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I have a simple question. Let

$$P(\theta;K) = \left(1-\theta\right)^K\left[\frac{1-(1-\theta)^K-\theta^K}{(1-\theta)^K+\theta^K}-\sum_{i=1}^{\frac{K-1}{2}}\left(\begin{array}{l} K \\ i \end{array}\right)\left(\frac{\theta}{1-\theta}\right)^i\right] $$

What are $$\displaystyle \lim_{K\rightarrow \infty}\max_\theta \; P(\theta;K)$$ and $$\displaystyle \DeclareMathOperator*{\argmax}{arg\,max} \lim_{K\rightarrow \infty}\argmax_\theta \; P(\theta;K) $$ where $\theta \in [0,\frac12]$ ?

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    I assume the maximum is to be taken with respect to $\theta$ (and have edited your question accordingly), but what's the range of $\theta$? $\mathbb R \setminus \{1\}$? $(0,1)$? Something else?2012-07-06
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    $\theta\in[0,0.5]$2012-07-06
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    Thank you Norbert and Ilmari2012-07-06
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    I assume $K$ should be odd, too, otherwise summing up to $\frac{K-1}2$ doesn't make much sense.2012-07-06
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    yes Ilmari K should be odd.2012-07-06

1 Answers 1

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Both limits are $\frac12$ (but not with the same meaning).

As in your other question, introducing a random variable $X_K$ with binomial $(K,\theta)$ distribution, one first rewrite $P(K;\theta)$ as $$ P(K;\theta)=\mathrm P(X_K\geqslant\tfrac12K)-\frac1{1+t_K(\theta)},\qquad t_K(\theta)=\left(\frac{1-\theta}\theta\right)^K. $$ For every $\theta\leqslant\frac12$, $\mathrm P(X_K\geqslant\frac12K)\leqslant\frac12$ hence $P(K;\theta)\lt\frac12$.

On the other hand, assume that $\theta=\theta_K(x)$ where $\theta_K(x)=\frac12\left(1-\frac{x}{\sqrt{K}}\right)$, for some fixed positive $x$. Then, $\frac12K=\mathrm E(X_K)+x_K\cdot\sigma(X_K)$ with $x_K=x/\sqrt{4\theta_K(x)(1-\theta_K(x))}\sim x$. The central limit theorem implies that $\mathrm P(X_K\geqslant\frac12K)\to\gamma(x)$ when $K\to\infty$, where $\gamma(x)$ is the probability that a standard normal random variable is $\geqslant x$. Since $t_K(\theta_K(x))\to\infty$ when $K\to\infty$, one gets $$ \liminf\limits_{K\to\infty}\left(\max\limits_{\theta\leqslant1/2}P(K;\theta)\right)\geqslant\lim\limits_{K\to\infty}P(K;\theta_K(x))=\gamma(x). $$ Since $\gamma(x)\to\frac12$ when $x\to0^+$, this proves the claim.

Edit: About the fact that $\lim\limits_{K\to\infty}\left(\max\limits_{\theta\leqslant1/2}P(K;\theta)\right)\ne0$ although $\lim\limits_{K\to\infty}P(K;\theta)=0$, for every $\theta\leqslant1/2$, which seems to bother the OP, here is a simpler, analogous situation: for every $0\leqslant\theta\leqslant1$, consider $$ Q(K;\theta)=K\cdot\theta^K\cdot(1-\theta). $$ Then $\lim\limits_{K\to\infty}\left(\max\limits_{0\leqslant\theta\leqslant1}Q(K;\theta)\right)=1/\mathrm e$ although $\lim\limits_{K\to\infty}Q(K;\theta)=0$ for every $\theta$ in $[0,1]$ (and in this case, all the computations can be made explicit). The graph of the functions $Q$ is quite similar to the one you draw for $P(K;\theta)$, where one sees distinctly the argument of the maximum of each function $P(K;\ )$ accumulate at $\theta=1$ when $K\to\infty$.

