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‎P/mgf-lincomb.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: 2020-08-19 08:36:00
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title: "Moment-generating function of linear combination of independent random variables"
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chapter: "General Theorems"
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section: "Probability theory"
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topic: "Probability functions"
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theorem: "Moment-generating function of linear combination"
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sources:
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- authors: "ProofWiki"
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year: 2020
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title: "Moment Generating Function of Linear Combination of Independent Random Variables"
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in: "ProofWiki"
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pages: "retrieved on 2020-08-19"
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url: "https://proofwiki.org/wiki/Moment_Generating_Function_of_Linear_Combination_of_Independent_Random_Variables"
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proof_id: "P155"
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shortcut: "mgf-lincomb"
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username: "JoramSoch"
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---
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**Theorem:** Let $X_1, \ldots, X_n$ be $n$ [independent](/D/ind) [random variables](/D/rvar) with [moment-generating functions](/D/mgf) $M_{X_i}(t)$. Then, the moment-generating function of the linear combination $X = \sum_{i=1}^{n} a_i X_i$ is given by
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$$ \label{eq:mgf-lincomb}
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M_X(t) = \prod_{i=1}^{n} M_{X_i}(a_i t)
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$$
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where $a_1, \ldots, a_n$ are $n$ real numbers.
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**Proof:** The [moment-generating function of a random variable](/D/mgf) $X_i$ is
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$$ \label{eq:mfg-vect}
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M_{X_i}(t) = \mathrm{E} \left( \exp \left[ t X_i \right] \right)
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$$
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and therefore the moment-generating function of the linear combination $X$ is given by
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$$ \label{eq:mgf-lincomb-s1}
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\begin{split}
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M_X(t) &= \mathrm{E} \left( \exp \left[ t X \right] \right) \\
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&= \mathrm{E} \left( \exp \left[ t \sum_{i=1}^{n} a_i X_i \right] \right) \\
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&= \mathrm{E} \left( \prod_{i=1}^{n} \exp \left[ t \, a_i X_i \right] \right) \; .
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\end{split}
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$$
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Because the [expected value is multiplicative for independent random variables](/P/mean-mult), we have
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$$ \label{eq:mgf-lincomb-s2}
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\begin{split}
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M_X(t) &= \prod_{i=1}^{n} \mathrm{E} \left( \exp \left[ (a_i t) X_i \right] \right) \\
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&= \prod_{i=1}^{n} M_{X_i}(a_i t) \; .
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\end{split}
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$$

‎P/mgf-ltt.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: 2020-08-19 08:09:00
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title: "Linear transformation theorem for the moment-generating function"
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chapter: "General Theorems"
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section: "Probability theory"
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topic: "Probability functions"
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theorem: "Moment-generating function of linear transformation"
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sources:
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- authors: "ProofWiki"
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year: 2020
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title: "Moment Generating Function of Linear Transformation of Random Variable"
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in: "ProofWiki"
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pages: "retrieved on 2020-08-19"
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url: "https://proofwiki.org/wiki/Moment_Generating_Function_of_Linear_Transformation_of_Random_Variable"
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proof_id: "P154"
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shortcut: "mgf-ltt"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be an $n \times 1$ [random vector](/D/rvec) with the [moment-generating function](/D/mgf) $M_X(t)$. Then, the moment-generating function of the linear transformation $Y = A X + b$ is given by
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$$ \label{eq:mgf-ltt}
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M_Y(t) = \exp \left[ t^\mathrm{T} b \right] \cdot M_X(At)
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$$
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where $A$ is an $m \times n$ matrix and $b$ is an $m \times 1$ vector.
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**Proof:** The [moment-generating function of a random vector](/D/mgf) $X$ is
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$$ \label{eq:mfg-vect}
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M_X(t) = \mathrm{E} \left( \exp \left[ t^\mathrm{T} X \right] \right)
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$$
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and therefore the moment-generating function of the [random vector](/D/rvec) $Y$ is given by
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$$ \label{eq:mgf-ltt-qed}
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\begin{split}
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M_Y(t) &= \mathrm{E} \left( \exp \left[ t^\mathrm{T} (AX + b) \right] \right) \\
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&= \mathrm{E} \left( \exp \left[ t^\mathrm{T} A X \right] \cdot \exp \left[ t^\mathrm{T} b \right] \right) \\
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&= \exp \left[ t^\mathrm{T} b \right] \cdot \mathrm{E} \left( \exp \left[ (A t)^\mathrm{T} X \right] \right) \\
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&= \exp \left[ t^\mathrm{T} b \right] \cdot M_X(At) \; .
