hcju/proofwikiqa huggingface dataset:
From Skewness in terms of Non-Central Moments, we have:  :$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$ where:  :$\mu$ is the expectation of $X$.  :$\sigma$ is the standard deviation of $X$. By Expectation of Exponential Distribution we have:  :$\mu = \beta$  By Variance of Exponential Distribution we have:  :$\sigma = \beta$  By Raw Moment of Exponential Distribution we also have:  :$\expect {X^3} = 3! \beta^3 = 6 \beta^3$ So:  {{begin-eqn}} {{eqn	| l = \gamma_1 	| r = \frac {6 \beta^3 - 3\beta^3 - \beta^3} {\beta^3} }} {{eqn	| r = \frac {2 \beta^3} {\beta^3} }} {{eqn	| r = 2 }} {{end-eqn}} {{qed}} Category:Skewness Category:Exponential Distribution ewsnznswr0tnmxqwdfyu1hgbrth0keq
We have that every rational number $\dfrac p q$ between $0$ and $1$ occurs infinitely often in $\sequence {a_n}$: :$\dfrac p q, \dfrac {2 p} {2 q}, \dfrac {3 p} {3 q}, \ldots$ Let $x \in \closedint 0 1$. From Between two Real Numbers exists Rational Number, a term $a_{n_1}$ of $\sequence {a_n}$ can be found such that: :$x - 1 < a_{n_1} < x + 1$ Then a term $a_{n_2}$ of $\sequence {a_n}$ such that $n_2 > n_1$ can be found such that: :$x - \dfrac 1 2 < a_{n_2} < x + \dfrac 1 2$ Hence we can create a subsequence $\sequence {a_{n_k} }$ such that: :$x - \dfrac 1 k < a_{n_k} < x + \dfrac 1 k$ and so by the Squeeze Theorem for Real Sequences: :$r_{n_k} \to x$ as $k \to \infty$ {{qed}}
{{begin-eqn}} {{eqn	| l = \expect X 	| r = \prod_{k \mathop = 0}^0 \paren {n + 2 k} 	| c = Raw Moment of Chi-Squared Distribution }} {{eqn	| r = n }} {{end-eqn}} {{qed}} Category:Expectation Category:Chi-Squared Distribution 5elkpwo87jmaekz3vfbsu94evuz4e5u
By Variance as Expectation of Square minus Square of Expectation, we have:  :$\var X = \expect {X^2} - \paren {\expect X}^2$ By Expectation of Chi-Squared Distribution, we have:  :$\expect X = n$ We also have:  {{begin-eqn}} {{eqn	| l = \expect {X^2} 	| r = \prod_{k \mathop = 0}^1 \paren {n + 2 k} 	| c = Raw Moment of Chi-Squared Distribution }} {{eqn	| r = n \paren {n + 2} }} {{eqn	| r = n^2 + 2n }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \var X 	| r = n^2 + 2n - n^2 }} {{eqn	| r = 2n }} {{end-eqn}} {{qed}} Category:Variance Category:Chi-Squared Distribution jn4fuad6z738a6tu2sjffwr1d9j6iai
From Skewness in terms of Non-Central Moments, we have:  :$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$ where:  :$\mu$ is the expectation of $X$.  :$\sigma$ is the standard deviation of $X$. By Expectation of Chi-Squared Distribution we have:  :$\mu = n$ By Variance of Chi-Squared Distribution we have:  :$\sigma = \sqrt {2 n}$ We also have:  {{begin-eqn}} {{eqn	| l = \expect {X^3} 	| r = \prod_{k \mathop = 0}^2 \paren {n + 2 k} 	| c = Raw Moment of Chi-Squared Distribution }} {{eqn	| r = n \paren {n + 2} \paren {n + 4} }} {{eqn	| r = n^3 + 6 n^2 + 8 n }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \gamma_1 	| r = \frac {n^3 + 6 n^2 + 8 n - 6 n^2 - n^3} {\paren {\sqrt {2 n} }^3} }} {{eqn	| r = \frac {\sqrt {\paren {8 n}^2} } {\sqrt {\paren {2 n}^3} } }} {{eqn	| r = \frac {\sqrt {64 n^2} } {\sqrt {8 n^3} }  }} {{eqn	| r = \sqrt {\frac 8 n} }} {{end-eqn}} {{qed}} Category:Skewness Category:Chi-Squared Distribution 15ar3wso9nvwv3wsx9sst0s4vp1k21a
