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I'm interested in estimating the following quantity

\begin{equation}\lim_{k \to \infty} \sum \limits_{i=1}^l \sum \limits_{l_1 = 0}^{i} {{nk-k}\choose{l_1}} \Big( \frac{1}{k}\Big)^{l_1} \Big( \frac{k-1}{k}\Big)^{nk - k -l_1} {{k}\choose{i-l_1}} \Big( \frac{l}{nk+1}\Big)^{i-l_1} \Big( \frac{nk+1-l}{nk+1}\Big)^{k-i + l_1} \end{equation}

where $l$ satisfies $1 \leq l \leq nk$ and $l_1, l_2$ satisfy $l=l_1+l_2$. The above sum arises in a problem concerning the sum of independent random variables, where $nk-k$ variables have image $\{0,k\}$ and $k$ variables have image $\{0, \frac{nk+1}{l} \}$ . Given that $l$ of these variables are non zero, we assume that there are $l_1$ non zero variables of the first kind and $l_2$ variables of the second kind, hence $l_1+l_2=l$. The probability of a variable with the first image of being non zero is $p_1=\frac{1}{k}$ and of the seond image $p_2=\frac{l}{nk+1}$. So the above quantity is the probability that at most $l$ variables are non zero.

Question: Is the above quantity greater or equal to $1/e$ ?

Any help or suggestions of techniques that might crack the problem would be much appreciated.

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  • $\begingroup$ @enthdegree could you please elaborate on your comment. $\endgroup$
    – sigmatau
    Commented Mar 14, 2020 at 9:44
  • $\begingroup$ It was just a term expansion I couldn't do anything with. (It also had a typo in the last exponent) $\endgroup$ Commented Mar 16, 2020 at 15:42

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