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‎P/bin-pgf.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: 2022-10-11 09:25:00
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title: "Probability-generating function of the binomial distribution"
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chapter: "Probability Distributions"
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section: "Univariate discrete distributions"
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topic: "Binomial distribution"
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theorem: "Probability-generating function"
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sources:
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- authors: "ProofWiki"
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year: 2022
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title: "Probability Generating Function of Binomial Distribution"
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in: "ProofWiki"
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pages: "retrieved on 2022-10-11"
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url: "https://proofwiki.org/wiki/Probability_Generating_Function_of_Binomial_Distribution"
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proof_id: "P363"
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shortcut: "bin-pgf"
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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 [binomial distribution](/D/bin):
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$$ \label{eq:bin}
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X \sim \mathrm{Bin}(n,p) \; .
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$$
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Then, the [probability-generating function](/D/pgf) of $X$ is
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$$ \label{eq:bin-pgf}
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G_X(z) = (q + pz)^n
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$$
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where $q = 1-p$.
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**Proof:** The [probability-generating function](/D/pgf) of $X$ is defined as
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$$ \label{eq:pgf}
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G_X(z) = \sum_{x=0}^{\infty} f_X(x) \, z^x
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$$
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With the [probability mass function of the binomial distribution](/P/bin-pmf)
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$$ \label{eq:bin-pmf}
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f_X(x) = {n \choose x} \, p^x \, (1-p)^{n-x} \; ,
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$$
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we obtain:
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$$ \label{eq:pgf-zero-s1}
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\begin{split}
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G_X(z) &= \sum_{x=0}^{n} {n \choose x} \, p^x \, (1-p)^{n-x} \, z^x \\
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&= \sum_{x=0}^{n} {n \choose x} \, (pz)^x \, (1-p)^{n-x} \; .
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\end{split}
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$$
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According to the binomial theorem
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$$ \label{eq:bin-th}
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(x+y)^n = \sum_{k=0}^{n} {n \choose k} \, x^{n-k} \, y^k \; ,
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$$
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the sum in equation \eqref{eq:pgf-zero-s1} equals
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$$ \label{eq:pgf-zero-s2}
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G_X(z) = \left( (1-p) + (pz) \right)^n
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$$
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which is equivalent to the result in \eqref{eq:bin-pgf}.