From the definition of excess kurtosis, we have:  :$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$ where:  :$\mu$ is the expectation of $X$.  :$\sigma$ is the standard deviation of $X$. By Expectation of Chi-Squared Distribution: :$\mu = n$ By Variance of Chi-Squared Distribution: :$\sigma = \sqrt {2 n}$ As is shown in Skewness of Chi-Squared Distribution, we have:  :$\expect {X^3} = n^3 + 6 n^2 + 8 n$ We also have:  {{begin-eqn}} {{eqn	| l = \expect {X^4} 	| r = \prod_{k \mathop = 0}^3 \paren {n + 2 k} 	| c = Raw Moment of Chi-Squared Distribution }} {{eqn	| r = n \paren {n + 2} \paren {n + 4} \paren {n + 6} }} {{eqn	| r = n \paren {n^2 + 6 n + 8} \paren {n + 6} }} {{eqn	| r = n^4 + 12 n^3 + 44 n^2 + 48 n }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = \gamma_2 	| r = \frac {\expect {X^4 - 4 X^3 n + 6 X^2 n^2 - 4 X n^3 + n^4} } {\paren {\sqrt {2 n} }^4} - 3 	| c = Fourth Power of Sum, Linearity of Expectation Function }} {{eqn	| r = \frac {\expect {X^4} - 4 n \expect {X^3} + 6 n^2 \expect {X^2} - 4 n^3 \expect X + n^4} {4 n^2} - 3 	| c = Linearity of Expectation Function }} {{eqn	| r = \frac {n^4 + 12 n^3 + 44 n^2 + 48 n - 4 n \paren {n^3 + 6 n^2 + 8 n} + 6 n^2 \paren {n^2 + 2 n} - 4 n^4 + n^4} {4 n^2} - 3 	| c = Variance as Expectation of Square minus Square of Expectation }} {{eqn	| r = \frac {n^4 + 12 n^3 + 44 n^2 + 48 n - 4n^4 - 24 n^3 - 32 n^2 + 6 n^4 + 12 n^3 - 4 n^4 + n^4 - 12 n^2} {4 n^2} }} {{eqn	| r = \frac {48 n} {4 n^2} }} {{eqn	| r = \frac {12} n }} {{end-eqn}} {{qed}} Category:Kurtosis Category:Chi-Squared Distribution ait1fa9weoei4doy8fr2pkbccewemw5
From the definition of the chi-squared distribution, $X$ has probability density function: :$\displaystyle \map {f_X} x = \dfrac 1 {2^{n / 2} \map \Gamma {n / 2} } x^{\paren {n / 2} - 1} e^{- x / 2}$ From the definition of the expected value of a continuous random variable:  :$\displaystyle \expect {X^m} = \int_0^\infty x^m \map {f_X} x \rd x$ So:  {{begin-eqn}} {{eqn	| l = \expect {X^m} 	| r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty x^{\paren {n / 2} + m - 1} e^{- x / 2} \rd x }} {{eqn	| r = \frac 2 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty \paren {2 u}^{\paren {n / 2} + m - 1} e^{- u} \rd u 	| c = substituting $x = 2 u$ }} {{eqn	| r = \frac {2^{\paren {n / 2} + m - 1 + 1} } {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty u^{\paren {n / 2} + m - 1} e^{- u} \rd u }} {{eqn	| r = \frac {2^m} {\map \Gamma {n / 2} } \map \Gamma {\paren {n / 2} + m} 	| c = {{Defof|Gamma Function}} }} {{eqn	| r = \frac {\map \Gamma {n / 2} } {\map \Gamma {n / 2} } 2^m \prod_{k \mathop = 0}^{m - 1} \paren {\frac n 2 + k} 	| c = Gamma Difference Equation }} {{eqn	| r = \prod_{k \mathop = 0}^{m - 1} \paren {n + 2 k} }} {{end-eqn}} {{qed}} Category:Raw Moments Category:Chi-Squared Distribution 2nds18bmb9uqlqh4p5bzv4sf272uv0n
{{AimForCont}} $\sequence {x_n}$ is bounded. Let $H \in \R$ be an upper bound of $\sequence {x_n}$ By the Archimedean Principle, we have that: :$\exists N \in \N: N > H^2$ and so: :$\exists N \in \N: \sqrt N > H$ But $\sqrt N$ is a term of $\sequence {x_n}$. So $H$ cannot be an upper bound of $\sequence {x_n}$ Hence by Proof by Contradiction, $\sequence {x_n}$ is not bounded. From Real Cauchy Sequence is Bounded, it follows by the Rule of Transposition that $\sequence {x_n}$ is not a Cauchy sequence So, while we have the case that: :$\size {\sqrt {n + 1} - \sqrt n} \to 0$ as $n \to \infty$ it is not the case that: :$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \size {\sqrt n - \sqrt m} < \epsilon$ {{qed}}
