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‎P/dir-ep.md‎

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---
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layout: proof
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mathjax: true
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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: 2020-10-22 08:04:00
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title: "Exceedance probabilities for the the Dirichlet distribution"
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chapter: "Probability Distributions"
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section: "Multivariate continuous distributions"
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topic: "Dirichlet distribution"
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theorem: "Exceedance probabilities"
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sources:
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- authors: "Soch J, Allefeld C"
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year: 2016
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title: "Exceedance Probabilities for the Dirichlet Distribution"
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in: "arXiv stat.AP"
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pages: "1611.01439"
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url: "https://arxiv.org/abs/1611.01439"
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proof_id: "P181"
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shortcut: "dir-ep"
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username: "JoramSoch"
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---
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**Theorem:** Let $r = [r_1, \ldots, r_k]$ be a [random vector](/D/rvec) following a [Dirichlet distribution](/D/dir) with concentration parameters $\alpha = [\alpha_1, \ldots, \alpha_k]$:
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$$ \label{eq:r-Dir}
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r \sim \mathrm{Dir}(\alpha) \; .
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$$
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<br>
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1) If $k = 2$, then the [exceedance probability](/D/prob-exc) for $r_1$ is
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$$ \label{eq:Dir2-EP}
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\varphi_1 = 1 - \frac{\mathrm{B}\left( \frac{1}{2};\alpha_1,\alpha_2 \right)}{\mathrm{B}(\alpha_1,\alpha_2)}
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$$
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where $\mathrm{B}(x,y)$ is the beta function and $\mathrm{B}(x;a,b)$ is the incomplete beta function.
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<br>
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2) If $k > 2$, then the [exceedance probability](/D/prob-exc) for $r_i$ is
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$$ \label{eq:Dir-EP}
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\varphi_i = \int_0^\infty \prod_{j \neq i} \left( \frac{\gamma(\alpha_j,q_i)}{\Gamma(\alpha_j)} \right) \, \frac{q_i^{\alpha_i-1} \exp[-q_i]}{\Gamma(\alpha_i)} \, \mathrm{d}q_i \; .
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$$
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where $\Gamma(x)$ is the gamma function and $\gamma(s,x)$ is the lowerr incomplete gamma function.
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**Proof:** In the context of the [Dirichlet distribution](/D/dir), the [exceedance probability](/D/prob-exc) for a particular $r_i$ is defined as:
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$$ \label{eq:Dir-EP-def}
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\begin{split}
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\varphi_i &= p \Bigl( \forall j \in \left\lbrace 1, \ldots, k \Bigm| j \neq i \right\rbrace: \, r_i > r_j |\alpha \bigr) \\
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&= p \Bigl( \bigwedge_{j \neq i} r_i > r_j \Bigm| \alpha \Bigr) \; .
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\end{split}
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$$
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The [probability density function of the Dirichlet distribution](/P/dir-pdf) is given by:
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$$ \label{eq:Dir-pdf}
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\mathrm{Dir}(r; \alpha) = \frac{\Gamma\left( \sum_{i=1}^k \alpha_i \right)}{\prod_{i=1}^k \Gamma(\alpha_i)} \, \prod_{i=1}^k {r_i}^{\alpha_i-1} \; .
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$$
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Note that the probability density function is only calculated, if
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$$ \label{eq:Dir-req}
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r_i \in [0,1] \quad \text{for} \quad i = 1,\ldots,k \quad \text{and} \quad \sum_{i=1}^k r_i = 1 \; ,
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$$
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and [defined to be zero otherwise](/D/dir).