‎P/mean-prodsqr.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: 2022-10-11 01:39:00
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title: "Square of expectation of product is less than or equal to product of expectation of squares"
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chapter: "General Theorems"
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section: "Probability theory"
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topic: "Expected value"
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theorem: "Squared expectation of product"
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sources:
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- authors: "ProofWiki"
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year: 2022
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title: "Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares"
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in: "ProofWiki"
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pages: "retrieved on 2022-10-11"
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url: "https://proofwiki.org/wiki/Square_of_Expectation_of_Product_is_Less_Than_or_Equal_to_Product_of_Expectation_of_Squares"
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proof_id: "P359"
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shortcut: "mean-prodsqr"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ and $Y$ be two [random variables](/D/rvar) with [expected values](/D/mean) $\mathrm{E}(X)$ and $\mathrm{E}(Y)$ and let $\mathrm{E}(XY)$ exist and be finite. Then, the square of the expectation of the product of $X$ and $Y$ is less than or equal to the product of the expectation of the squares of $X$ and $Y$:
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$$ \label{eq:EXY2-EX2-EY2}
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\left[ \mathrm{E}(XY) \right]^2 \leq \mathrm{E}\left( X^2 \right) \mathrm{E}\left( Y^2 \right) \; .
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$$
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**Proof:** Note that $Y^2$ is a non-negative [random variable](/D/rvar) whose [expected value is also non-negative](/P/mean-nonneg):
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$$ \label{eq:mean-Y2-nonneg}
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\mathrm{E}\left( Y^2 \right) \geq 0 \; .
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$$
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1) First, consider the case that $\mathrm{E}\left( Y^2 \right) > 0$. Define a new random variable Z as
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$$ \label{eq:Z}
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Z = X - Y \, \frac{\mathrm{E}(XY)}{\mathrm{E}\left( Y^2 \right)} \; .
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$$
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Once again, because $Z^2$ is always non-negative, we have the expected value:
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$$ \label{eq:mean-Z2-nonneg}
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\mathrm{E}\left( Z^2 \right) \geq 0 \; .
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$$
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Thus, using the [linearity of the expected value](/P/mean-lin), we have
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$$ \label{eq:mean-prodsqr-v1}
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\begin{split}
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0 &\leq \mathrm{E}\left( Z^2 \right) \\
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&\leq \mathrm{E}\left( \left( X - Y \, \frac{\mathrm{E}(XY)}{\mathrm{E}\left( Y^2 \right)} \right)^2 \right) \\
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&\leq \mathrm{E}\left( X^2 - 2 X Y \, \frac{\mathrm{E}(XY)}{\mathrm{E}\left( Y^2 \right)} + Y^2 \, \frac{\left[ \mathrm{E}(XY) \right]^2}{\left[ \mathrm{E}\left( Y^2 \right) \right]^2} \right) \\
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&\leq \mathrm{E}\left( X^2 \right) - 2 \, \mathrm{E}(XY) \, \frac{\mathrm{E}(XY)}{\mathrm{E}\left( Y^2 \right)} + \mathrm{E}\left( Y^2 \right) \, \frac{\left[ \mathrm{E}(XY) \right]^2}{\left[ \mathrm{E}\left( Y^2 \right) \right]^2} \\
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&\leq \mathrm{E}\left( X^2 \right) - 2 \, \frac{\left[ \mathrm{E}(XY) \right]^2}{\mathrm{E}\left( Y^2 \right)} + \frac{\left[ \mathrm{E}(XY) \right]^2}{\mathrm{E}\left( Y^2 \right)} \\
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&\leq \mathrm{E}\left( X^2 \right) - \frac{\left[ \mathrm{E}(XY) \right]^2}{\mathrm{E}\left( Y^2 \right)} \; , \\
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\end{split}
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$$
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giving
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$$ \label{eq:EXY2-EX2-EY2-qed-v1}
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\left[ \mathrm{E}(XY) \right]^2 \leq \mathrm{E}\left( X^2 \right) \mathrm{E}\left( Y^2 \right)
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$$
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as required.
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2) Next, consider the case that $\mathrm{E}\left( Y^2 \right) = 0$. In this case, $Y$ [must be a constant](/D/const) with [mean](/D/mean) $\mathrm{E}(Y) = 0$ and [variance](/D/var) $\mathrm{Var}(Y) = 0$, thus we have
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$$ \label{eq:Pr-Y-0}
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\mathrm{Pr}(Y = 0) = 1 \; .
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$$
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This implies
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$$ \label{eq:Pr-XY-0}
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\mathrm{Pr}(XY = 0) = 1 \; ,
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$$
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such that
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$$ \label{eq:EXY}
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\mathrm{E}(XY) = 0 \; .
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$$
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Thus, we can write
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$$ \label{eq:mean-prodsqr-v2}
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0 = \left[ \mathrm{E}(XY) \right]^2 = \mathrm{E}\left( X^2 \right) \mathrm{E}\left( Y^2 \right) = 0 \; ,
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$$
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giving
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$$ \label{eq:EXY2-EX2-EY2-qed-v2}
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\left[ \mathrm{E}(XY) \right]^2 \leq \mathrm{E}\left( X^2 \right) \mathrm{E}\left( Y^2 \right)
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$$
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as required.

‎P/pgf-mean.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: 2022-10-11 02:01:00
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title: "Probability-generating function is expectation of function of random variable"
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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: "Probability-generating function in terms of expected value"
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sources:
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- authors: "ProofWiki"
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year: 2022
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title: "Probability Generating Function as Expectation"
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in: "ProofWiki"
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pages: "retrieved on 2022-10-11"
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url: "https://proofwiki.org/wiki/Probability_Generating_Function_as_Expectation"
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proof_id: "P360"
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shortcut: "pgf-mean"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [discrete](/D/rvar-disc) [random variable](/D/rvar) whose set of possible values $\mathcal{X}$ is a subset of the natural numbers $\mathbb{N}$. Then, the [probability-generating function](/D/pgf) of $X$ can be expressed in terms of an [expected value](/D/mean) of a function of $X$
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$$ \label{eq:pgf-mean}
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G_X(z) = \mathrm{E}\left[ z^X \right]
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$$
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where $z \in \mathbb{C}$.
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**Proof:** The [law of the unconscious statistician](/P/mean-lotus) states that
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$$ \label{eq:mean-lotus}
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\mathrm{E}[g(X)] = \sum_{x \in \mathcal{X}} g(x) f_X(x)
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$$
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where $f_X(x)$ is the [probability mass function](/D/pmf) of $X$. Here, we have $g(X) = z^X$, such that
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$$ \label{eq:E-zX-s1}
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\mathrm{E}\left[ z^X \right] = \sum_{x \in \mathcal{X}} z^x f_X(x) \; .
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$$
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By the definition of $X$, this is equal to
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$$ \label{eq:E-zX-s2}
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\mathrm{E}\left[ z^X \right] = \sum_{x=0}^{\infty} f_X(x) \, z^x \; .
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$$
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The right-hand side is equal to the [probability-generating function](/D/pgf) of $X$:
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$$ \label{eq:pgf-mean-qed}
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\mathrm{E}\left[ z^X \right] = G_X(z) \; .
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$$

