hcju/proofwikiqa huggingface dataset:
From the definition of Variance as Expectation of Square minus Square of Expectation: :$\var X = \expect {X^2} - \paren {\expect X}^2$ From Expectation of Function of Discrete Random Variable: :$\displaystyle \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \, \map \Pr {X = x}$ To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$. Thus: {{begin-eqn}} {{eqn | l = \expect {X^2}        | r = \sum_{k \mathop \ge 0} k^2 p q^{k - 1}       | c = {{Defof|Shifted Geometric Distribution}}, with $p + q = 1$ }} {{eqn | r = \sum_{k \mathop \ge 1} k^2 p q^{k - 1}       | c = The term in $k=0$ is zero, so we change the limits }} {{eqn | r = \sum_{k \mathop \ge 1} k \paren {k + 1} p q^{k - 1} - \sum_{k \mathop \ge 1} k p q^{k - 1}       | c = splitting sum up into two }} {{eqn | r = \sum_{k \mathop \ge 1} k \paren {k + 1} p q^{k - 1} - \frac 1 p       | c = Second term is Expectation of Shifted Geometric Distribution }} {{eqn | r = p \frac 2 {\paren {1 - q}^3} - \frac 1 p       | c = Derivative of Geometric Sequence: Corollary }} {{eqn | r = \frac 2 {p^2} - \frac 1 p       | c = putting $p = 1 - q$ back in and simplifying }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \var X       | r = \expect {X^2} - \paren {\expect X}^2       | c =  }} {{eqn | r = \frac 2 {p^2} - \frac 1 p - \frac 1 {p^2}       | c = Expectation of Shifted Geometric Distribution: $\expect X = \dfrac 1 p$ }} {{eqn | r = \frac 1 {p^2} - \frac 1 p       | c =  }} {{eqn | r = \frac {1 - p} {p^2}       | c =  }} {{end-eqn}} {{qed}}
From Variance of Discrete Random Variable from PGF, we have: :$\var X = \map {\Pi''_X} 1 + \mu - \mu^2$ where $\mu = \expect X$ is the expectation of $X$. From the Probability Generating Function of Shifted Geometric Distribution, we have: :$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$ where $q = 1 - p$. From Expectation of Shifted Geometric Distribution, we have: :$\mu = \dfrac 1 p$ From Derivatives of PGF of Shifted Geometric Distribution, we have: :$\map {\Pi''_X} s = \dfrac {p q} {\paren {1 - q s}^3}$ Putting $s = 1$ using the formula $\map {\Pi''_X} 1 + \mu - \mu^2$: :$\var X = \dfrac {p q} {\paren {1 - q}^3} + \dfrac 1 p - \paren {\dfrac 1 p}^2$ and hence the result, after some algebra. {{qed}}
From the definition of expectation: :$\displaystyle \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$ By definition of geometric distribution: :$\displaystyle \expect X = \sum_{k \mathop \in \Omega_X} k p^k \paren {1 - p}$ Let $q = 1 - p$: {{begin-eqn}} {{eqn | l = \expect X       | r = q \sum_{k \mathop \ge 0} k p^k       | c = as $\Omega_X = \N$ }} {{eqn | r = q \sum_{k \mathop \ge 1} k p^k       | c = as the $k = 0$ term is zero }} {{eqn | r = q p \sum_{k \mathop \ge 1} k p^{k - 1} }} {{eqn | r = q p \frac 1 {\paren {1 - p}^2}       | c = Derivative of Geometric Sequence }} {{eqn | r = \frac p {1 - p}       | c = as $q = 1 - p$ }} {{end-eqn}} {{qed}}
From the Probability Generating Function of Geometric Distribution, we have: :$\Pi_X \left({s}\right) = \dfrac q {1 - ps}$ where $q = 1 - p$. From Expectation of Discrete Random Variable from PGF, we have: :$E \left({X}\right) = \Pi'_X \left({1}\right)$ We have: {{begin-eqn}} {{eqn | l = \Pi'_X \left({s}\right)       | r = \frac {\mathrm d} {\mathrm d s} \left({\frac q {1 - p s} }\right) }} {{eqn | r = \frac {q p} {\left({1 - p s}\right)^2}       | c = Derivatives of PGF of Geometric Distribution }} {{end-eqn}} Plugging in $s = 1$: {{begin-eqn}} {{eqn | l = \Pi'_X \left({1}\right)       | r = \frac {q p} {\left({1 - p}\right)^2} }} {{eqn | r = \frac p {1 - p}       | c = as $q = 1 - p$ }} {{end-eqn}} Hence the result. {{qed}}