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<br>
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1) If $k = 2$, the [probability density function of the Dirichlet distribution](/P/dir-pdf) reduces to
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$$ \label{eq:Dir2-pdf}
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p(r) = \frac{\Gamma(\alpha_1 + \alpha_2)}{\Gamma(\alpha_1) \, \Gamma(\alpha_2)} \, r_1^{\alpha_1-1} \, r_2^{\alpha_2-1}
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$$
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which is equivalent to the [probability density function of the beta distribution](/P/beta-pdf)
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$$ \label{eq:Beta-pdf}
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p(r_1) = \frac{r_1^{\alpha_1-1} \, (1-r_1)^{\alpha_2-1}}{\mathrm{B}(\alpha_1,\alpha_2)}
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$$
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with the beta function given by
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$$ \label{eq:beta-fct}
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\mathrm{B}(x,y) = \frac{\Gamma(x) \, \Gamma(y)}{\Gamma(x + y)} \; .
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$$
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With \eqref{eq:Dir-req}, the exceedance probability for this bivariate case simplifies to
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$$ \label{eq:Dir2-EP-def}
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\varphi_1 = p(r_1 > r_2) = p(r_1 > 1 - r_1) = p(r_1 > 1/2) = \int_{\frac{1}{2}}^1 p(r_1) \, \mathrm{d}r_1 \; .
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$$
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Using the [cumulative distribution function of the beta distribution](/P/beta-cdf), it evaluates to
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$$ \label{eq:Dir2-EP-qed}
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\varphi_1 = 1 - \int_0^{\frac{1}{2}} p(r_1) \, \mathrm{d}r_1 = 1 - \frac{\mathrm{B}\left( \frac{1}{2};\alpha_1,\alpha_2 \right)}{\mathrm{B}(\alpha_1,\alpha_2)}
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$$
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with the incomplete beta function
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$$ \label{eq:inc-beta-fct}
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\mathrm{B}(x; a, b) = \int_0^x x^{a-1} \, (1-x)^{b-1} \, \mathrm{d}x \; .
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$$
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<br>
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2) If $k > 2$, there is no similarly simple expression, because in general
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$$ \label{eq:Dir-EP-ineq}
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\varphi_i = p(r_i = \mathrm{max}(r)) > p(r_i > 1/2) \quad \text{for} \quad i = 1, \ldots, k \; ,
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$$
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i.e. exceedance probabilities cannot be evaluated using a simple threshold on $r_i$, because $r_i$ might be the maximal element in $r$ without being larger than $1/2$. Instead, we make use of the [relationship between the Dirichlet and the gamma distribution](/P/gam-dir) which states that
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$$ \label{eq:Gam-Dir}
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\begin{split}
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& Y_1 \sim \mathrm{Gam}(\alpha_1,\beta), \, \ldots, \, Y_k \sim \mathrm{Gam}(\alpha_k,\beta), \, Y_s = \sum_{i=1}^k Y_j \\
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\Rightarrow \; & X = (X_1, \ldots, X_k) = \left( \frac{Y_1}{Y_s}, \ldots, \frac{Y_k}{Y_s} \right) \sim \mathrm{Dir}(\alpha_1, \ldots, \alpha_k) \; .
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\end{split}
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$$
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The [probability density function of the gamma distribution](/P/gam-pdf) is given by
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$$ \label{eq:Gam-pdf}
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\mathrm{Gam}(x; a, b) = \frac{{b}^{a}}{\Gamma(a)} \, x^{a-1} \, \exp[-b x] \quad \text{for} \quad x > 0 \; .
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$$
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Consider the [gamma random variables](/D/gam)
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$$ \label{eq:Gam-Dir-A}
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q_1 \sim \mathrm{Gam}(\alpha_1,1), \, \ldots, \, q_k \sim \mathrm{Gam}(\alpha_k,1), \, q_s = \sum_{j=1}^k q_j
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$$
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and the [Dirichlet random vector](/D/dir)
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$$ \label{eq:Gam-Dir-B}
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r = (r_1, \ldots, r_k) = \left( \frac{q_1}{q_s}, \ldots, \frac{q_k}{q_s} \right) \sim \mathrm{Dir}(\alpha_1, \ldots, \alpha_k) \; .