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\end{split}
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$$

‎P/mom-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: 2020-08-19 07:51:00
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title: "Moment in terms of moment-generating function"
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chapter: "General Theorems"
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section: "Probability theory"
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topic: "Further moments"
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theorem: "Moment in terms of moment-generating function"
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sources:
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- authors: "ProofWiki"
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year: 2020
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title: "Moment in terms of Moment Generating Function"
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in: "ProofWiki"
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pages: "retrieved on 2020-08-19"
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url: "https://proofwiki.org/wiki/Moment_in_terms_of_Moment_Generating_Function"
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proof_id: "P153"
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shortcut: "mom-mgf"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a scalar [random variable](/D/rvar) with the [moment-generating function](/D/mgf) $M_X(t)$. Then, the $n$-th [moment](/D/mom) of $X$ can be calculated from the moment-generating function via
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$$ \label{eq:mom-mgf}
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\mathrm{E}(X^n) = M_X^{(n)}(0)
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$$
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where $n$ is a positive integer and $M_X^{(n)}(t)$ is the $n$-th derivative of $M_X(t)$.
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**Proof:** Using the [definition of the moment-generating function](/D/mgf), we can write:
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$$ \label{eq:mom-mgf-s1}
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M_X^{(n)}(t) = \frac{\mathrm{d}^n}{\mathrm{d}t^n} \mathrm{E}(e^{tX}) \; .
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$$
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Using the power series expansion of the exponential function
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$$ \label{eq:exp-ps}
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e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \; ,
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$$
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equation \eqref{eq:mom-mgf-s1} becomes
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$$ \label{eq:mom-mgf-s2}
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M_X^{(n)}(t) = \frac{\mathrm{d}^n}{\mathrm{d}t^n} \mathrm{E}\left( \sum_{m=0}^\infty \frac{t^m X^m}{m!} \right) \; .
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$$
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Because the [expected value is a linear operator](/P/mean-lin), we have:
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$$ \label{eq:mom-mgf-s3}
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\begin{split}
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M_X^{(n)}(t) &= \frac{\mathrm{d}^n}{\mathrm{d}t^n} \sum_{m=0}^\infty \mathrm{E}\left( \frac{t^m X^m}{m!} \right) \\
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&= \sum_{m=0}^\infty \frac{\mathrm{d}^n}{\mathrm{d}t^n} \frac{t^m}{m!} \mathrm{E}\left( X^m \right) \; .
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\end{split}
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$$
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Using the $n$-th derivative of the $m$-th power
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$$ \label{eq:dndx-xm}
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\frac{\mathrm{d}^n}{\mathrm{d}x^n} x^m = \left\{
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\begin{array}{rl}
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m^{\underline{n}} \, x^{m-n} \; , & \text{if} \; n \leq m \\
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0 \; , & \text{if} \; n > m \; .
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\end{array}
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\right.
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$$
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with the falling factorial
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$$ \label{eq:fact-fall}
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m^{\underline{n}} = \prod_{i=0}^{n-1} (m-i) = \frac{m!}{(m-n)!} \; ,
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$$
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equation \eqref{eq:mom-mgf-s3} becomes
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$$ \label{eq:mom-mgf-s4}
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\begin{split}
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M_X^{(n)}(t) &= \sum_{m=n}^\infty \frac{m^{\underline{n}} \, t^{m-n}}{m!} \mathrm{E}\left( X^m \right) \\
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&\overset{\eqref{eq:fact-fall}}{=} \sum_{m=n}^\infty \frac{m! \, t^{m-n}}{(m-n)! \, m!} \mathrm{E}\left( X^m \right) \\
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&= \sum_{m=n}^\infty \frac{t^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \\
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&= \frac{t^{n-n}}{(n-n)!} \mathrm{E}\left( X^n \right) + \sum_{m=n+1}^\infty \frac{t^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \\
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&= \frac{t^0}{0!} \, \mathrm{E}\left( X^n \right) + \sum_{m=n+1}^\infty \frac{t^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \\
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&= \mathrm{E}\left( X^n \right) + \sum_{m=n+1}^\infty \frac{t^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \; .
