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Sep 2, 2015 at 19:38 history edited user76479 CC BY-SA 3.0
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Aug 22, 2015 at 0:50 comment added user76479 Instead of 1000 I took 499 or 498. I used mathematica.
Aug 19, 2015 at 15:40 comment added Max Alekseyev @Arul: I do not understand what is 499 in here. Can you give numerical (integer) values of each of the parameters when the expression becomes negative? Just in case here is my PARI/GP code: { pr(T,S,N,k) = sum(i=0,T, (-1)^i * binomial(T,i) * binomial(T-i+S,N)^k ) / binomial(T+S,N)^k }
Aug 19, 2015 at 2:26 comment added user76479 It is $>0$ at $499$, $<0$ at $498$.
Aug 19, 2015 at 2:19 comment added user76479 I tried $k=Log2[1000]$, $T=N=(S+T)/k=1000/k$ in mathematica. Infact it goes to $<0$ if $10000$ replaces $1000$.
Aug 19, 2015 at 2:06 comment added Max Alekseyev @Arul: What are your numerical values? The formula follows from the inclusion-exclusion principle.
Aug 19, 2015 at 2:01 history edited user76479 CC BY-SA 3.0
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Aug 19, 2015 at 1:55 history edited user76479 CC BY-SA 3.0
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Aug 19, 2015 at 1:44 comment added user76479 @MaxAlekseyev I am not getting prob <1? Could you write your proof?
Aug 19, 2015 at 1:32 history edited user76479 CC BY-SA 3.0
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Aug 19, 2015 at 1:21 comment added user76479 @Henry Yes. $N=1$ is standard. What about $N>1$?
Aug 18, 2015 at 20:16 comment added Henry If $N=1$ then the expectation is $(S+T)H_T$
Aug 18, 2015 at 18:42 comment added Max Alekseyev Not an asymptotic but exact answer for question 1: $$\binom{T+S}{N}^{-k} \sum_{i=0}^T (-1)^i \binom{T}{i} \binom{T-i+S}{N}^k.$$
Aug 18, 2015 at 15:28 history asked user76479 CC BY-SA 3.0