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    I am confused now. $$P(\theta;K) = \frac{1}{1+(\theta/(1-\theta))^K}-\left(1-\theta\right)^K\sum_{i=1}^{\frac{K-1}{2}}\left(\begin{array}{l} K \\ i \end{array}\right)\left(\frac{\theta}{1-\theta}\right)^i $$ The first term $\rightarrow 1$ as $K\rightarrow \infty$ and the second term is $P(X_k<(K-1)/2)\rightarrow 1$ as $K\rightarrow 1$. The difference is then $0$. Where is my mistake? I couldnt understand how you could write $P(K;\theta)=P(X_k\geq 0.5K)-\frac{1}{1+t_K(\theta)}$2012-07-08
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    Yes (except your sum should begin at $i=0$) and this is equivalent to the formula in my post: replace your first fraction by $1-\frac1{1+t_K(\theta)}$ and your sum on $i\leqslant\frac12(K-1)$ by $1$ minus the sum on $i\geqslant\frac12K$.2012-07-08
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    ok I understood this part, looking for the rest.2012-07-08
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    Ok I checked. I have the following link http://s14.directupload.net/file/d/2946/fa8ns99e_jpg.htm here I did some simulations.x- axis is not scaled it means $500\rightarrow 0.5$ Each curve corresponds to a specific $K$. If $K\rightarrow \infty$ then the maximum of the curve increases and corresponding $\theta$ also increases.The curve with smallest maximum is for $K=3$ and the largest for $K=100$.My problem is that when $K\rightarrow\infty$ if $\theta<0.5$ then $P(K;\theta)\rightarrow 0$.It seems like I will have 0.5 at $\theta=0.5$ when $K\rightarrow\infty$ but at $\theta=0.5$ $P(K;\theta)=0$2012-07-09
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    See Edit. $ $ $ $2012-07-09
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    Interesting. I am sorry. I don't know how to calculate such lim max. I am writing from mobile. Referring to your simple example $Q(K,\theta)$ could you please explain or give a easily understandable reference to read. I just learned limsup or liminf for sets. I guess it might be similar but still seems strange.2012-07-11
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    There is no limsup nor liminf (or limmax or limmin) in the picture, these are altogether other beasts. Here one computes a maximum $M_K$ (depending on K) then the limit of $M_K$ when $K\to\infty$. Surely you can compute $M_K=\max\limits_{0\leqslant\theta\leqslant1}Q(K;\theta)$ in the case at hand... (Added some parenthesis in the post.)2012-07-11
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    Okay I forgot the derivative of multiplication that's why I hadn't got a function based on K. Now I did it. The idea is that $\theta$ is no more constant when we have the maximum. That makes the difference. If I may ask. How long time have you been in this forum? I cannot imagine 50k points here. Sounds unbelievable.2012-07-11
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    $$\displaystyle \DeclareMathOperator*{\argmax}{arg\,max} \lim_{K\rightarrow \infty}\argmax_\theta \; P(\theta;K)=? $$2012-07-11
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    The post contains everything that is needed to show the limit of the argmax is $\frac12$. Try it...2012-07-11
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    $\mathrm P(X_K\geqslant\frac12K)\leqslant\frac12$ seems not exactly correct. See http://stattrek.com/online-calculator/binomial.aspx .2012-07-31
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    Good catch, when $K$ is even one needs to replace the upper bound $\frac12$ by $\frac12+\mathrm P(X_K=\frac12K)\leqslant\frac12+{K\choose K/2}2^{-K}$. This does not change the overall conclusion and I am sure you can modify the proof to take this corrected upper bound into account.2012-07-31
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    Probability of success on a single trial= 0.499, Number of trials =101 , Number of successes (x) = 50, Cumulative Probability: P(X >= 50)= 0.5709 but Cumulative Probability: P(X > 50) = 0.492.2012-07-31
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    And 0.492 is less than 0.5. We happy.2012-07-31
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    Your claim is not > but >= then 0.57 and we are unhappy.2012-07-31
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    Read better: the number of trials is 101, this is $K$, hence integer $\geqslant\frac12K$ means integer $\geqslant51$ or, equivalently, integer $\gt50$. We happy.2012-07-31
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    You re right ok. Thx.2012-07-31
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/4331/discussion-between-seyhmus-gungoren-and-did)2012-07-31
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    Ok finally I understood the proof completely. For me it needs too many intermediate steps. You @did so well;)2012-07-31
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    This is to retract my *Good catch* comment above: since you specified yourself that $K$ should be odd (Jul 6 at 20:23), your *seems not exactly correct* comment is in fact wrong.2012-08-03
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    yes. It is true.2012-08-04