‎P/pgf-one.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: 2022-10-11 08:17:00
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title: "Value of the probability-generating function for argument one"
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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: "Probability-generating function of one"
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sources:
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- authors: "ProofWiki"
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year: 2022
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title: "Probability Generating Function of One"
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in: "ProofWiki"
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pages: "retrieved on 2022-10-11"
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url: "https://proofwiki.org/wiki/Probability_Generating_Function_of_One"
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proof_id: "P362"
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shortcut: "pgf-one"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [random variable](/D/rvar) with [probability-generating function](/D/pgf) $G_X(z)$ and set of possible values $\mathcal{X}$. Then, the value of the probability-generating function at one is equal to one:
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$$ \label{eq:pgf-one}
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G_X(1) = 1 \; .
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$$
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**Proof:** The [probability-generating function](/D/pgf) of $X$ is defined as
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$$ \label{eq:pgf}
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G_X(z) = \sum_{x=0}^{\infty} f_X(x) \, z^x
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$$
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where $f_X(x)$ is the [probability mass function](/D/pmf) of $X$. Setting $z = 1$, we obtain:
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$$ \label{eq:pgf-zero-s1}
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\begin{split}
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G_X(1) &= \sum_{x=0}^{\infty} f_X(x) \cdot 1^x \\
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&= \sum_{x=0}^{\infty} f_X(x) \cdot 1 \\
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&= \sum_{x=0}^{\infty} f_X(x) \; .
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\end{split}
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$$
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Because the [probability mass function](/D/pmf) sums up to one, this becomes:
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$$ \label{eq:pgf-zero-s2}
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\begin{split}
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G_X(1) &= \sum_{x \in \mathcal{X}} f_X(x) \\
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&= 1 \; .
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\end{split}
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$$

‎P/pgf-zero.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: 2022-10-11 08:06:00
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title: "Value of the probability-generating function for argument zero"
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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: "Probability-generating function of zero"
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sources:
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- authors: "ProofWiki"
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year: 2022
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title: "Probability Generating Function of Zero"
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in: "ProofWiki"
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pages: "retrieved on 2022-10-11"
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url: "https://proofwiki.org/wiki/Probability_Generating_Function_of_Zero"
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proof_id: "P361"
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shortcut: "pgf-zero"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [random variable](/D/rvar) with [probability-generating function](/D/pgf) $G_X(z)$ and [probability mass function](/D/pmf) $f_X(x)$. Then, the value of the probability-generating function at zero is equal to the value of the probability mass function at zero:
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$$ \label{eq:pgf-zero}
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G_X(0) = f_X(0) \; .
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$$
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**Proof:** The [probability-generating function](/D/pgf) of $X$ is defined as
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$$ \label{eq:pgf}
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G_X(z) = \sum_{x=0}^{\infty} f_X(x) \, z^x
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$$
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where $f_X(x)$ is the [probability mass function](/D/pmf) of $X$. Setting $z = 0$, we obtain:
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$$ \label{eq:pgf-zero-qed}
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\begin{split}
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G_X(0) &= \sum_{x=0}^{\infty} f_X(x) \cdot 0^x \\
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&= f_X(0) + 0^1 \cdot f_X(1) + 0^2 \cdot f_X(2) + \ldots \\
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&= f_X(0) + 0 + 0 + \ldots \\
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&= f_X(0) \; .
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\end{split}
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$$

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