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$$
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Obviously, it holds that
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$$ \label{eq:Gam-Dir-eq}
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r_i > r_j \; \Leftrightarrow \; q_i > q_j \quad \text{for} \quad i,j = 1, \ldots, k \quad \text{with} \quad j \neq i \; .
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$$
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Therefore, consider the probability that $q_i$ is larger than $q_j$, given $q_i$ is known. This probability is equal to the probability that $q_j$ is smaller than $q_i$, given $q_i$ is known
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$$ \label{eq:Gam-EP0}
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p(q_i > q_j|q_i) = p(q_j < q_i|q_i)
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$$
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which can be expressed in terms of the [cumulative distribution function of the gamma distribution](/P/gam-cdf) as
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$$ \label{eq:Gam-EP1}
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p(q_j < q_i|q_i) = \int_0^{q_i} \mathrm{Gam}(q_j;\alpha_j,1) \, \mathrm{d}q_j = \frac{\gamma(\alpha_j,q_i)}{\Gamma(\alpha_j)}
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$$
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where $\Gamma(x)$ is the gamma function and $\gamma(s,x)$ is the lower incomplete gamma function. Since the gamma variates are independent of each other, these probabilties factorize:
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$$ \label{eq:Gam-EP2}
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p(\forall_{j \neq i} \left[ q_i > q_j \right]|q_i) = \prod_{j \neq i} p(q_i > q_j|q_i) = \prod_{j \neq i} \frac{\gamma(\alpha_j,q_i)}{\Gamma(\alpha_j)} \; .
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$$
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In order to obtain the exceedance probability $\varphi_i$, the dependency on $q_i$ in this probability still has to be removed. From equations (\ref{eq:Dir-EP-def}) and (\ref{eq:Gam-Dir-eq}), it follows that
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$$ \label{eq:Dir-EP2a}
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\varphi_i = p(\forall_{j \neq i} \left[ r_i > r_j \right]) = p(\forall_{j \neq i} \left[ q_i > q_j \right]) \; .
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$$
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Using the [law of marginal probability](/D/prob-marg), we have
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$$ \label{eq:Dir-EP2b}
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\varphi_i = \int_0^\infty p(\forall_{j \neq i} \left[ q_i > q_j \right]|q_i) \, p(q_i) \, \mathrm{d}q_i \; .
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$$
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With (\ref{eq:Gam-EP2}) and (\ref{eq:Gam-Dir-A}), this becomes
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$$ \label{eq:Dir-EP2c}
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\varphi_i = \int_0^\infty \prod_{j \neq i} \left( p(q_i > q_j|q_i) \right) \cdot \mathrm{Gam}(q_i;\alpha_i,1) \, \mathrm{d}q_i \; .
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$$
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And with (\ref{eq:Gam-EP1}) and (\ref{eq:Gam-pdf}), it becomes
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$$ \label{eq:Dir-EP-qed}
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\varphi_i = \int_0^\infty \prod_{j \neq i} \left( \frac{\gamma(\alpha_j,q_i)}{\Gamma(\alpha_j)} \right) \cdot \frac{q_i^{\alpha_i-1} \exp[-q_i]}{\Gamma(\alpha_i)} \, \mathrm{d}q_i \; .
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$$
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In other words, the [exceedance probability](/D/prob-exc) for one element from a [Dirichlet-distributed](/D/dir) [random vector](/D/rvec) is an integral from zero to infinity where the first term in the integrand conforms to a product of [gamma](/D/gam) [cumulative distribution functions](/D/cdf) and the second term is a [gamma](/D/gam) [probability density function](/D/pdf).

