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**Definition:** Given a [random experiment](/D/rexp), an event space $\mathcal{E}$ is any set of events, where an [event](/D/reve) is any set of zero or more elements from the [sample space](/D/samp-spc) $\Omega$ of this experiment.
**Definition:** Given a [random experiment](/D/rexp), a probability space $(\Omega, \mathcal{E}, P)$ is a triple consisting of
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* the [sample space](/D/samp-spc) $\Omega$, i.e. the set of all possible outcomes from this experiment;
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* an [event space](/D/eve-spc) $\mathcal{E} \subseteq 2^\Omega$, i.e. a set of subsets from the sample space, called [events](/D/reve);
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* a [probability measure](/D/prob-meas) $P: \; \mathcal{E} \rightarrow [0,1]$, i.e. a function mapping from the [event space](/D/eve-spc) to the real numbers, observing the [axioms of probability](/D/prob-ax).
**Definition:** Given a [random experiment](/D/rexp), the set of all possible outcomes from this experiment is called the sample space of the experiment. A sample space is usually denoted as $\Omega$ and specified using set notation.
**Theorem:** Let $X$, $Y$ and $Z$ be [random variables](/D/rvar) defined on the same [probability space](/D/prob-spc) and assume that the [covariance](/D/cov) of $X$ and $Y$ is finite. Then, the sum of the [expectation](/D/mean) of the conditional covariance and the [covariance](/D/cov) of the conditional expectations of $X$ and $Y$ given $Z$ is equal to the [covariance](/D/cov) of $X$ and $Y$:
**Theorem:** Let $X$ be a [random variable](/D/rvar) with [expected value](/D/mean) $\mathrm{E}(X)$ and let $Y$ be any [random variable](/D/var) defined on the same [probability space](/D/prob-spc). Then, the [expected value](/D/mean) of the [conditional expectation](/D/mean-cond) of $X$ given $Y$ is the same as the [expected value](/D/mean) of $X$:
**Proof:** Let $X$ and $Y$ be [discrete random variables](/D/rvar-disc) with sets of possible outcomes $\mathcal{X}$ and $\mathcal{Y}$. Then, the expectation of the conditional expetectation can be rewritten as:
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$$ \label{eq:mean-tot-s1}
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\begin{split}
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\mathrm{E}[\mathrm{E}(X \vert Y)] &= \mathrm{E}\left[ \sum_{x \in \mathcal{X}} x \cdot \mathrm{Pr}(X = x \vert Y) \right] \\
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&= \sum_{y \in \mathcal{Y}} \left[ \sum_{x \in \mathcal{X}} x \cdot \mathrm{Pr}(X = x \vert Y = y) \right] \cdot \mathrm{Pr}(Y = y) \\
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&= \sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} x \cdot \mathrm{Pr}(X = x \vert Y = y) \cdot \mathrm{Pr}(Y = y) \; .
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\end{split}
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$$
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Using the [law of conditional probability](/D/prob-cond), this becomes:
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$$ \label{eq:mean-tot-s2}
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\begin{split}
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\mathrm{E}[\mathrm{E}(X \vert Y)] &= \sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} x \cdot \mathrm{Pr}(X = x, Y = y) \\
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&= \sum_{x \in \mathcal{X}} x \sum_{y \in \mathcal{Y}} \mathrm{Pr}(X = x, Y = y) \; .
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\end{split}
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$$
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Using the [law of marginal probability](/D/prob-marg), this becomes:
**Theorem:** Let $X$ and $Y$ be [random variables](/D/rvar) defined on the same [probability space](/D/prob-spc) and assume that the [variance](/D/var) of $Y$ is finite. Then, the sum of the [expectation](/D/mean) of the conditional variance and the [variance](/D/var) of the conditional expectation of $Y$ given $X$ is equal to the [variance](/D/var) of $Y$:
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