added 4 proofs

Author: JoramSoch

P/gam-cdf.md

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+---
+layout: proof
+mathjax: true
+
+author: "Joram Soch"
+affiliation: "BCCN Berlin"
+e_mail: "joram.soch@bccn-berlin.de"
+date: 2020-10-15 12:34:00
+
+title: "Cumulative distribution function of the gamma distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Gamma distribution"
+theorem: "Cumulative distribution function"
+
+sources:
+
+proof_id: "P178"
+shortcut: "gam-cdf"
+username: "JoramSoch"
+---
+
+
+**Theorem:** Let $X$ be a positive [random variable](/D/rvar) following a [gamma distribution](/D/gam):
+
+$$ \label{eq:gam}
+X \sim \mathrm{Gam}(a, b) \; .
+$$
+
+Then, the [cumulative distribution function](/D/cdf) of $X$ is
+
+$$ \label{eq:gam-cdf}
+F_X(x) = \frac{\gamma(a,bx)}{\Gamma(a)}
+$$
+
+where $\Gamma(x)$ is the gamma function and $\gamma(s,x)$ is the lower incomplete gamma function.
+
+
+**Proof:** The [probability density function of the gamma distribution](/P/gam-pdf) is:
+
+$$ \label{eq:gam-pdf}
+f_X(x) = \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \; .
+$$
+
+Thus, the [cumulative distribution function](/D/cdf) is:
+
+$$ \label{eq:gam-cdf-s1}
+\begin{split}
+F_X(x) &= \int_{0}^{x} \mathrm{Gam}(z; a, b) \, \mathrm{d}z \\
+&= \int_{0}^{x} \frac{b^a}{\Gamma(a)} z^{a-1} \exp[-b z] \, \mathrm{d}z \\
+&= \frac{b^a}{\Gamma(a)} \int_{0}^{x} z^{a-1} \exp[-b z] \, \mathrm{d}z \; .
+\end{split}
+$$
+
+Substituting $t = b z$, i.e. $z = t/b$, this becomes:
+
+$$ \label{eq:gam-cdf-s2}
+\begin{split}
+F_X(x) &= \frac{b^a}{\Gamma(a)} \int_{b \cdot 0}^{b x} \left(\frac{t}{b}\right)^{a-1} \exp\left[-b \left(\frac{t}{b}\right)\right] \, \mathrm{d}\left(\frac{t}{b}\right) \\
+&= \frac{b^a}{\Gamma(a)} \cdot \frac{1}{b^{a-1}} \cdot \frac{1}{b} \int_{0}^{b x} t^{a-1} \exp[-t] \, \mathrm{d}t \\
+&= \frac{1}{\Gamma(a)} \int_{0}^{b x} t^{a-1} \exp[-t] \, \mathrm{d}t \; .
+\end{split}
+$$
+
+With the definition of the lower incomplete gamma function
+
+$$ \label{eq:li-gam-fct}
+\gamma(s,x) = \int_{0}^{x} t^{s-1} \exp[-t] \, \mathrm{d}t \; ,
+$$
+
+we arrive at the final result given by equation \eqref{eq:gam-cdf}:
+
+$$ \label{eq:gam-cdf-qed}
+F_X(x) = \frac{\gamma(a,bx)}{\Gamma(a)} \; .
+$$