From the definition of Variance as Expectation of Square minus Square of Expectation: :$\var X = \expect {X^2} - \paren {\expect X}^2$ From Expectation of Function of Discrete Random Variable: :$\displaystyle \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$ To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$. Thus: {{begin-eqn}} {{eqn | l = \expect {X^2}       | r = \sum_{k \mathop \ge 1} k^2 q p^k       | c = {{Defof|Geometric Distribution}}, with $p + q = 1$ }} {{eqn | r = p \sum_{k \mathop \ge 1} k^2 q p^{k-1}       | c =  }} {{eqn | r = p \paren {\frac 2 {q^2} - \frac 1 q}       | c = Variance of Shifted Geometric Distribution: Proof 1 }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \var X       | r = \expect {X^2} - \paren {\expect X}^2       | c =  }} {{eqn | r = p \paren {\frac 2 {\paren {1 - p}^2} - \frac 1 {1 - p} } - \frac {p^2} {\paren {1 - p}^2}       | c = Expectation of Geometric Distribution: $\expect X = \dfrac p {1 - p}$ }} {{eqn | r = \frac p {\paren {1 - p}^2}       | c = after some algebra }} {{end-eqn}} {{qed}}
From Variance of Discrete Random Variable from PGF, we have: :$\operatorname{var} \left({X}\right) = \Pi''_X \left({1}\right) + \mu - \mu^2$ where $\mu = E \left({x}\right)$ is the expectation of $X$. From the Probability Generating Function of Geometric Distribution, we have: :$\Pi_X \left({s}\right) = \dfrac q {1 - ps}$ where $q = 1 - p$. From Expectation of Geometric Distribution, we have: :$\mu = \dfrac p q$ From Derivatives of PGF of Geometric Distribution, we have: :$\Pi''_X \left({s}\right) = \dfrac {2 q p^2} {\left({1 - ps}\right)^3}$ Putting $s = 1$ using the formula $\Pi''_X \left({1}\right) + \mu - \mu^2$: :$\operatorname{var} \left({X}\right) = \dfrac {2 q p^2} {\left({1-p}\right)^3} + \dfrac p q - \left({\dfrac p q}\right)^2$ and hence the result, after some algebra. {{qed}}
=== Proof of Bounds of CDF === This follows directly from the definition of $\Pr$. {{begin-eqn}} {{eqn | o =        | r = S \in \Sigma       | c =  }} {{eqn | o = \leadsto       | r = \O \subseteq S \subseteq \Omega       | c =  }} {{eqn | o = \leadsto       | r = 0 \le \map \Pr S \le 1       | c = Probability Measure is Monotone }} {{end-eqn}} {{qed}} === Proof that CDF is Increasing === Let $x, y \in \R: x \le y$. Let $\map X \omega \le x$. Then: :$\map X \omega \le y$ and so: :$\set {\omega \in \Omega: \map X \omega \le x} \subseteq \set {\omega \in \Omega: \map X \omega \le y}$ Hence the result. {{Qed}} === Proof of Limits of CDF === As $x \to -\infty$, $\hointl \gets x \to \O$. So: :$\map {X^{-1} } {\hointl \gets x} \to \O$ and so: :$\map F x \to 0$ {{explain|this definitely needs more attention}} Similarly, as $x \to +\infty$, $\hointl \gets x \to \R$. So: :$\map {X^{-1} } {\hointl \gets x} \to \Omega$ and so: :$\map F x \to 1$ {{qed}} Category:Cumulative Distribution Functions 9035007i4x30p5byq0l4nki6caodga6
