## 统计代写|随机过程代写stochastic process代考|STATS217

2023年4月3日

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## 统计代写|随机过程代写stochastic process代考|Bayesian Tests of Hypotheses

The above example gives an opportunity to present an elementary version of the theory of hypothesis testing. We now interpret the set ${1,2, \ldots, k}$ as the possible states of Nature. The variable $Y$ is the actual random state of nature that is not directly available to the observer, who can however see the vector $X$. The observer devises a guessing strategy in terms of this observation. More precisely, this strategy consists in choosing a partition $\left{A_1, A_2, \ldots, A_k\right}$ of the range space $\mathbb{R}^n$ of $X$, the observer deciding that the state of Nature is $\widehat{Y}=i$ if $X \in A_i(1 \geq i \geq k)$. Of course there is a possibility that $\widehat{Y} \neq Y$, in which case one says that an error has occurred. The goal is to find the strategy (the partition) that minimizes this probability of error $P_E$ or, equivalently, maximizes $1-P_E$. The latter is given by a straightforward application of Bayes’ rules:
\begin{aligned} 1-P_E & =P(\widehat{Y}=Y) \ & =\sum_{i=1}^k P(\widehat{Y}=i \mid Y=i) P(Y=i) \ & =\sum_{i=1}^k P\left(X \in A_i \mid Y=i\right) P(Y=i) \ & =\sum_{i=1}^k \int_{A_i} f_i(x) \mathrm{d} x \times \pi_i \ & =\int_{\mathbb{R}^n} \sum_{i=1}^k 1_{A_i}(x) \pi_i f_i(x) \mathrm{d} x \end{aligned}
It is clear from this expression that any partition such that
$$x \in A_i \Longleftarrow \pi_i f_i(x) \geq \max _k \pi_k f_k(x)$$
is optimal.

## 统计代写|随机过程代写stochastic process代考|The General Theory

Previously, the conditioning was with respect to random variables or vectors. We now condition with respect to $\sigma$-fields.

Definition 3.3.16 Let $Y$ be an integrable (resp. finite non-negative) random variable, and let $\mathcal{G}$ be a sub- $\sigma$-field of $\mathcal{F}$. A version of the conditional expectation of $Y$ given $\mathcal{G}$ is any integrable (resp. finite non-negative) $\mathcal{G}$-measurable random variable $Z$ such that
$$\mathrm{E}[Y U]=\mathrm{E}[Z U]$$
for all bounded (resp. bounded non-negative) $\mathcal{G}$-measurable random variables $U$.
Theorem 3.3.17 Let $Y$ and $\mathcal{G}$ be as above. There exists at least one version of the conditional expectation of $Y$ given $\mathcal{G}$, and it is essentially unique, that is, if $Z^{\prime}$ is another version of the conditional expectation of $Y$ given $\mathcal{G}$, then $Z=Z^{\prime}, P$-a.s.

There will be no problem in representing two versions of this conditional expectation by the same symbol, since, as we just saw, they are $P$-almost surely equal. We choose the symbol $\mathrm{E}[Y \mid \mathcal{G}]$ or $\mathrm{E}^{\mathcal{G}}[Y]$ indifferently. From now on we say: $\mathrm{E}^{\mathcal{G}}[Y]$ (or $\mathrm{E}[Y \mid \mathcal{G}]$ ) is the conditional expectation of $Y$ given $\mathcal{G}$. The defining equality (3.40) reads
$$\mathrm{E}[Y U]=\mathrm{E}\left[\mathrm{E}^{\mathcal{G}}[Y] U\right]$$
for all bounded (resp. bounded non-negative) $\mathcal{G}$-measurable random variables $U$.
Proof. Uniqueness. The integrable case will be treated, the other case being similar. First observe that
$$0=\mathrm{E}[Z U]-\mathrm{E}\left[Z^{\prime} U\right]=\mathrm{E}\left[\left(Z-Z^{\prime}\right) U\right]$$
for all bounded non-negative $\mathcal{G}$-measurable $U$. In particular, with $U=1_{\left{Z>Z^{\prime}\right}}$,
$$\mathrm{E}\left[\left(Z-Z^{\prime}\right) 1_{\left{Z>Z^{\prime}\right}}\right]=0$$
Since the random variable in the expectation is non-negative, it can have a null expectation only if it is $P$-a.s. null, that is if $P$-a.s., $Z \leq Z^{\prime}$. By symmetry, $P$-a.s., $Z \geq Z^{\prime}$, and therefore, as announced, $Z=Z^{\prime}$, P-a.s.

# 随机过程代考

## 统计代写|随机过程代写stochastic process代考|Bayesian Tests of Hypotheses

mathrm ${E}[Y U]=I m a t h r m{E}[Z U]$
$\$ \$$对于所有有界 (分别为有界非负) \\mathcal{G} \$$-可测 量随机变量 $\$ U \$。$

$$\mathrm{E}[Y U]=\mathrm{E}\left[\mathrm{E}^{\mathcal{G}}[Y] U\right]$$

$$0=\mathrm{E}[Z U]-\mathrm{E}\left[Z^{\prime} U\right]=\mathrm{E}\left[\left(Z-Z^{\prime}\right) U\right]$$

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