑ i = 0 ∞ n i .Then this is simply evaluated as S 0 = 1 + n + n 2 + n 3 + ⋯ = 1 1 − n which is perfectly valid as n = 1 2 . If, then, I start this sequence at i = 1, is it valid to say that S 1 = ∑ i = 1 ∞ n i = 1 1 − n − 1 = 1 − ( 1 − n ) 1 − n = n 1 − n and as such that, as k → ∞, we have something like m n 1 − n + m n 1 − m = 1.Have I made any ridiculous logical jumps or is this valid?

Question

Let   m  =      1    2   and   n  =      1    2   be the probabilities of success and failure respectively and let p(k) be a probability mass function such that   p  (  k  )  =  m      ∑          i      =      1        k        n    i    +  n      ∑          i      =      1        k        m    i    .Is this a valid probability mass function?Say we carry out this test infinitely many times, does   p  (  k  )  =  1?We recognise that the probability mass function is a sum of two slightly shifted geometric series. Let&#039;s consider one of the two (the other will follow the exact same argument).Say we are summing from 0 to   ∞, that is let      S    0    =      ∑          i      =      0              ∞            n    i    .Then this is simply evaluated as       S    0    =  1  +  n  +      n    2    +      n    3    +  ⋯  =      1          1      −      n      which is perfectly valid as   n  =      1    2  . If, then, I start this sequence at   i  =  1, is it valid to say that       S    1    =      ∑          i      =      1        ∞        n    i    =      1          1      −      n        −  1  =            1      −      (      1      −      n      )              1      −      n        =      n          1      −      n      and as such that, as   k  →  ∞, we have something like            m      n              1      −      n        +            m      n              1      −      m        =  1.Have I made any ridiculous logical jumps or is this valid?

ahem37 · Accepted Answer

Step 1Using your formula we obtain   p  (  1  )  =      1    2   and   p  (  2  )  =      3    4  . Thus   p  (  1  )  +  p  (  2  )  &amp;gt;  1, which is impossible.Step 2If we define the distribution of a random variable X by saying that   Pr  (  X  ≤  k  ) is given by your sum, then we will have defined a legitimate distribution. The probability mass function is then given by   p  (  k  )  =  m      n    k    +  n      m    k  . Since   m  =  n  =      1    2  , it would be clearer to say   p  (  k  )  =      1          2      k       for every positive integer k.

Let m=1/2 and n=1/2 be the probabilities of success and failure respectively and let p(k) be a probability mass function such that p(k)=m \sum_{i=1}^{k}n^i+n \sum_{i=1}^{k} m^i.

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