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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 经济代写|博弈论代写Game Theory代考|Statistical frequencies

Assume that the gambler of the previous sections has observed:

• The event $A_i$ has popped up $s_i$ times in $n$ consecutive instances of the bet.

So, under the strategy $a$, the original portfolio $B$ of unit size $b=1$ would have developed into size
$$B_n(a)=\left(a_0 r_0\right)^{s_0}\left(a_1 r_1\right)^{s_1} \cdots\left(a_{k-1} r_{k-1}\right)^{s_{k-1}}$$ with the logarithmic utility
\begin{aligned} U_n(a)=\ln B_n(a) &=\sum_{i=0}^{k-1} s_i \ln \left(a_i r_i\right) \ &=\sum_{i=0}^{k-1} s_i \ln a_i+\sum_{i=0}^{k-1} s_i \ln r_i . \end{aligned}
As in the proof of Theorem 4.1, the gambler finds in hindsight:
COROLLARY 4.1. The strategy $a^=\left(s_0 / n, \ldots, s_{k-1} / n\right)$ would have led to the maximal logarithmic utility value $$U_n\left(a^\right)=\sum_{i=0}^{k-1} s_i \ln \left(s_i / t\right)+\sum_{i=0}^{k-1} p_i \ln r_i$$
and hence to the growth with the maximal geometric growth rate
$$B_n\left(a^*\right)=\frac{\left(s_0 r_0\right)^{s_0}\left(s_1 r_1\right)^{s_1} \cdots\left(s_{k-1} r_{k-1}\right)^{s_{k-1}}}{n^n} .$$
Based on the observed frequencies $s_i$ the gambler might reasonably estimate the events $A_i$ to occur with probabilities according to the relative frequencies
$$p_i=s_i / n \quad(i=0,1, \ldots, k-1)$$
and then expect optimal geometric growth.

## 经济代写|博弈论代写Game Theory代考|Betting and information

Assuming a betting situation with the $k$ alternatives $A_0, A_1, \ldots$, $A_{k-1}$ and the odds $\rho_x: 1$ (for $x=0,1, \ldots, k-1$ ) as before, suppose, however, that the event $A_x$ is actually already established – but that the bettor does not have this information before placing the bet.
Suppose further that information now arrives through some (human or technical) communication channel $K$ so that the outcome $A_x$ is reported to the bettor (perhaps incorrectly) as $A_y:$
$$x \rightarrow K^{\prime} \rightarrow y .$$
The question is:

• Having received the (“insider”) information ” $y$ “, how should the bettor place the bet?
$p(x \mid y)=$ probability for the true result to be $x$ when $y$ is received.
REMARK 4.4. The parameters $p(x \mid y)$ are typically subjective evaluations of the bettor’s trust in the channel $K$.

A betting strategy in this setting of information transmission becomes a $(k \times k)$-matrix $A$ with coefficients $a(x \mid y) \geq 0$ which satisfy
$$\sum_{x=0}^{k-1} a(x \mid y)=1 \quad \text { for } y=0,1, \ldots, k-1 .$$
According to strategy $A, a(x \mid y)$ would be the fraction of the budget that is bet on the event $A_x$ when $y$ is received. In particular, the bettor’s trust matrix $P$ with the coefficients $p(x \mid y)$ is a strategy. In the case where $A_x$ is the true result, one therefore expects the logarithmic utility
$$U_x(A)=\sum_{y=0}^{k-1} p(x \mid y) \ln \left[a(x \mid y) r_x\right]=\sum_{y=0}^{k-1} p(x \mid y) \ln a(x \mid y)+\ln r_x$$

## 经济代写|博弈论代写博弈论代考|统计频率

• 事件$A_i$在连续$n$实例中出现了$s_i$次。

$$B_n(a)=\left(a_0 r_0\right)^{s_0}\left(a_1 r_1\right)^{s_1} \cdots\left(a_{k-1} r_{k-1}\right)^{s_{k-1}}$$，对数效用
\begin{aligned} U_n(a)=\ln B_n(a) &=\sum_{i=0}^{k-1} s_i \ln \left(a_i r_i\right) \ &=\sum_{i=0}^{k-1} s_i \ln a_i+\sum_{i=0}^{k-1} s_i \ln r_i . \end{aligned}

，因此以最大的几何增长率增长
$$B_n\left(a^*\right)=\frac{\left(s_0 r_0\right)^{s_0}\left(s_1 r_1\right)^{s_1} \cdots\left(s_{k-1} r_{k-1}\right)^{s_{k-1}}}{n^n} .$$

$$p_i=s_i / n \quad(i=0,1, \ldots, k-1)$$
，然后期望最佳的几何增长率

## 经济代写|博弈论代写博弈论代考|博彩和信息

.

$$x \rightarrow K^{\prime} \rightarrow y .$$

• 在收到(“内幕”)信息“$y$”后，投注者应该如何投注?
要回答这个问题，让
$p(x \mid y)=$当接收到$y$时，真实结果为$x$的概率。4.4.
参数$p(x \mid y)$通常是投注者对频道$K$的信任的主观评价。

$$\sum_{x=0}^{k-1} a(x \mid y)=1 \quad \text { for } y=0,1, \ldots, k-1 .$$
。根据策略，$A, a(x \mid y)$将是当$y$被接收时对事件$A_x$下注的预算的一部分。特别是，投注者的信任矩阵$P$，其系数为$p(x \mid y)$是一种策略。在$A_x$是真实结果的情况下，人们因此期望对数效用
$$U_x(A)=\sum_{y=0}^{k-1} p(x \mid y) \ln \left[a(x \mid y) r_x\right]=\sum_{y=0}^{k-1} p(x \mid y) \ln a(x \mid y)+\ln r_x$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师