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‎P/beta-mgf.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: 2020-11-25 06:55:00
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title: "Moment-generating function of the beta distribution"
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chapter: "Probability Distributions"
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section: "Univariate continuous distributions"
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topic: "Beta distribution"
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theorem: "Moment-generating function"
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sources:
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- authors: "Wikipedia"
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year: 2020
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title: "Beta distribution"
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in: "Wikipedia, the free encyclopedia"
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pages: "retrieved on 2020-11-25"
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url: "https://en.wikipedia.org/wiki/Beta_distribution#Moment_generating_function"
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- authors: "Wikipedia"
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year: 2020
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title: "Confluent hypergeometric function"
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in: "Wikipedia, the free encyclopedia"
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pages: "retrieved on 2020-11-25"
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url: "https://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Kummer's_equation"
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proof_id: "P198"
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shortcut: "beta-mgf"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a positive [random variable](/D/rvar) following a [beta distribution](/D/gam):
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$$ \label{eq:beta}
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X \sim \mathrm{Bet}(\alpha, \beta) \; .
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$$
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Then, the [moment-generating function](/D/mgf) of $X$ is
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$$ \label{eq:beta-mgf}
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M_X(t) = 1 + \sum_{n=1}^{\infty} \left( \prod_{m=0}^{n-1} \frac{\alpha + m}{\alpha + \beta + m} \right) \frac{t^n}{n!} \; .
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$$
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**Proof:** The [probability density function of the beta distribution](/P/beta-pdf) is
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$$ \label{eq:beta-pdf}
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f_X(x) = \frac{1}{\mathrm{B}(\alpha, \beta)} \, x^{\alpha-1} \, (1-x)^{\beta-1}
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$$
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and the [moment-generating function](/D/mgf) is defined as
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$$ \label{eq:mgf-var}
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M_X(t) = \mathrm{E} \left[ e^{tX} \right] \; .
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$$
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Using the [expected value for continuous random variables](/D/mean), the moment-generating function of $X$ therefore is
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$$ \label{eq:beta-mgf-s1}
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\begin{split}
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M_X(t) &= \int_{0}^{1} \exp[tx] \cdot \frac{1}{\mathrm{B}(\alpha, \beta)} \, x^{\alpha-1} \, (1-x)^{\beta-1} \, \mathrm{d}x \\
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&= \frac{1}{\mathrm{B}(\alpha, \beta)} \int_{0}^{1} e^{tx} \, x^{\alpha-1} \, (1-x)^{\beta-1} \, \mathrm{d}x \; .
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\end{split}
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$$
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With the relationship between beta function and gamma function
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$$ \label{eq:beta-gam-fct}
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\mathrm{B}(\alpha, \beta) = \frac{\Gamma(\alpha) \, \Gamma(\beta)}{\Gamma(\alpha+\beta)}
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$$
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and the integral representation of the confluent hypergeometric function (Kummer's function of the first kind)
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$$ \label{eq:con-hyp-geo-fct-int}
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{}_1 F_1(a,b,z) = \frac{\Gamma(b)}{\Gamma(a) \, \Gamma(b-a)} \int_{0}^{1} e^{zu} \, u^{a-1} \, (1-u)^{(b-a)-1} \, \mathrm{d}u \; ,
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$$
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the moment-generating function can be written as
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$$ \label{eq:beta-mgf-s2}
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M_X(t) = {}_1 F_1(\alpha,\alpha+\beta,t) \; .
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$$
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Note that the series equation for the confluent hypergeometric function (Kummer's function of the first kind) is
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$$ \label{eq:con-hyp-geo-fct-ser}
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{}_1 F_1(a,b,z) = \sum_{n=0}^{\infty} \frac{a^{\overline{n}}}{b^{\overline{n}}} \, \frac{z^n}{n!}
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$$
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where $m^{\overline{n}}$ is the rising factorial
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$$ \label{eq:fact-rise}
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m^{\overline{n}} = \prod_{i=0}^{n-1} (m+i) \; ,
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$$
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so that the moment-generating function can be written as
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$$ \label{eq:beta-mgf-s3}
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M_X(t) = \sum_{n=0}^{\infty} \frac{\alpha^{\overline{n}}}{(\alpha+\beta)^{\overline{n}}} \, \frac{t^n}{n!} \; .
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$$
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Applying the rising factorial equation \eqref{eq:fact-rise} and using $m^{\overline{0}} = x^0 = 0! = 1$, we finally have:
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$$ \label{eq:beta-mgf-s4}
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M_X(t) = 1 + \sum_{n=1}^{\infty} \left( \prod_{m=0}^{n-1} \frac{\alpha + m}{\alpha + \beta + m} \right) \frac{t^n}{n!} \; .
