Discrepância entre cara e coroa

Considere uma sequência de $n$ lançamentos de uma moeda imparcial. Deixe $H_i$ denotar o valor absoluto do excesso do número de cabeças sobre caudas vistas nos primeiros $i$ flips. Defina $H=\text{max}_i H_i$ . Mostre que $E[H_i]=\Theta ( \sqrt{i} )$ e $E[H]=\Theta( \sqrt{n} )$ .

Esse problema aparece no primeiro capítulo de `Algoritmos aleatórios 'de Raghavan e Motwani, portanto, talvez haja uma prova elementar da afirmação acima. Não consigo resolvê-lo, por isso gostaria de receber qualquer ajuda.

probability-theory randomized-algorithms

— Plummer
fonte

Respostas:

Seu coin flips formar uma caminhada unidimensional aleatória a partir de , com , cada uma das opções com probabilidade . Agora e então . É fácil calcular $X_0,X_1,\ldots$ $X_0 = 0$ $X_{i+1} = X_i \pm 1$ $1/2$ $H_i = |X_i|$ $H_i^2 = X_i^2$ (esta é apenas a variação), e então $E[X_i^2] = i$ de convexidade. Sabemos também queé distribuído aproximadamente normal com média zero e variância, e assim você pode calcular $E[H_i] \leq \sqrt{E[H_i^2]} = \sqrt{i}$ $X_i$ $i$ . $E[H_i] \approx \sqrt{(2/\pi)i}$

Quanto a , temos a lei do logaritmo iterado , que (talvez) nos leva a esperar algo um pouco maior que $E[\max_{i \leq n} H_i]$ . Se você é bom com um limite superior de $\sqrt{n}$ , você pode usar um grande desvio com destino a cadae depois a união obrigado, no entanto, que ignora o fato de que oestão relacionados. $\tilde{O}(\sqrt{n})$ $X_i$ $X_i$

Edit: Por acaso, devido ao princípio de reflexão, consulte esta pergunta . Então $\Pr[\max_{i \leq n} X_i = k] = \Pr[X_n = k] + \Pr[X_n = k+1]$ uma vez que. Agora

\begin{aligned} E [max_{i \leq n} X_{i}] & = \sum_{k \geq 0} k (Pr [X_{n} = k] + Pr [X_{n} = k + 1]) \\ = \sum_{k \geq 1} (2 k - 1) Pr [X_{n} = k] \\ = \sum_{k \geq 1} 2 k Pr [X_{n} = k] - \frac{1}{2} + \frac{1}{2} Pr [X_{n} = 0] \\ = E [H_{n}] + Θ (1), \end{aligned}

$\begin{align*} E[\max_{i \leq n} X_i] &= \sum_{k \geq 0} k(\Pr[X_n = k] + \Pr[X_n = k+1]) \\ &= \sum_{k \geq 1} (2k-1) \Pr[X_n = k] \\ &= \sum_{k \geq 1} 2k \Pr[X_n = k] - \frac{1}{2} + \frac{1}{2}\Pr[X_n = 0] \\ &= E[H_n] + \Theta(1), \end{align*}$

Pr [H_{n} = k] = Pr [X_{n} = k] + Pr [X_{n} = - k] = 2 Pr [X_{n} = k]

$\Pr[H_n = k] = \Pr[X_n = k] + \Pr[X_n = -k] = 2\Pr[X_n = k]$

portanto,

\frac{max_{i \leq n} X_{i} + max_{i \leq n} (- X_{i})}{2} \leq max_{i \leq n} H_{i} \leq max_{i \leq n} X_{i} + max_{i \leq n} (- X_{i}),

$\frac{\max_{i \leq n} X_i + \max_{i \leq n} (-X_i)}{2} \leq \max_{i \leq n} H_i \leq \max_{i \leq n} X_i + \max_{i \leq n} (-X_i),$

. A outra direção é semelhante.

E [max_{i \leq n} H_{i}] \leq 2 E [H_{n}] + Θ (1) = O (\sqrt{n})

$E[\max_{i \leq n} H_i] \leq 2E[H_n] + \Theta(1) = O(\sqrt{n})$

— Yuval Filmus
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E [H_{i}] = Θ (\sqrt{i})

$E[H_{i}] = \Theta ( \sqrt{i} )$ , couldn't we say that for

i = n

$i=n$ we have the second result, i.e. no

E [H_{i}]

$E[H_{i}]$ is greater than

Θ (\sqrt{n})

$\Theta ( \sqrt{n} )$ .

