条件期望：Conditional Expectation 举例详解之入门之入门之草履虫都说听懂了( 二 ) _生活百科

\[\mathbb{E}\left[ X ~ | ~ \mathcal{G} \right](w) = \mathbb{E}\left[ X ~ | ~ G_{n} \right] = \frac{\mathbb{E}\left[ X \cdot \mathbb{I}_{G_{n}} \right]}{P(G_{n})} \qquad \qquad \mbox{if } w \in G_{n}\]首先，\(\mathbb{I}_{G_{n}}\)是一个随机变量，或者说函数：
\[\mathbb{I}_{G_{n}}: \Omega \longrightarrow \left\{ 0, 1 \right\}, \quad x \longrightarrow \mathbb{I}_{G_{n}}(x) = \begin{cases}1 \qquad \mbox{if } x \in G_{n}\\0 \qquad \mbox{otherwise}\end{cases}\]因此则可以判定，Conditional Expectation \(\mathbb{E}\left[ X ~ | ~ \mathcal{G} \right]\) 算出来也是一个随机变量，而并非常数。最后，我们可以发现一旦假设 \(w \in G_{n}\)，那么一定意味着 \(w \notin G_{k}, ~ \forall k \in \mathbb{N}^{+}\setminus\left\{n\right\}\) 。
回到扔硬币的例子。这里显然我们有：\(G_{1} = \left\{ HH, HT \right\}, ~ G_{2} = \left\{ TT, TH \right\}\)，且 \(G_{1} \cup G_{2} = \Omega\) 。那么。我们现在只需要依次：假设 \(w \in G_{n}\) 并求 \(\frac{\mathbb{E}\left[ X \cdot \mathbb{I}_{G_{n}} \right]}{P(G_{n})}\)，最后将所有所求结果相加即可。
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假设 \(w \in G_{1} = \left\{ HH, HT \right\}\)，

\[ \begin{align*} \mathbb{E}\left[ X ~ | ~ \mathcal{G} \right](w) &= \frac{\mathbb{E}\left[ X \cdot \mathbb{I}_{G_{1}}, ~ w \in G_{1} \right]}{P(G_{1})}\\ &= \frac{\sum\limits_{w \in G_{1}}\mathbb{E}\left[ X \cdot \mathbb{I}_{G_{1}} ~ | ~ w \in G_{1} \right] \cdot P\big(\left\{ w \right\}\big)}{P(G_{1})}\\ &= \frac{\sum\limits_{w \in G_{1}} X(w) \cdot P\big(\left\{ w \right\}\big)}{P(G_{1})}\\ & = \frac{X(HH) \cdot P\big( \left\{ HH \right\} \big) + X(HT) \cdot P\big( \left\{ HT \right\} \big)}{P\big( \left\{ HH, HT \right\} \big)}\\ & = \frac{\frac{1}{4} \cdot a + \frac{1}{4} \cdot b}{\frac{1}{2}}\\ & = \frac{a + b}{2} \end{align*}\]

假设 \(w \in G_{2} = \left\{ TT, TH \right\}\)，

\[ \begin{align*} \mathbb{E}\left[ X ~ | ~ \mathcal{G} \right](w) &= \frac{\mathbb{E}\left[ X \cdot \mathbb{I}_{G_{2}}, ~ w \in G_{2} \right]}{P(G_{2})}\\ &=\frac{\sum\limits_{w \in G_{2}}\mathbb{E}\left[ X \cdot \mathbb{I}_{G_{2}} ~ | ~ w \in G_{2} \right] \cdot P\big(\left\{ w \right\}\big)}{P(G_{2})}\\ &= \frac{\sum\limits_{w \in G_{2}} X(w) \cdot P\big(\left\{ w \right\}\big)}{P(G_{2})}\\ & = \frac{X(TT) \cdot P\big( \left\{ TT \right\} \big) + X(TH) \cdot P\big( \left\{ TH \right\} \big)}{P\big( \left\{ TT, TH \right\} \big)}\\ & = \frac{\frac{1}{4} \cdot c + \frac{1}{4} \cdot d}{\frac{1}{2}}\\ & = \frac{c + d}{2} \end{align*}\]综上所述：
\[\mathbb{E}\left[ X ~ | ~ \mathcal{G} \right](w) = \begin{cases}\frac{a + b}{2} \qquad \mbox{if } ~ w \in \left\{ HH, HT \right\}\\\frac{c + d}{2} \qquad \mbox{if } ~ w \in \left\{ TT, TH \right\}\\\end{cases}\]