Let $X$ be a discrete random variable. Then by definition: :$\forall x \in \R: \left\{{\omega \in \Omega: X \left({\omega}\right) = x}\right\} \in \Sigma$ But see that: :$\displaystyle \left\{{\omega \in \Omega: X \left({\omega}\right) \le x}\right\} = \bigcup_{{y \in \Omega_X} \atop{y \le x}} \left\{{\omega \in \Omega: X \left({\omega}\right) = y}\right\}$ This is the countable union of events in $\Sigma$. Hence, as $\Sigma$ is a sigma-algebra, $\left\{{\omega \in \Omega: X \left({\omega}\right) \le x}\right\} \in \Sigma$ as required. {{qed}} Category:Probability Theory 5zgur2mr0ewgk0mt41umw47h6uu4h3h
From the definition of expectation: :$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$ Thus: {{begin-eqn}} {{eqn | l = \expect X       | r = \sum_{k \mathop = 1}^n k \paren {\frac 1 n}       | c = {{Defof|Discrete Uniform Distribution}} }} {{eqn | r = \frac 1 n \sum_{k \mathop = 1}^n k       | c =  }} {{eqn | r = \frac 1 n \frac {n \paren {n + 1} } 2       | c = from Closed Form for Triangular Numbers }} {{eqn | r = \frac {n + 1} 2       | c =  }} {{end-eqn}} {{qed}} Category:Expectation Category:Discrete Uniform Distribution jl5q0i8jkbrykmwfb72baf7c2vlfj4v
From the definition of Variance as Expectation of Square minus Square of Expectation: :$\var X = \expect {X^2} - \paren {\expect X}^2$ From Expectation of Function of Discrete Random Variable: :$\displaystyle \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$ So: {{begin-eqn}} {{eqn | l = \expect {X^2}       | r = \sum_{k \mathop = 1}^n k^2 \paren {\frac 1 n}       | c = {{Defof|Discrete Uniform Distribution}} }} {{eqn | r = \frac 1 n \sum_{k \mathop = 1}^n k^2       | c =  }} {{eqn | r = \frac 1 n \frac {n \paren {n + 1} \paren {2 n + 1} } 6       | c = Sum of Sequence of Squares }} {{eqn | r = \frac {\paren {n + 1} \paren {2 n + 1} } 6       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \var X       | r = \expect {X^2} - \paren {\expect X}^2       | c =  }} {{eqn | r = \frac {\paren {n + 1} \paren {2 n + 1} } 6 - \frac {\paren {n + 1}^2} 4       | c = Expectation of Discrete Uniform Distribution: $\expect X = \dfrac {n + 1} 2$ }} {{eqn | r = \frac {2 \paren {2 n^2 + 3 n + 1} - 3 \paren {n^2 + 2 n + 1} } {12}       | c =  }} {{eqn | r = \frac {n^2 - 1} {12}       | c =  }} {{end-eqn}} {{qed}} Category:Variance Category:Discrete Uniform Distribution t5km9pce5d2nng1hedk146beqnjnd7c
Follows directly from the definition: :$\displaystyle \Pi_X \left({s}\right) = \sum_{x \ge 0} p_X \left({x}\right) s^x$ As $p_X \left({x}\right) \ne 0$ for only one value of $x$, all the terms vanish except that one. Hence the result. {{qed}} Category:Probability Generating Functions Category:Degenerate Distribution 392gp80iskxe4zo00lz9wh0ruzzhtxb
From the definition of p.g.f: :$\displaystyle \Pi_X \left({s}\right) = \sum_{x \mathop \ge 0} p_X \left({x}\right) s^x$ From the definition of the Bernoulli distribution: :$p_X \left({x}\right) = \begin{cases} p & : x = a \\ 1 - p & : x = b \\ 0 & : x \notin \left\{{a, b}\right\} \\ \end{cases}$ So: {{begin-eqn}} {{eqn | l=\Pi_X \left({s}\right)       | r=p_X \left({0}\right) s^0 + p_X \left({1}\right) s^1       | c= }} {{eqn | r=\left({1 - p}\right) + p s       | c= }} {{end-eqn}} Hence the result. {{qed}}
