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I would like to compute the limit of CDF for a Binomial distribution as $n \rightarrow \infty$, \begin{equation*} \lim_{n \rightarrow \infty} F( \theta;n,q) = \lim_{n \rightarrow \infty }\sum_{k=0}^{\lfloor \theta n \rfloor} \binom{n}{k} q^k (1-q)^{n-k} \end{equation*}

An hint for this problem can be obtained in this link : The asymptotic behavior of the CDF of Binomial distribution $$ \lim_{n\to\infty}F(\theta;n,q) = \begin{cases}0 & \text{if } \theta < q\\1 & \text{if } \theta > q\end{cases}. $$ My question 1 is : Where and how to apply the Weak Law of Large Number to obtain the solution?

Further more, let $$G_n(\theta) = \sum_{k=0}^{\lfloor \theta n \rfloor} \binom{n}{k} \int_0^1 q^k (1-q)^{n-k} d W(q). $$

My Question 2 is how to prove that, $$ G_{\infty}(\theta) = W(\theta) \text{ in distribution.}$$

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James Foo
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  • weak law is aabout convergence in probability. strong law is about convergence almost surely. does convergence as imply convergence in distribution? Might this be helpful? https://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem – BCLC Dec 17 '15 at 13:33
  • @BCLC: thanks for your hints. I got that binomial can be approximated by normal. I have no idea how to apply LLN here to evaluate the sum. From my impression, LLN said that the average of a series iid r.v converges to their expect value. But, where to use it here? – James Foo Dec 17 '15 at 14:49
  • which sum? dunno about the second question hehehehe. why don't you just ask another question? – BCLC Dec 17 '15 at 14:55
  • I mean the $\sum_k P(Y=k)$, and how to use LLN to obtain its limit as n goes to infinity – James Foo Dec 17 '15 at 17:07
  • i guess you mean the weak LLN. Well cdf if $P(X \le x)$ but weak LLN has $P(X > x)$. How about $1 - P(X \le x)$? – BCLC Dec 18 '15 at 10:16

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