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\end{split}
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$$
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Setting $t = 0$ in \eqref{eq:mom-mgf-s4} yields
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$$ \label{eq:mom-mgf-s5}
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\begin{split}
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M_X^{(n)}(0) &= \mathrm{E}\left( X^n \right) + \sum_{m=n+1}^\infty \frac{0^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \\
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&= \mathrm{E}\left( X^n \right)
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\end{split}
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$$
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which conforms to equation \eqref{eq:mom-mgf}.

‎P/norm-fwhm.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: 2020-08-19 06:39:00
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title: "Full width at half maximum for the 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: "Full width at half maximum"
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sources:
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- authors: "Wikipedia"
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year: 2020
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title: "Full width at half maximum"
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in: "Wikipedia, the free encyclopedia"
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pages: "retrieved on 2020-08-19"
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url: "https://en.wikipedia.org/wiki/Full_width_at_half_maximum"
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proof_id: "P152"
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shortcut: "norm-fwhm"
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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):
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$$ \label{eq:norm}
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X \sim \mathcal{N}(\mu, \sigma^2) \; .
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$$
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Then, the [full width at half maximum](/D/fwhm) (FWHM) of $X$ is
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$$ \label{eq:norm-fwhm}
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\mathrm{FWHM}(X) = 2 \sqrt{2 \ln 2} \sigma \; .
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$$
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**Proof:** The [probability density function of the normal distribution](/P/norm-pdf) is
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$$ \label{eq:norm-pdf}
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f_X(x) = \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right]
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$$
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and the [mode of the normal distribution](/P/norm-mode) is
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$$ \label{eq:norm-mode}
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\mathrm{mode}(X) = \mu \; ,
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$$
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such that
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$$ \label{eq:norm-pdf-max}
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f_\mathrm{max} = f_X(\mathrm{mode}(X)) \overset{\eqref{eq:norm-mode}}{=} f_X(\mu) \overset{\eqref{eq:norm-pdf}}{=} \frac{1}{\sqrt{2 \pi} \sigma} \; .
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$$
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The FWHM bounds satisfy the equation
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$$ \label{eq:x-FHWM}
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f_X(x_\mathrm{FWHM}) = \frac{1}{2} f_\mathrm{max} \overset{\eqref{eq:norm-pdf-max}}{=} \frac{1}{2 \sqrt{2 \pi} \sigma} \; .
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$$
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Using \eqref{eq:norm-pdf}, we can develop this equation as follows:
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$$ \label{eq:x-FHWM-s1}
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\begin{split}
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\frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x_\mathrm{FWHM}-\mu}{\sigma} \right)^2 \right] &= \frac{1}{2 \sqrt{2 \pi} \sigma} \\
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\exp \left[ -\frac{1}{2} \left( \frac{x_\mathrm{FWHM}-\mu}{\sigma} \right)^2 \right] &= \frac{1}{2} \\
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-\frac{1}{2} \left( \frac{x_\mathrm{FWHM}-\mu}{\sigma} \right)^2 &= \ln \frac{1}{2} \\
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\left( \frac{x_\mathrm{FWHM}-\mu}{\sigma} \right)^2 &= -2 \ln \frac{1}{2} \\
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\frac{x_\mathrm{FWHM}-\mu}{\sigma} &= \pm \sqrt{2 \ln 2} \\
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x_\mathrm{FWHM}-\mu &= \pm \sqrt{2 \ln 2} \sigma \\
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x_\mathrm{FWHM} &= \pm \sqrt{2 \ln 2} \sigma + \mu \; .
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\end{split}
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$$
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This implies the following two solutions for $x_\mathrm{FWHM}$
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$$ \label{eq:x-FHWM-s2}
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\begin{split}
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x_1 &= \mu - \sqrt{2 \ln 2} \sigma \\
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x_2 &= \mu + \sqrt{2 \ln 2} \sigma \; ,
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\end{split}
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$$
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such that the [full width at half maximum](/D/fwhm) of $X$ is
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$$ \label{eq:norm-fwhm-qed}
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\begin{split}
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\mathrm{FWHM}(X) &= \Delta x = x_2 - x_1 \\
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&\overset{\eqref{eq:x-FHWM-s2}}{=} \left( \mu + \sqrt{2 \ln 2} \sigma \right) - \left( \mu - \sqrt{2 \ln 2} \sigma \right) \\
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&= 2 \sqrt{2 \ln 2} \sigma \; .
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\end{split}
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$$

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