From the definition of p.g.f: :$\displaystyle \Pi_X \left({s}\right) = \sum_{k \mathop \ge 0} p_X \left({k}\right) s^k$ From the definition of the binomial distribution: :$\displaystyle p_X \left({k}\right) = \binom n k p^k \left({1 - p}\right)^{n-k}$ So: {{begin-eqn}} {{eqn | l=\Pi_X \left({s}\right)       | r=\sum_{k \mathop = 0}^n \binom n k p^k \left({1 - p}\right)^{n - k} s^k       | c= }} {{eqn | r=\sum_{k \mathop = 0}^n \binom n k \left({p s}\right)^k \left({1 - p}\right)^{n - k}       | c= }} {{eqn | r=\left({\left({p s}\right) + \left({1 - p}\right)}\right)^n       | c=by the Binomial Theorem }} {{end-eqn}} Hence the result. {{qed}}
From the definition of p.g.f: :$\displaystyle \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$ From the definition of the shifted geometric distribution: :$\forall k \in \N, k \ge 1: \map {p_X} k = p q^{k - 1}$ So: {{begin-eqn}} {{eqn | l = \map {\Pi_X} s       | r = \sum_{k \mathop \ge 1} p q^{k - 1} s^k       | c =  }} {{eqn | r = p s \sum_{k \mathop \ge 1} \paren {q s}^{k - 1}       | c =  }} {{eqn | r = p s \sum_{j \mathop \ge 0} \paren {q s}^j       | c = change of index }} {{eqn | r = p s \frac 1 {1 - q s}       | c = Sum of Infinite Geometric Sequence }} {{end-eqn}} Hence the result. {{qed}}
From the definition of p.g.f: :$\displaystyle \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$ From the definition of the geometric distribution: :$\forall k \in \N, k \ge 0: \map {p_X} k = q p^k$ So: {{begin-eqn}} {{eqn | l = \map {\Pi_X} s       | r = \sum_{k \mathop \ge 0} q p^k s^k       | c =  }} {{eqn | r = q \sum_{k \mathop \ge 0} \paren {p s}^k       | c =  }} {{eqn | r = q \frac 1 {1 - p s}       | c = Sum of Infinite Geometric Sequence }} {{end-eqn}} Hence the result. {{qed}} Category:Probability Generating Functions Category:Geometric Distribution 92qvfwunjq08tkn6sxsc8v745sflady
From the definition of p.g.f: :$\displaystyle \Pi_X \left({s}\right) = \sum_{x \mathop \ge 0} p_X \left({x}\right) s^x$ From the definition of the Poisson distribution: :$\displaystyle \forall k \in \N, k \ge 0: p_X \left({k}\right) = \frac {e^{-\lambda} \lambda^k} {k!}$ So: {{begin-eqn}} {{eqn | l=\Pi_X \left({s}\right)       | r=\sum_{k \mathop \ge 0} \frac {e^{-\lambda} \lambda^k} {k!} s^k       | c= }} {{eqn | r=e^{-\lambda} \sum_{k \mathop \ge 0} \frac {\left({\lambda s}\right)^k} {k!}       | c= }} {{eqn | r=e^{-\lambda} e^{\lambda s}       | c=by Taylor Series Expansion for Exponential Function }} {{eqn | r=e^{-\lambda + \lambda s}       | c= }} {{end-eqn}} Hence the result. {{qed}}
For ease of notation, write $\map p x$ for $\map \Pr {X = x}$. From the definition of the probability generating function: :$\displaystyle \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map p x s^x = \map p 0 + \map p 1 s + \map p 2 s^2 + \map p 3 s^3 + \cdots$ Differentiate this {{WRT|Differentiation}} $s$: {{begin-eqn}} {{eqn | l = \map {\Pi'_X} s       | r = \frac \d {\d s} \sum_{k \mathop = 0}^\infty \map \Pr {X = k} s^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac \d {\d s} \map \Pr {X = k} s^k       | c = Abel's Theorem }} {{eqn | r = \sum_{k \mathop = 0}^\infty k \, \map \Pr {X = k} s^{k - 1}       | c = Power Rule for Derivatives }} {{end-eqn}} Plugging in $s = 1$ gives: :$\displaystyle \map {\Pi'_X} 1 = \sum_{k \mathop = 0}^\infty k \, \map \Pr {X = k} \, 1^{k - 1} = \map p 1 + 2 \map p 2 + 3 \map p 3 + \cdots$ But: :$\displaystyle \sum_{k \mathop = 0}^\infty k \, \map \Pr {X = k} \, 1^{k - 1} = \sum_{k \mathop = 0}^\infty k \, \map \Pr {X = k}$ is precisely the definition of the expectation. {{qed}}