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$$

‎P/chi2-pdf.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: 2020-11-25 05:56:00
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title: "Probability density function of the chi-square distribution"
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chapter: "Probability Distributions"
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section: "Univariate continuous distributions"
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topic: "Chi-square distribution"
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theorem: "Probability density function"
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sources:
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- authors: "Wikipedia"
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year: 2020
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title: "Proofs related to chi-squared distribution"
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in: "Wikipedia, the free encyclopedia"
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pages: "retrieved on 2020-11-25"
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url: "https://en.wikipedia.org/wiki/Proofs_related_to_chi-squared_distribution#Derivation_of_the_pdf_for_k_degrees_of_freedom"
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- authors: "Wikipedia"
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year: 2020
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title: "n-sphere"
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in: "Wikipedia, the free encyclopedia"
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pages: "retrieved on 2020-11-25"
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url: "https://en.wikipedia.org/wiki/N-sphere#Volume_and_surface_area"
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proof_id: "P197"
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shortcut: "chi2-pdf"
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username: "JoramSoch"
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---
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**Theorem:** Let $Y$ be a [random variable](/D/rvar) following a [chi-square distribution](/D/chi2):
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$$ \label{eq:chi2}
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Y \sim \chi^{2}(k) \; .
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$$
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Then, the [probability density function](/D/pdf) of $X$ is
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$$ \label{eq:chi2-pdf}
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f_Y(y) = \frac{1}{2^{k/2} \, \Gamma (k/2)} \, y^{k/2-1} \, e^{-y/2} \; .
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$$
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**Proof:** A [chi-square-distributed random variable](/D/chi2) with $k$ degrees of freedom is defined as the sum of $k$ squared [standard normal random variables](/D/snorm):
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$$ \label{eq:chi2-def}
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X_1, \ldots, X_k \sim \mathcal{N}(0,1) \quad \Rightarrow \quad Y = \sum_{i=1}^{k} X_i^2 \sim \chi^{2}(k) \; .
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$$
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Let $x_1, \ldots, x_k$ be values of $X_1, \ldots, X_k$ and consider $x = \left( x_1, \ldots, x_k \right)$ to be a point in $k$-dimensional space. Define
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$$ \label{eq:y-x}
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y = \sum_{i=1}^{k} x_i^2
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$$
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and let $f_Y(y)$ and $F_Y(y)$ be the [probability density function](/D/pdf) and [cumulative distribution function](/D/cdf) of $Y$. Because [the PDF is the first derivative of the CDF](/P/pdf-cdf), we can write:
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$$ \label{eq:y-pdf-s0}
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F_Y(y) = \frac{F_Y(y)}{\mathrm{d}y} \, \mathrm{d}y = f_Y(y) \, \mathrm{d}y \; .
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$$
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Then, the [cumulative distribution function](/D/cdf) of $Y$ can be expressed as
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$$ \label{eq:y-cdf-s1}
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f_Y(y) \, \mathrm{d}y = \int_{V} \prod_{i=1}^{k} \left( \mathcal{N}(x_i; 0, 1) \, \mathrm{d}x_i \right)
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$$
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where $\mathcal{N}(x_i; 0, 1)$ is the [probability density function](/D/pdf) of the [standard normal distribution](/D/snorm) and $V$ is the elemental shell volume at $y(x)$, which is proportional to the $(k-1)$-dimensional surface in $k$-space for which equation \eqref{eq:y-x} is fulfilled. Using the [probability density function of the normal distribution](/D/norm-pdf), equation \eqref{eq:y-cdf-s1} can be developed as follows:
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$$ \label{eq:y-cdf-s2}
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\begin{split}
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f_Y(y) \, \mathrm{d}y &= \int_{V} \prod_{i=1}^{k} \left( \frac{1}{\sqrt{2 \pi}} \cdot \exp \left[ -\frac{1}{2} x_i^2 \right] \, \mathrm{d}x_i \right) \\
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&= \int_{V} \frac{\exp \left[ -\frac{1}{2} \left( x_1^2 + \ldots + x_k^2 \right) \right]}{(2 \pi)^{k/2}} \; \mathrm{d}x_1 \, \ldots \, \mathrm{d}x_k \\
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&= \frac{1}{(2 \pi)^{k/2}} \int_{V} \exp \left[ -\frac{y}{2} \right] \; \mathrm{d}x_1 \, \ldots \, \mathrm{d}x_k \; .
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\end{split}
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$$
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Because $y$ is constant within the set $V$, it can be moved out of the integral:
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$$ \label{eq:y-cdf-s3}
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f_Y(y) \, \mathrm{d}y = \frac{\exp \left[ -y/2 \right]}{(2 \pi)^{k/2}} \int_{V} \; \mathrm{d}x_1 \, \ldots \, \mathrm{d}x_k \; .