‎P/dir-mle.md‎

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---
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layout: proof
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mathjax: true
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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: 2020-10-22 09:31:00
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title: "Maximum likelihood estimation for Dirichlet-distributed data"
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chapter: "Statistical Models"
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section: "Probability data"
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topic: "Dirichlet-distributed data"
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theorem: "Maximum likelihood estimation"
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sources:
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- authors: "Minka TP"
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year: 2012
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title: "Estimating a Dirichlet distribution"
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in: "Papers by Tom Minka"
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pages: "retrieved on 2020-10-22"
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url: "https://tminka.github.io/papers/dirichlet/minka-dirichlet.pdf"
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proof_id: "P182"
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shortcut: "dir-mle"
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username: "JoramSoch"
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---
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**Theorem:** Let there be a [Dirichlet-distributed data](/D/dir-data) set $y = \left\lbrace y_1, \ldots, y_n \right\rbrace$:
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$$ \label{eq:Dir}
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y_i \sim \mathrm{Dir}(\alpha), \quad i = 1, \ldots, n \; .
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$$
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Then, the [maximum likelihood estimate](/D/mle) for the concentration parameters $\alpha$ can be obtained by iteratively computing
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$$ \label{eq:Dir-MLE}
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\alpha_j^{\text{(new)}} = \psi^{-1}\left[ \psi\left( \sum_{j=1}^k \alpha_j^{\text{(old)}} \right) + \log \bar{y}_j \right]
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$$
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where $\psi(x)$ is the digamma function and $\log \bar{y}_j$ is given by:
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$$ \label{eq:log-pi}
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\log \bar{y}_j = \frac{1}{n} \sum_{i=1}^n \log y_{ij} \; .
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$$
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**Proof:** The [likelihood function](/D/lf) for each observation is given by the [probability density function of the Dirichlet distribution](/P/dir-pdf)
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$$ \label{eq:Dir-yi}
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p(y_i|\alpha) = \frac{\Gamma\left( \sum_{j=1}^k \alpha_j \right)}{\prod_{j=1}^k \Gamma(\alpha_j)} \, \prod_{j=1}^k {y_{ij}}^{\alpha_j-1}
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$$
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and because observations are [independent](/D/ind), the likelihood function for all observations is the product of the individual ones:
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$$ \label{eq:Dir-LF}
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p(y|\alpha) = \prod_{i=1}^n p(y_i|\alpha) = \prod_{i=1}^n \left[ \frac{\Gamma\left( \sum_{j=1}^k \alpha_j \right)}{\prod_{j=1}^k \Gamma(\alpha_j)} \, \prod_{j=1}^k {y_{ij}}^{\alpha_j-1} \right] \; .
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$$
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Thus, the [log-likelihood function](/D/llf) is
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$$ \label{eq:Dir-LL}
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\mathrm{LL}(\alpha) = \log p(y|\alpha) = \log \prod_{i=1}^n \left[ \frac{\Gamma\left( \sum_{j=1}^k \alpha_j \right)}{\prod_{j=1}^k \Gamma(\alpha_j)} \, \prod_{j=1}^k {y_{ij}}^{\alpha_j-1} \right]
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$$
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which can be developed into
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$$ \label{eq:Dir-LL-der}
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\begin{split}
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\mathrm{LL}(\alpha) &= \sum_{i=1}^n \log \Gamma\left( \sum_{j=1}^k \alpha_j \right) - \sum_{i=1}^n \sum_{j=1}^k \log \Gamma(\alpha_j) + \sum_{i=1}^n \sum_{j=1}^k (\alpha_j-1) \log y_{ij} \\
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&= n \log \Gamma\left( \sum_{j=1}^k \alpha_j \right) - n \sum_{j=1}^k \log \Gamma(\alpha_j) + n \sum_{j=1}^k (\alpha_j-1) \, \frac{1}{n} \sum_{i=1}^n \log y_{ij} \\
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&= n \log \Gamma\left( \sum_{j=1}^k \alpha_j \right) - n \sum_{j=1}^k \log \Gamma(\alpha_j) + n \sum_{j=1}^k (\alpha_j-1) \, \log \bar{y}_j
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\end{split}
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$$
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where we have specified
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$$ \label{eq:log-pi-reit}
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\log \bar{y}_j = \frac{1}{n} \sum_{i=1}^n \log y_{ij} \; .