From the definition of the probability generating function: :$\displaystyle \Pi_X \left({s}\right) = \sum_{x \mathop \ge 2} p \left({x}\right) s^x$ From Derivatives of Probability Generating Function at One: :$\displaystyle \Pi''_X \left({s}\right) = \sum_{x \mathop \ge 2} x \left({x-1}\right) p \left({x}\right) s^{x-2}$ But it also holds when you include $x = 0$ and $x = 1$ in the sum, as in both cases the term evaluates to zero and therefore vanishes. So: :$\displaystyle \Pi''_X \left({s}\right) = \sum_{x \mathop \ge 0} x \left({x-1}\right) p \left({x}\right) s^{x-2}$ Plugging in $s = 1$ gives: {{begin-eqn}} {{eqn | l=\Pi''_X \left({1}\right)       | r=\sum_{x \mathop \ge 0} x \left({x-1}\right) p \left({x}\right) 1^{x-2}       | c= }} {{eqn | r=\sum_{x \mathop \ge 0} x^2 p \left({x}\right) - \sum_{x \mathop \ge 0} x p \left({x}\right)       | c= }} {{eqn | r=E \left({X^2}\right) - E \left({X}\right)       | c= }} {{end-eqn}} The result follows from the definition of variance: :$\operatorname{var} \left({X}\right) = E \left({X^2}\right) - \left({E \left({X}\right)}\right)^2$ after a little algebra. {{qed}}
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\dfrac {\d^n} {\d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$ === Basis for the Induction === $\map P 1$ is true, as this just says: :$\dfrac {\d} {\d x} x^m = m x^{m - 1}$ This follows by Power Rule for Derivatives, which also includes the case: :$\dfrac {\d} {\d x} x^0 = 0$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\dfrac {\d^k} {\d x^k} x^m = \begin{cases} m^{\underline k} x^{m - k} & : k \le m \\ 0 & : k > m \end{cases}$ Then we need to show: :$\dfrac {\d^{k + 1} } {\d x^{k + 1} } x^m = \begin{cases} m^{\underline {k + 1} } x^{m - \paren {k + 1} } & : k + 1 \le m \\ 0 & : k + 1 > m \end{cases}$ === Induction Step === This is our induction step: First, let $k < m$. Then we have: {{begin-eqn}} {{eqn | l = \frac {\d^{k + 1} } {\d x^{k + 1} } x^m       | r = \map {\frac \d {\d x} } {\frac {\d^k} {\d x^k} x^m}       | c =  }} {{eqn | r = \map {\frac \d {\d x} } {m^{\underline k} x^{m - k} }       | c = Induction Hypothesis }} {{eqn | r = m^{\underline k} \map {\frac \d {\d x} } {x^{m - k} }       | c = Derivative of Constant Multiple }} {{eqn | r = m^{\underline k} \paren {m - k} \paren {x^{m - k - 1} }       | c = Basis for the Induction }} {{eqn | r = m^{\underline {k + 1} } x^{m - \paren {k + 1} }       | c = Definition of Falling Factorial }} {{end-eqn}} At this stage, as $k < m$, we have that $k+1 \le m$. So far so good. Now suppose that $k = m$. Then by the induction hypothesis: :$\dfrac {\d^k} {\d x^k} x^m = m^{\underline k} x^0 = m!$ Then by Derivative of Constant: :$\dfrac {\d^{k + 1} } {\d x^{k + 1} } x^m = \dfrac \d {\d x} m! = 0$ At this stage, as $k = m$, we have that $k + 1 > m$. Again, so far so good. Finally, suppose that $k > m$. Then by the induction hypothesis: :$\dfrac {\d^k} {\d x^k} x^m = 0$ Then by Derivative of Constant: :$\dfrac {\d^{k + 1} } {\d x^{k + 1} } = 0$ At this stage, as $k > m$, we have that $k + 1 > m$. This is all that is needed. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\dfrac {\d^n} {\d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$ {{Qed}} Category:Derivatives e2c6fy0r9f2jj55j5i701dchu32jjej
