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Copy file name to clipboardExpand all lines: D/exc.md
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**Definition:** Generally speaking, [random events](/D/reve) are mutually exclusive, if the [probability](/D/prob) of their disjunction can be expressed in terms of their [marginal probabilities](/D/prob-marg).
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**Definition:** Generally speaking, [random events](/D/reve) are mutually exclusive, if they cannot occur together, such that their intersection is equal to the [empty set](/P/prob-emp).
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<br>
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More precisely, a set of statements $A_1, \ldots, A_n$ is called mutually exclusive, if
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$$ \label{eq:exc}
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p(A_1 \vee \ldots \vee A_n) = \sum_{i=1}^n p(A_i)
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p(A_1, \ldots, A_n) = 0
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$$
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where $p(A_1, \ldots, A_n)$ is the [probability](/D/prob) of the disjunction of $A_1, \ldots, A_n$ and $p(A_i)$ is the [marginal probability](/D/prob-marg) of $A_i$, for all $i = 1, \ldots, n$.
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where $p(A_1, \ldots, A_n)$ is the [joint probability](/D/prob-joint) of the statements $A_1, \ldots, A_n$.
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**Definition:** (law of marginal probability, also called "sum rule") Let $A$ and $X$ be two arbitrary statements about [random variables](/D/rvar), such as statements about the presence or absence of an event or about the value of a scalar, vector or matrix. Furthermore, assume a [joint probability](/D/prob-joint) distribution $p(A,X)$. Then, $p(A)$ is called the marginal probability of $A$ and,
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1) if $X$ is a discrete [random variable](/D/rvar) with domain $\mathcal{X}$, is given by
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1) if $X$ is a [discrete](/D/rvar-disc)[random variable](/D/rvar) with domain $\mathcal{X}$, is given by
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$$ \label{eq:prob-marg-disc}
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p(A) = \sum_{x \in \mathcal{X}} p(A,x) \; ;
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$$
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2) if $X$ is a continuous [random variable](/D/rvar) with domain $\mathcal{X}$, is given by
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2) if $X$ is a [continuous](/D/rvar-disc)[random variable](/D/rvar) with domain $\mathcal{X}$, is given by
Copy file name to clipboardExpand all lines: P/f-pdf.md
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author: "Joram Soch"
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affiliation: "BCCN Berlin"
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e_mail: "joram.soch@bccn-berlin.de"
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date: 2021-10-12 09:00
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date: 2021-10-12 09:00:00
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title: "Probability density function of the F-distribution"
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chapter: "Probability Distributions"
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**Proof:** An [F-distributed random variable](/D/f) is defined as the ratio of two [chi-squared random variables](/D/chi2), divided by their [degrees of freedom](/D/dof)
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$$ \label{eq:f-def}
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X \sim \chi^2(u), \; Y \sim \chi^2(v) \quad \Rightarrow \quad F = \frac{X/u}{\sqrt{Y/v}} \sim F(u,v)
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X \sim \chi^2(u), \; Y \sim \chi^2(v) \quad \Rightarrow \quad F = \frac{X/u}{Y/v} \sim F(u,v)
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$$
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where $X$ and $Y$ are [independent of each other](/D/ind).
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\end{split}
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$$
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The [marginal density](/D/dist-marg) of $F$ can now be [obtained by integrating out](/D/dist-marg) $W$:
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The [marginal density](/D/dist-marg) of $F$ can now be [obtained by integrating out](/D/prob-marg) $W$:
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**Theorem:** Let $A$ and $B$ be two statements about [random variables](/D/rvar). Then, if $A$ and $B$ are [mutually exclusive](/D/exc), their [joint probability](/D/prob-joint)is zero:
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**Theorem:** Let $A$ and $B$ be two statements about [random variables](/D/rvar). Then, if $A$ and $B$ are [mutually exclusive](/D/exc), the [probability](/D/prob) of their disjunction is equal to the sum of the [marginal probabilities](/D/prob-marg):
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$$ \label{eq:prob-exc}
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p(A,B) = 0 \; .
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p(A \vee B) = p(A) + p(B) \; .
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$$
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**Proof:** If $A$ and $B$ are [mutually exclusive](/D/exc), then the [probability](/D/prob) of their disjunction is the sum of the [marginal probabilities](/D/prob-marg):
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**Proof:** If $A$ and $B$ are [mutually exclusive](/D/exc), then their [joint probability](/D/prob-joint)is zero:
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$$ \label{eq:exc}
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p(A \vee B) = p(A) + p(B) \; .
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$$
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The [law of marginal probability](/D/prob-marg) implies that
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$$ \label{eq:prob-marg}
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\begin{split}
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p(A) &= p(A,B) + p(A,\overline{B}) \\
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p(B) &= p(A,B) + p(\overline{A},B)
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\end{split}
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p(A,B) = 0 \; .
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$$
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where $\overline{A}$ and $\overline{B}$ are the complements of $A$ and $B$. The probability of the disjunction $p(A \vee B)$ can also be expressed as the probability of a disjunction of three mutually exclusive statements
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The [addition law of probability](/D/prob-marg) states that
Since $A$ and $B$ are [mutually exclusive](/D/exc), we obtain:
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Because the [union of mutually exclusive events is the empty set](/D/exc) and the [probability of the empty set is zero](/P/prob-emp), the [joint probability](/D/prob-joint) term cancels out:
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$$ \label{eq:prob-exc-qed}
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\begin{split}
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p(A \vee B) &\overset{\eqref{eq:prob-exc-s2}}{=} p(A) + p(B) - p(A,B) \\
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p(A \vee B) &\overset{\eqref{eq:exc}}{=} p(A \vee B) - p(A,B) \\
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