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$$
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The derivative of the log-likelihood with respect to a particular parameter $\alpha_j$ is
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$$ \label{eq:Dir-dLLdaj}
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\begin{split}
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\frac{\mathrm{d}\mathrm{LL}(\alpha)}{\mathrm{d}\alpha_j} &= \frac{\mathrm{d}}{\mathrm{d}\alpha_j} \left[ n \log \Gamma\left( \sum_{j=1}^k \alpha_j \right) - n \sum_{j=1}^k \log \Gamma(\alpha_j) + n \sum_{j=1}^k (\alpha_j-1) \, \log \bar{y}_j \right] \\
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&= \frac{\mathrm{d}}{\mathrm{d}\alpha_j} \left[ n \log \Gamma\left( \sum_{j=1}^k \alpha_j \right) \right] - \frac{\mathrm{d}}{\mathrm{d}\alpha_j} \left[ n \log \Gamma(\alpha_j) \right] + \frac{\mathrm{d}}{\mathrm{d}\alpha_j} \left[ n (\alpha_j-1) \, \log \bar{y}_j \right] \\
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&= n \psi\left( \sum_{j=1}^k \alpha_j \right) - n \psi(\alpha_j) + n \log \bar{y}_j
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\end{split}
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$$
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where we have used the digamma function
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$$ \label{eq:digamma-fct}
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\psi(x) = \frac{\mathrm{d}\log \Gamma(x)}{\mathrm{d}x} \; .
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$$
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Setting this derivative to zero, we obtain:
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$$ \label{eq:Dir-dLLdaj-0}
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\begin{split}
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\frac{\mathrm{d}\mathrm{LL}(\alpha)}{\mathrm{d}\alpha_j} &= 0 \\
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0 &= n \psi\left( \sum_{j=1}^k \alpha_j \right) - n \psi(\alpha_j) + n \log \bar{y}_j \\
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0 &= \psi\left( \sum_{j=1}^k \alpha_j \right) - \psi(\alpha_j) + \log \bar{y}_j \\
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\psi(\alpha_j) &= \psi\left( \sum_{j=1}^k \alpha_j \right) + \log \bar{y}_j \\
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\alpha_j &= \psi^{-1} \left[ \psi\left( \sum_{j=1}^k \alpha_j \right) + \log \bar{y}_j \right] \; .
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\end{split}
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$$
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In the following, we will use a fixed-point iteration to maximize $\mathrm{LL}(\alpha)$. Given an initial guess for $\alpha$, we construct a lower bound on the likelihood function \eqref{eq:Dir-LL-der} which is tight at $\alpha$. The maximum of this bound is computed and it becomes the new guess. Because the [Dirichlet distribution](/D/dir) belongs to the [exponential family](/D/dist-expfam), the log-likelihood function is convex in $\alpha$ ánd the maximum is the only stationary point, such that the procedure is guaranteed to converge to the maximum.