Define: :$\displaystyle \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$ From Radius of Convergence of Derivative of Complex Power Series, it follows that $g$ has radius of convergence $R$.  Fix an $\epsilon > 0$ satisfying $\epsilon < R - \cmod {z - \xi}$. Define: :$\displaystyle M = \paren {R - \epsilon - \cmod {z - \xi} }^{-2} \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^n$ Suppose that $\cmod h \le R - \epsilon - \cmod {z - \xi}$. Then, by the Binomial Theorem and the Triangle Inequality: {{begin-eqn}} {{eqn | l = \cmod {\dfrac {\map f {z + h} - \map f z} h - \map g z}       | r = \cmod {\dfrac 1 h \paren {\sum_{n \mathop = 0}^\infty a_n \paren {z + h - \xi}^n - \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n} - \sum_{n \mathop = 1}^\infty n  a_n \paren {z - \xi}^{n - 1} } }} {{eqn | r = \cmod {\sum_{n \mathop = 0}^\infty a_n \sum_{k \mathop = 0}^n \binom n k \paren {z - \xi}^k h^{n - k - 1} - \sum_{n \mathop = 0}^\infty \dfrac {a_n} h \paren {z - \xi}^n - \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1} }       | c = Binomial Theorem }} {{eqn | r = \cmod {\sum_{n \mathop = 1}^\infty a_n \paren {\sum_{k \mathop = 0}^n \binom n k \paren {z - \xi}^k h^{n - k - 1} - \dfrac 1 h \paren {z - \xi}^n - n \paren {z - \xi}^{n - 1} } }        | c = Difference of Absolutely Convergent Series }} {{eqn | r = \cmod {\sum_{n \mathop = 2}^\infty a_n \sum_{k \mathop = 0}^{n - 2} \binom n k \paren {z - \xi}^k h^{n - k - 1} }       | c = by algebraic manipulations }} {{eqn | o = \le       | r = \cmod h \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{k \mathop = 0}^{n - 2} \binom n k \cmod {z - \xi}^k \cmod h^{n-k-2}       | c = Triangle Inequality }} {{eqn | o = \le       | r = \cmod h \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{k \mathop = 0}^{n - 2} \binom n k \cmod {z - \xi}^k \paren {R - \epsilon - \cmod {z - \xi} }^{n - k - 2}       | c = by assumption }} {{eqn | o = \le       | r = \cmod h \paren {R - \epsilon - \cmod {z - \xi} }^{-2} \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{k \mathop = 0}^n \binom n k \cmod {z - \xi}^k \paren {R - \epsilon - \cmod {z - \xi} }^{n - k} }} {{eqn | r = \cmod h \paren {R - \epsilon - \cmod {z - \xi} }^{-2} \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^n       | c = by the binomial theorem }} {{eqn | r = M \cmod h }} {{end-eqn}} Letting $h\to 0$ we see that $\map {f'} z$ exists and $\map {f'} z = \map g z$, as desired. {{qed}}
From the definition of p.g.f: :$\displaystyle \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$ From the definition of the discrete uniform distribution: :$\forall k \in \N, 1 \le k \le n: \map {p_X} k = \dfrac 1 n$ So: {{begin-eqn}} {{eqn | l = \map {\Pi_X} s       | r = \sum_{k \mathop = 1}^n \frac 1 n s^k       | c =  }} {{eqn | r = \frac s n \sum_{k \mathop = 0}^{n - 1} s^k       | c =  }} {{eqn | r = \frac s n \paren {\frac {1 - s^n} {1 - s} }       | c = Sum of Geometric Sequence }} {{end-eqn}} Hence the result. {{qed}} Category:Probability Generating Functions Category:Discrete Uniform Distribution 5vg5nqiib46llfc39erbmrnjn7ph7yt
