Binomial Sum limit

Calculus Level 5

$\large 100 \lim_{n \to \infty} \sum_{k=1}^{\left \lfloor \frac{n-\sqrt{n}}{2} \right \rfloor} \frac{\binom{50n}{k}\binom{50n}{n-k}}{\binom{100n}{n}} = \, ?$

The answer is 16.00.

1 solution

M M
Apr 25, 2019

Intuitive approach (not a proof):

The inner expression is the hypergeometric probability distribution with a population size 100n, where the population consists of 50n individuals from one category and 50n individuals from the other category, where (n) individuals have been drawn from the population.

In the limit, the hypergeometric distribution with the ratios of successes and failures fixed, approaches a Binomial distribution with n draws, with p=.5, which has a standard deviation of sqrt(n)/2. But the Binomial distribution, scaled, approaches the standard normal distribution. Thus this question is asking for the probability that an observation in the standard normal distribution falls one standard deviation below the mean (multiplied by 100). The probability associated with the z-score of -1 is approximately 15.8%, so the answer should be around 15.8.

The one place I'm not 100% on is the fact that the ratio of observations to population size stays fixed; I know that this approximation works for adequately small k, but I'd actually have to verify this with characteristic functions or something, which I'm not in the mood to do.

Binomial Sum limit

The answer is 16.00.

1 solution

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