P/gam-sgam2.md

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+---
+layout: proof
+mathjax: true
+
+author: "Joram Soch"
+affiliation: "BCCN Berlin"
+e_mail: "joram.soch@bccn-berlin.de"
+date: 2020-10-15 12:04:00
+
+title: "Relation between gamma distribution and standard gamma distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Gamma distribution"
+theorem: "Relation to standard gamma distribution"
+
+sources:
+
+proof_id: "P177"
+shortcut: "gam-sgam2"
+username: "JoramSoch"
+---
+
+
+**Theorem:** Let $X$ be a [random variable](/D/rvar) following a [gamma distribution](/D/gam) with shape $a$ and rate $b$:
+
+$$ \label{eq:X-gam}
+X \sim \mathrm{Gam}(a,b) \; .
+$$
+
+Then, the quantity $Y = b X$ will have a [standard gamma distribution](/D/sgam) with shape $a$ and rate $1$:
+
+$$ \label{eq:Y-snorm}
+Y = b X \sim \mathrm{Gam}(a,1) \; .
+$$
+
+
+**Proof:** Note that $Y$ is a function of $X$
+
+$$ \label{eq:Y-X}
+Y = g(X) = b X
+$$
+
+with the inverse function
+
+$$ \label{eq:X-Y}
+X = g^{-1}(Y) = \frac{1}{b} Y \; .
+$$
+
+Because $b$ is positive, $g(X)$ is strictly increasing and we can calculate the [probability density function of a strictly increasing function](/P/pdf-sifct) as
+
+$$ \label{eq:pdf-sifct}
+f_Y(y) = \left\{
+\begin{array}{rl}
+f_X(g^{-1}(y)) \, \frac{\mathrm{d}g^{-1}(y)}{\mathrm{d}y} \; , & \text{if} \; y \in \mathcal{Y} \\
+0 \; , & \text{if} \; y \notin \mathcal{Y}
+\end{array}
+\right.
+$$
+
+where $\mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace$. With the [probability density function of the gamma distribution](/P/gam-pdf), we have
+
+$$ \label{eq:pdf-Y}
+\begin{split}
+f_Y(y) &= \frac{b^a}{\Gamma(a)} [g^{-1}(y)]^{a-1} \exp[-b \, g^{-1}(y)] \cdot \frac{\mathrm{d}g^{-1}(y)}{\mathrm{d}y} \\
+&= \frac{b^a}{\Gamma(a)} \left(\frac{1}{b} y\right)^{a-1} \exp\left[-b \left(\frac{1}{b} y\right)\right] \cdot \frac{\mathrm{d}\left(\frac{1}{b} y\right)}{\mathrm{d}y} \\
+&= \frac{b^a}{\Gamma(a)} \, \frac{1}{b^{a-1}} \, y^{a-1} \exp[-y] \cdot \frac{1}{b} \\
+&= \frac{1}{\Gamma(a)} \, y^{a-1} \exp[-y]
+\end{split}
+$$
+
+which is the [probability density function](/D/pdf) of the [standard gamma distribution](/D/sgam).

P/gam-xlogx.md

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+---
+layout: proof
+mathjax: true
+
+author: "Joram Soch"
+affiliation: "BCCN Berlin"
+e_mail: "joram.soch@bccn-berlin.de"
+date: 2020-10-15 13:02:00
+
+title: "Expected value of x times ln(x) for a gamma distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Gamma distribution"
+theorem: "Expectation of x ln x"
+
+sources:
+  - authors: "gunes"
+    year: 2020
+    title: "What is the expected value of x log(x) of the gamma distribution?"
+    in: "StackExchange CrossValidated"
+    pages: "retrieved on 2020-10-15"
+    url: "https://stats.stackexchange.com/questions/457357/what-is-the-expected-value-of-x-logx-of-the-gamma-distribution"
+
+proof_id: "P179"
+shortcut: "gam-xlogx"
+username: "JoramSoch"
+---
+
+
+**Theorem:** Let $X$ be a [random variable](/D/rvar) following a [gamma distribution](/D/gam):
+
+$$ \label{eq:gam}
+X \sim \mathrm{Gam}(a, b) \; .
+$$
+
+Then, the [mean or expected value](/D/mean) of $(X \cdot \ln X)$ is
+
+$$ \label{eq:gam-xlogx}
+\mathrm{E}(X \ln X) = \frac{a}{b} \left[ \psi(a) - \ln(b) \right] \; .
+$$
+
+
+**Proof:** With the [definition of the expected value](/D/mean) and the [probability density function of the gamma distribution](/P/gam-pdf), we have:
+
+$$ \label{eq:gam-xlogx-s1}
+\begin{split}
+\mathrm{E}(X \ln X) &= \int_{0}^{\infty} x \ln x \cdot \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \, \mathrm{d}x \\
+&= \frac{1}{\Gamma(a)} \int_{0}^{\infty} \ln x \cdot \frac{b^{a+1}}{b} x^{a} \exp[-b x] \, \mathrm{d}x \\
+&= \frac{\Gamma(a+1)}{\Gamma(a) \, b} \int_{0}^{\infty} \ln x \cdot \frac{b^{a+1}}{\Gamma(a+1)} x^{(a+1)-1} \exp[-b x] \, \mathrm{d}x \\
+\end{split}
+$$
+
+The integral now corresponds to the [logarithmic expectation of a gamma distribution](/P/gam-logmean) with shape $a+1$ and rate $b$
+
+$$ \label{eq:logmean-a+1}
+\mathrm{E}(\ln Y) \quad \text{where} \quad Y \sim \mathrm{Gam}(a+1,b)
+$$
+
+which [is given by](/P/gam-logmean)
+
+$$ \label{eq:gam-logmean}
+\mathrm{E}(\ln Y) = \psi(a+1) - \ln(b)
+$$
+
+where $\psi(x)$ is the digamma function. Additionally employing the relation
+
+$$ \label{eq:gam-fct}
+\Gamma(x+1) = \Gamma(x) \cdot x \quad \Leftrightarrow \quad \frac{\Gamma(x+1)}{\Gamma(x)} = x \; ,
+$$
+
+the expression in equation \eqref{eq:gam-mean-s1} develops into:
+
+$$ \label{eq:gam-xlogx-qed}
+\mathrm{E}(X \ln X) = \frac{a}{b} \left[ \psi(a) - \ln(b) \right] \; .
+$$