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$$
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Now, the integral is simply the surface area of the $(k-1)$-dimensional sphere with radius $r = \sqrt{y}$, which is
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$$ \label{eq:A}
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A = 2 r^{k-1} \, \frac{\pi^{k/2}}{\Gamma(k/2)} \; ,
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$$
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times the infinitesimal thickness of the sphere, which is
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$$ \label{eq:dR}
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\frac{\mathrm{d}r}{\mathrm{d}y} = \frac{1}{2} y^{-1/2} \quad \Leftrightarrow \quad \mathrm{d}r = \frac{\mathrm{d}y}{2 y^{1/2}} \; .
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$$
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Substituting \eqref{eq:A} and \eqref{eq:dR} into \eqref{eq:y-cdf-s3}, we have:
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$$ \label{eq:y-cdf-s4}
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\begin{split}
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f_Y(y) \, \mathrm{d}y &= \frac{\exp \left[ -y/2 \right]}{(2 \pi)^{k/2}} \cdot A \, \mathrm{d}r \\
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&= \frac{\exp \left[ -y/2 \right]}{(2 \pi)^{k/2}} \cdot 2 r^{k-1} \, \frac{\pi^{k/2}}{\Gamma(k/2)} \cdot \frac{\mathrm{d}y}{2 y^{1/2}} \\
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&= \frac{1}{2^{k/2} \, \Gamma(k/2)} \cdot \frac{2 \sqrt{y}^{k-1}}{2 \sqrt{y}} \cdot \exp \left[ -y/2 \right] \, \mathrm{d}y \\
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&= \frac{1}{2^{k/2} \, \Gamma(k/2)} \cdot y^{\frac{k}{2}-1} \cdot \exp \left[ -\frac{y}{2} \right] \, \mathrm{d}y \; .
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\end{split}
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$$
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From this, we get the final result in \eqref{eq:chi2-pdf}:
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$$ \label{eq:y-cdf-s5}
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f_Y(y) = \frac{1}{2^{k/2} \, \Gamma (k/2)} \, y^{k/2-1} \, e^{-y/2} \; .
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$$

‎P/norm-gi.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: 2020-11-25 04:47:00
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title: "Gaussian integral"
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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: "Gaussian integral"
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sources:
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- authors: "ProofWiki"
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year: 2020
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title: "Gaussian Integral"
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in: "ProofWiki"
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pages: "retrieved on 2020-11-25"
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url: "https://proofwiki.org/wiki/Gaussian_Integral"
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- authors: "ProofWiki"
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year: 2020
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title: "Integral to Infinity of Exponential of -t^2"
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in: "ProofWiki"
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pages: "retrieved on 2020-11-25"
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url: "https://proofwiki.org/wiki/Integral_to_Infinity_of_Exponential_of_-t%5E2"
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proof_id: "P196"
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shortcut: "norm-gi"
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username: "JoramSoch"
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---
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**Theorem:** The definite integral of $\mathrm{exp}\left[ -x^2 \right]$ from $-\infty$ to $+\infty$ is equal to the square root of $\pi$:
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$$ \label{eq:norm-gi}
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\int_{-\infty}^{+\infty} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x = \sqrt{\pi} \; .
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$$
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**Proof:** Let
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$$ \label{eq:I}
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I = \int_{0}^{\infty} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x
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$$
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and
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$$ \label{eq:IP}
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I_P = \int_{0}^{P} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x = \int_{0}^{P} \mathrm{exp}\left[ -y^2 \right] \, \mathrm{d}y \; .
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$$
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Then, we have
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$$ \label{eq:IP-I}
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\lim\limits_{P \rightarrow \infty} I_P = I
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$$
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and
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$$ \label{eq:IP2-I2}
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\lim\limits_{P \rightarrow \infty} I_P^2 = I^2 \; .