The Probability Generating Function of Bernoulli Distribution is: :$\Pi_X \left({s}\right) = q + ps$ where $q = 1 - p$. We have that for a given Bernoulli distribution, $p$ and $q$ are constant. So, from Derivative of Constant, Sum Rule for Derivatives, Derivative of Identity Function and Derivative of Constant Multiple: :$\dfrac {\mathrm d} {\mathrm d s} \Pi_X \left({s}\right) = p$ Again, $p$ is constant, so from Derivative of Constant: :$\dfrac {\mathrm d} {\mathrm d s} p = 0$ Higher derivatives are also of course zero, also from Derivative of Constant. {{qed}}
Follows directly from Derivatives of PGF of Binomial Distribution, setting $n = 1$. {{qed}}
The Probability Generating Function of Binomial Distribution is: :$\Pi_X \left({s}\right) = \left({q + ps}\right)^n$ where $q = 1 - p$. From Derivatives of Function of $a x + b$: :$\dfrac {\mathrm d^k} {\mathrm d s^k} \left({f \left({q + ps}\right)}\right) = p^k \dfrac {\mathrm d^k} {\mathrm d z^k} \left({f \left({z}\right)}\right)$ where $z = q + ps$. Here we have that $f \left({z}\right) = z^n$. From Nth Derivative of Mth Power: :$\dfrac {\mathrm d^k} {\mathrm d z^k}  z^n = \begin{cases} n^{\underline k} z^{n-k} & : k \le n \\ 0 & : k > n \end{cases}$ So putting it together: :$\dfrac {\mathrm d^k} {\mathrm d s^k} \Pi_X \left({s}\right) = \begin{cases} n^{\underline k} p^k \left({q + ps}\right)^{n-k} & : k \le n \\ 0 & : k > n \end{cases}$ {{qed}} Category:Binomial Distribution Category:Derivatives of PGFs 9zfxdnno1e3s0lydmaj8m5f3v8854q9
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$ === Basis for the Induction === $\map P 1$ is true, as this is the case proved in Derivative of Exponential Function: :$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\map {\dfrac {\d^k} {\d x^k} } {\exp x} = \exp x$ Then we need to show: :$\map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\exp x} = \exp x$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\exp x}       | r = \map {\dfrac \d {\d x} } {\dfrac {\d^k} {\d x^k} } {\exp x}        | c = {{Defof|Higher Derivatives}} }} {{eqn | r = \map {\dfrac \d {\d x} } {\exp x}       | c = Induction Hypothesis }} {{eqn | r = \exp x       | c = Basis for the Induction }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>0}: \map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$ {{qed}} Category:Derivatives involving Exponential Function qfqbo04vck83ilu68rgwh81rktmnglf
The Probability Generating Function of Poisson Distribution is: {{begin-eqn}} {{eqn | l = \map {\Pi_X} s       | r = e^{-\lambda \paren {1 - s} }       | c =  }} {{eqn | r = e^{-\lambda + \lambda s}       | c =  }} {{eqn | r = e^{-\lambda} e^{\lambda s}       | c = Exponential of Sum }} {{end-eqn}} We have that for a given Poisson distribution, $\lambda$ is constant. From Higher Derivatives of Exponential Function, we have that: :$\dfrac {\d^k} {\d s^k} \paren {e^{\lambda s} } = \lambda^k e^{\lambda s}$ Thus we have: {{begin-eqn}} {{eqn | l = \frac {\d^k} {\d s^k} \map {\Pi_X} s       | r = \frac {\d^k} {\d s^k} e^{-\lambda} \paren {e^{\lambda s} }       | c =  }} {{eqn | r = e^{-\lambda} \frac {\d^k} {\d s^k} \paren {e^{\lambda s} }       | c = Derivative of Constant Multiple }} {{eqn | r = e^{-\lambda} \lambda^k e^{\lambda s}       | c = Higher Derivatives of Exponential Function }} {{end-eqn}} Hence the result. {{qed}} Category:Poisson Distribution Category:Derivatives of PGFs 1dz559evrctwt2gzlccthlb7r3awxph