P/norm-snorm2.md

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+---
+layout: proof
+mathjax: true
+
+author: "Joram Soch"
+affiliation: "BCCN Berlin"
+e_mail: "joram.soch@bccn-berlin.de"
+date: 2020-10-15 11:42:00
+
+title: "Relation between normal distribution and standard normal distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Normal distribution"
+theorem: "Relation to standard normal distribution"
+
+sources:
+
+proof_id: "P176"
+shortcut: "norm-snorm2"
+username: "JoramSoch"
+---
+
+
+**Theorem:** Let $X$ be a [random variable](/D/rvar) following a [normal distribution](/D/norm) with mean $\mu$ and variance $\sigma^2$:
+
+$$ \label{eq:X-norm}
+X \sim \mathcal{N}(\mu, \sigma^2) \; .
+$$
+
+Then, the quantity $Z = (X-\mu)/\sigma$ will have a [standard normal distribution](/D/snorm) with mean $0$ and variance $1$:
+
+$$ \label{eq:Z-snorm}
+Z = \frac{X-\mu}{\sigma} \sim \mathcal{N}(0, 1) \; .
+$$
+
+
+**Proof:** Note that $Z$ is a function of $X$
+
+$$ \label{eq:Z-X}
+Z = g(X) = \frac{X-\mu}{\sigma}
+$$
+
+with the inverse function
+
+$$ \label{eq:X-Z}
+X = g^{-1}(Z) = \sigma Z + \mu \; .
+$$
+
+Because $\sigma$ is positive, $g(X)$ is strictly increasing and we can calculate the [probability density function of a strictly increasing function](/P/pdf-sifct) as
+
+$$ \label{eq:pdf-sifct}
+f_Y(y) = \left\{
+\begin{array}{rl}
+f_X(g^{-1}(y)) \, \frac{\mathrm{d}g^{-1}(y)}{\mathrm{d}y} \; , & \text{if} \; y \in \mathcal{Y} \\
+0 \; , & \text{if} \; y \notin \mathcal{Y}
+\end{array}
+\right.
+$$
+
+where $\mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace$. With the [probability density function of the normal distribution](/P/norm-pdf), we have
+
+$$ \label{eq:pdf-Z}
+\begin{split}
+f_Z(z) &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{g^{-1}(z)-\mu}{\sigma} \right)^2 \right] \cdot \frac{\mathrm{d}g^{-1}(z)}{\mathrm{d}z} \\
+&= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{(\sigma z + \mu)-\mu}{\sigma} \right)^2 \right] \cdot \frac{\mathrm{d}(\sigma Z + \mu)}{\mathrm{d}z} \\
+&= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} z^2 \right] \cdot \sigma \\
+&= \frac{1}{\sqrt{2 \pi}} \cdot \exp \left[ -\frac{1}{2} z^2 \right]
+\end{split}
+$$
+
+which is the [probability density function](/D/pdf) of the [standard normal distribution](/D/snorm).