$\sum_{i=0}^{\lceil zk \rceil}{k\choose i} p^i(1-p)^{k-i}$
$z, p \in [0,1]$
I am looking to find the limit as $k$ tends to infinity but don't know how I would do this
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1$\begingroup$ What does this mean if $z$ is, say, $\frac1\pi$? $\endgroup$– José Carlos SantosCommented Jul 20, 2017 at 12:40
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$\begingroup$ Yes, I should add the ciel function onto zk $\endgroup$– Henry McKayCommented Jul 20, 2017 at 12:48
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$\begingroup$ It would help to look at a derivation for the cumulative distribution function of the binomial distribution. Edit: Or approximations of the binomial distribution for large k. $\endgroup$– HughCommented Jul 20, 2017 at 13:03
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$\begingroup$ I'm not sure how rigorous of a proof this would lead to, but Newton's binomial theorem implies that for $z=1$ the sum converges to 1. That might have already been known to you, but it's a start maybe. $\endgroup$– Chessnerd321Commented Jul 20, 2017 at 13:20
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2 Answers
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If $\{X_i\}$ is an iid sequence of Bernoulli random variables with $P(X_i=1)=p$ and $S_k=\sum_{i=1}^kX_i$, then
$$\sum_{i=0}^{\lceil zk\rceil}{k\choose i}p^k(1-p)^{k-i}=P(S_k\le\lceil zk\rceil).$$
By the law of large numbers, $S_k/k\to p$ (almost surely or in probability). This implies that the limit is zero if $z<p$ and one if $z>p$. The case $z=p$ is more delicate and may require a local central limit theorem.
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$\begingroup$ The case $z=p$ is indeed more delicate but the simplest version of the CLT (not "local") suffices to solve it. $\endgroup$– DidCommented Jul 24, 2017 at 12:39
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Just a possible hint, too long for a comment. The cumulative distribution function for the binomial distribution $P(s; k, p)$ (probability of $s$ or less successes over $k$ trials, each with a probability $p$ of success), is given by $$ P(s; k, p) = \sum_{i=0}^{\lfloor s \rfloor} = {k \choose i} p^{i} (1-p)^{k-1}$$ Your expression has then a straightforward interpretation: the probability of no more than $\lfloor zk \rfloor$ successes over $k$ trials. One then considers the fact that the cumulative binomial tends to a normal distribution with mean $kp$ and variance $kp(1-p)$ in the limit $k \to \infty$. Hope it is of some help.