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$$
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Moreover, we can write
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$$ \label{eq:IP2}
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\begin{split}
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I_P^2 &\overset{\eqref{eq:IP}}{=} \left( \int_{0}^{P} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x \right) \left( \int_{0}^{P} \mathrm{exp}\left[ -y^2 \right] \, \mathrm{d}y \right) \\
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&= \int_{0}^{P} \int_{0}^{P} \mathrm{exp}\left[ - \left( x^2 + y^2 \right) \right] \, \mathrm{d}x \, \mathrm{d}y \\
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&= \iint_{S_P} \mathrm{exp}\left[ - \left( x^2 + y^2 \right) \right] \, \mathrm{d}x \, \mathrm{d}y
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\end{split}
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$$
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where $S_P$ is the square with corners $(0,0)$, $(0,P)$, $(P,P)$ and $(P,0)$. For this integral, we can write down the following inequality
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$$ \label{eq:IP2-ineq}
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\iint_{C_1} \mathrm{exp}\left[ - \left( x^2 + y^2 \right) \right] \, \mathrm{d}x \, \mathrm{d}y \leq I_P^2 \leq \iint_{C_2} \mathrm{exp}\left[ - \left( x^2 + y^2 \right) \right] \, \mathrm{d}x \, \mathrm{d}y
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$$
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where $C_1$ and $C_2$ are the regions in the first quadrant bounded by circles with center at $(0,0)$ and going through the points $(0,P)$ and $(P,P)$, respectively. The radii of these two circles are $r_1 = \sqrt{P^2} = P$ and $r_2 = \sqrt{2 P^2} = P \sqrt{2}$, such that we can rewrite equation \eqref{eq:IP2-ineq} using polar coordinates as
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$$ \label{eq:IP2-ineq-PC}
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\int_{0}^{\frac{\pi}{2}} \int_{0}^{r_1} \mathrm{exp}\left[ -r^2 \right] \, r \, \mathrm{d}r \, \mathrm{d}\theta \leq I_P^2 \leq \int_{0}^{\frac{\pi}{2}} \int_{0}^{r_2} \mathrm{exp}\left[ -r^2 \right] \, r \, \mathrm{d}r \, \mathrm{d}\theta \; .
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$$
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Solving the definite integrals yields:
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$$ \label{eq:IP2-ineq-PC-int}
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\begin{split}
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\int_{0}^{\frac{\pi}{2}} \int_{0}^{r_1} \mathrm{exp}\left[ -r^2 \right] r \, \mathrm{d}r \, \mathrm{d}\theta &\leq I_P^2 \leq \int_{0}^{\frac{\pi}{2}} \int_{0}^{r_2} \mathrm{exp}\left[ -r^2 \right] r \, \mathrm{d}r \, \mathrm{d}\theta \\
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\int_{0}^{\frac{\pi}{2}} \left[ -\frac{1}{2} \mathrm{exp}\left[ -r^2 \right] \right]_{0}^{r_1} \, \mathrm{d}\theta &\leq I_P^2 \leq \int_{0}^{\frac{\pi}{2}} \left[ -\frac{1}{2} \mathrm{exp}\left[ -r^2 \right] \right]_{0}^{r_2} \, \mathrm{d}\theta \\
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-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( \mathrm{exp}\left[ -r_1^2 \right] - 1 \right) \, \mathrm{d}\theta &\leq I_P^2 \leq -\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( \mathrm{exp}\left[ -r_2^2 \right] - 1 \right) \, \mathrm{d}\theta \\
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-\frac{1}{2} \left[ \left( \mathrm{exp}\left[ -r_1^2 \right] - 1 \right) \theta \right]_{0}^{\frac{\pi}{2}} &\leq I_P^2 \leq -\frac{1}{2} \left[ \left( \mathrm{exp}\left[ -r_2^2 \right] - 1 \right) \theta \right]_{0}^{\frac{\pi}{2}} \\
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\frac{1}{2} \left( 1 - \mathrm{exp}\left[ -r_1^2 \right] \right) \frac{\pi}{2} &\leq I_P^2 \leq \frac{1}{2} \left( 1 - \mathrm{exp}\left[ -r_2^2 \right] \right) \frac{\pi}{2} \\
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\frac{\pi}{4} \left( 1 - \mathrm{exp}\left[ -P^2 \right] \right) &\leq I_P^2 \leq \frac{\pi}{4} \left( 1 - \mathrm{exp}\left[ -2 P^2 \right] \right)
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\end{split}
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$$
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Calculating the limit for $P \rightarrow \infty$, we obtain
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$$ \label{eq:IP2-ineq-PC-int-lim}
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\begin{split}
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\lim\limits_{P \rightarrow \infty} \frac{\pi}{4} \left( 1 - \mathrm{exp}\left[ -P^2 \right] \right) \leq \lim\limits_{P \rightarrow \infty} I_P^2 &\leq \lim\limits_{P \rightarrow \infty} \frac{\pi}{4} \left( 1 - \mathrm{exp}\left[ -2 P^2 \right] \right) \\
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\frac{\pi}{4} \leq I^2 &\leq \frac{\pi}{4} \; ,
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\end{split}
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$$
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such that we have a preliminary result for $I$:
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$$ \label{eq:I-qed}
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I^2 = \frac{\pi}{4} \quad \Rightarrow \quad I = \frac{\sqrt{\pi}}{2} \; .
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$$
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Because the integrand in \eqref{eq:norm-gi} is an even function, we can calculate the final result as follows:
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$$ \label{eq:norm-gi-qed}
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\begin{split}
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\int_{-\infty}^{+\infty} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x &= 2 \int_{0}^{\infty} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x \\
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&\overset{\eqref{eq:I-qed}}{=} 2 \, \frac{\sqrt{\pi}}{2} \\
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&= \sqrt{\pi} \; .
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\end{split}
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$$

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