— chazisop

If the

H_{i}

$H_i$ were independent, then the conclusion won't be true, since you actually expect some of these values to be somewhat larger than the expectation. It's not true in general that

E [max (A, B)] = max (E [A], E [B])

$E[\max(A,B)] = \max(E[A],E[B])$ .

— Yuval Filmus

The law of the iterated logarithm doesn't apply here, since

n

$n$ is fixed and we're not normalizing by

i

$i$ . The answer for

E max_{i \leq n} H_{i}

$\mathrm{E} \max_{i\leq n} H_i$ is

θ (\sqrt{n})

$\theta(\sqrt{n})$ .

— Peter Shor

+1 for the first part. but I honestly dont understand the second part (can you elaborate more plz). This does not mean that it is not correct though.

— AJed

Nice proof. But I am stuck at how to prove

\sqrt{n}

$\sqrt{n}$ is the lower bound of

E (H_{i})

$E(H_i)$ ? It seems that the answer mentions no details about the lower bound.

— konjac

You can use the half-normal distribution to prove the answer.

The half-normal distribution states that if $X$ is a normal distribution with mean 0 and variance $\sigma^2$ , then $|X|$ follows a half distribution with mean $\sigma \sqrt{\frac{2}{\pi}}$ , and variance $\sigma^2 (1-2/\pi)$ . This gives the required answer, since the variance $\sigma^2$ of the normal walk is $n$ , and you can approximate the distribution of $X$ to a normal distribution using the central limit theorem.

$X$ is the sum of the random walk as Yuval Filmus mentioned.

— AJed
fonte

I dont prefer this that I posted.. . although it gives the lower bound, nothing can be told about the upper bound. I tried to use a maximum distribution argument to solve it, it turned out to be an ugly integration. But it is good to know all these distributions.

— AJed

In first $2i$ flips, suppose we get $k$ tails, then $H_{2i}=2|i-k|$ . Therefore,

\begin{aligned} E (H_{2 i}) & = 2 \sum_{k = 0}^{i} (\binom{2 i}{k}) (\frac{1}{2})^{2 i} \cdot 2 (i - k) \\ = (\frac{1}{2})^{2 i - 2} [i \sum_{k = 0}^{i} (\binom{2 i}{k}) - \sum_{k = 0}^{i} k (\binom{2 i}{k})] \\ = (\frac{1}{2})^{2 i - 2} [i (2^{2 i} + (\binom{2 i}{i})) / 2 - 2 i \sum_{k = 0}^{i - 1} (\binom{2 i - 1}{k})] \\ = (\frac{1}{2})^{2 i - 2} \cdot i [2^{2 i - 1} + (\binom{2 i}{i}) / 2 - 2 \cdot 2^{2 i - 1} / 2] \\ = 2 i \cdot (\binom{2 i}{i}) / 2^{2 i} . \end{aligned}

$\begin{align} E(H_{2i}) & =2\sum_{k=0}^{i}\binom{2i}{k}(\frac{1}{2})^{2i}\cdot 2(i-k)\\ & = (\frac{1}{2})^{2i-2}\left[i\sum_{k=0}^{i}\binom{2i}{k}-\sum_{k=0}^ik\binom{2i}{k}\right]\\ & = (\frac{1}{2})^{2i-2}\left[i\left(2^{2i}+\binom{2i}{i}\right)/2-2i\sum_{k=0}^{i-1}\binom{2i-1}{k}\right]\\ & = (\frac{1}{2})^{2i-2}\cdot i\left[2^{2i-1}+\binom{2i}{i}/2-2\cdot 2^{2i-1}/2\right]\\ & = 2i\cdot\binom{2i}{i}/2^{2i}. \end{align}$ Use Stirling's approximation, we know

E (H_{2 i}) = Θ (\sqrt{2 i})

$E(H_{2i})=\Theta(\sqrt{2i})$ .

— Aqua
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shouldn't we take into account cases where

i < k \leq 2 i

$i<k\leq 2i$ ? it seems that you miss a multiplication factor of 2, right?

— omerbp