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In our case, we use a bound on the gamma function
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$$ \label{eq:gamma-fct-bound}
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\begin{split}
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\Gamma(x) &\geq \Gamma(\hat{x}) \cdot \mathrm{exp}\left[(x-\hat{x}) \, \psi(\hat{x}) \right] \\
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\log \Gamma(x) &\geq \log \Gamma(\hat{x}) + (x-\hat{x}) \, \psi(\hat{x})
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\end{split}
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$$
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and apply it to $\Gamma\left( \sum_{j=1}^k \alpha_j \right)$ in \eqref{eq:Dir-LL-der} to yield
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$$ \label{eq:Dir-LL-bound}
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\begin{split}
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\frac{1}{n} \mathrm{LL}(\alpha) &= \log \Gamma\left( \sum_{j=1}^k \alpha_j \right) - \sum_{j=1}^k \log \Gamma(\alpha_j) + \sum_{j=1}^k (\alpha_j-1) \, \log \bar{y}_j \\
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\frac{1}{n} \mathrm{LL}(\alpha) &\geq \log \Gamma\left(\sum_{j=1}^k \hat{\alpha}_j\right) + \left(\sum_{j=1}^k \alpha_j - \sum_{j=1}^k \hat{\alpha}_j\right) \psi\left(\sum_{j=1}^k \hat{\alpha}_j\right) - \sum_{j=1}^k \log \Gamma(\alpha_j) + \sum_{j=1}^k (\alpha_j-1) \, \log \bar{y}_j \\
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\frac{1}{n} \mathrm{LL}(\alpha) &\geq \left(\sum_{j=1}^k \alpha_j\right) \psi\left(\sum_{j=1}^k \hat{\alpha}_j\right) - \sum_{j=1}^k \log \Gamma(\alpha_j) + \sum_{j=1}^k (\alpha_j-1) \, \log \bar{y}_j + \mathrm{const.}
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\end{split}
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$$
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which leads to the following fixed-point iteration using \eqref{eq:Dir-dLLdaj-0}:
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$$ \label{eq:Dir-MLE-qed}
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\alpha_j^{\text{(new)}} = \psi^{-1}\left[ \psi\left( \sum_{j=1}^k \alpha_j^{\text{(old)}} \right) + \log \bar{y}_j \right] \; .
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$$

‎P/norm-snorm3.md‎

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---
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layout: proof
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mathjax: true
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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: 2020-10-22 06:34:00
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title: "Relation between normal distribution and standard normal distribution"
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chapter: "Probability Distributions"
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section: "Univariate continuous distributions"
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topic: "Normal distribution"
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theorem: "Relation to standard normal distribution"
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sources:
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proof_id: "P180"
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shortcut: "norm-snorm3"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [random variable](/D/rvar) following a [normal distribution](/D/norm) with mean $\mu$ and variance $\sigma^2$:
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$$ \label{eq:X-norm}
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X \sim \mathcal{N}(\mu, \sigma^2) \; .
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$$
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Then, the quantity $Z = (X-\mu)/\sigma$ will have a [standard normal distribution](/D/snorm) with mean $0$ and variance $1$:
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$$ \label{eq:Z-snorm}
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Z = \frac{X-\mu}{\sigma} \sim \mathcal{N}(0, 1) \; .
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$$
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**Proof:** The [linear transformation theorem for multivariate normal distribution](/P/mvn-ltt) states
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$$ \label{eq:mvn-ltt}
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x \sim \mathcal{N}(\mu, \Sigma) \quad \Rightarrow \quad y = Ax + b \sim \mathcal{N}(A\mu + b, A \Sigma A^\mathrm{T})
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$$
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where $x$ is an $n \times 1$ [random vector](/D/rvec) following a [multivariate normal distribution](/D/mvn) with mean $\mu$ and covariance $\Sigma$, $A$ is an $m \times n$ matrix and $b$ is an $m \times 1$ vector. Note that
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$$ \label{eq:Z-X}
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Z = \frac{X-\mu}{\sigma} = \frac{X}{\sigma} - \frac{\mu}{\sigma}
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$$
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is a special case of \eqref{eq:mvn-ltt} with $x = X$, $\mu = \mu$, $\Sigma = \sigma^2$, $A = 1/\sigma$ and $b = \mu/\sigma$. Applying theorem \eqref{eq:mvn-ltt} to $Z$ as a function of $X$, we have
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$$ \label{eq:mvn-ltt-norm}
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X \sim \mathcal{N}(\mu, \sigma^2) \quad \Rightarrow \quad Z = \frac{X}{\sigma} - \frac{\mu}{\sigma} \sim \mathcal{N}\left( \frac{\mu}{\sigma} - \frac{\mu}{\sigma}, \frac{1}{\sigma} \cdot \sigma^2 \cdot \frac{1}{\sigma} \right)
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$$
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which results in the distribution:
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$$ \label{eq:Z-snorm-qed}
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Z \sim \mathcal{N}(0, 1) \; .
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$$

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