Let $Y \sim \chi^2_1$.  We aim to show that:  :$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$ for all real $x \ge 0$. We have:  {{begin-eqn}} {{eqn	| l = \map \Pr {Y < x^2} 	| r = \frac 1 {\sqrt 2 \map \Gamma {1 / 2} } \int_0^{x^2} t^{\paren {1 / 2} - 1} e^{-t / 2} \rd t 	| c = {{Defof|Chi-Squared Distribution}} }} {{eqn	| r = \frac 1 {\sqrt {2 \pi} } \int_0^{x^2} \frac {e^{-t / 2} } {\sqrt t} \rd t 	| c = Gamma Function of One Half }} {{eqn	| r = \frac 2 {\sqrt {2 \pi} } \int_0^x u \frac {e^{-u^2 / 2} } {\sqrt {u^2} } \rd u 	| c = substituting $t = u^2$ }} {{eqn	| r = \frac 2 {\sqrt {2 \pi} } \int_0^x e^{-u^2 / 2} \rd u }} {{eqn	| r = \frac 1 {\sqrt {2 \pi} } \int_{-x}^x e^{-u^2 / 2} \rd u 	| c = Definite Integral of Even Function }} {{eqn	| r = \map \Pr {-x < X < x} 	| c = {{Defof|Gaussian Distribution}} }} {{end-eqn}} {{qed}} Category:Gaussian Distribution Category:Chi-Squared Distribution gkaunz758ga94d7a250gglk5tc36xce
Let $Y \sim \chi^2_N$.  By Moment Generating Function of Chi-Squared Distribution, the moment generating function of $X_i$, $M_{X_i}$, is given by:  :$\map {M_{X_i} } t = \paren {1 - 2 t}^{-n_i / 2}$ Similarly, the moment generating function, $M_Y$, of $Y$ is given by:  :$\map {M_Y} t = \paren {1 - 2 t}^{-N / 2}$ By Moment Generating Function of Linear Combination of Independent Random Variables, the moment generating function of $X$, $M_X$, is given by:  :$\displaystyle \map {M_X} t = \prod_{i \mathop = 1}^k \map {M_{X_i} } t$ We aim to show that:  :$\displaystyle \map {M_X} t = \map {M_Y} t$ By Moment Generating Function is Unique, this ensures $X = Y$. We have: {{begin-eqn}} {{eqn	| l = \map {M_X} t 	| r = \prod_{i \mathop = 1}^k \paren {1 - 2 t}^{-n_i / 2} }} {{eqn	| r = \paren {1 - 2 t}^{-\paren {n_1 + n_2 + \ldots + n_k} / 2} }} {{eqn	| r = \paren {1 - 2 t}^{-N / 2} }} {{eqn	| r = \map {M_Y} t }} {{end-eqn}} {{qed}} Category:Chi-Squared Distribution nczwv335vmiwis9dnbwf8kc88vfz5x5
From the definition of the chi-squared distribution, $X$ has probability density function: :$\displaystyle \map {f_X} x = \dfrac 1 {2^{n / 2} \map \Gamma {n / 2} } x^{\paren {n / 2} - 1} e^{- x / 2}$ From the definition of a moment generating function: :$\displaystyle \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$ So:  {{begin-eqn}} {{eqn	| l = \map {M_X} t 	| r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty x^{\paren {n / 2} - 1} e^{t x - \paren {x / 2} } \rd x }} {{eqn	| r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty x^{\paren {n / 2} - 1} e^{-\paren {\paren {1 / 2} - t} x } \rd x }} {{end-eqn}} Note that if $\dfrac 1 2 - t < 0$, then $e^{-\paren {\paren {1 / 2} - t} x } \to \infty$ by Exponential Tends to Zero and Infinity, so the integral diverges in this case and the expectation fails to exist. If $\dfrac 1 2 - t = 0$, then $e^{-\paren {\paren {1 / 2} - t} x } = 1$ by Exponential of Zero, so the integral is equal to:  :$\displaystyle \int_0^\infty x^{\paren {n / 2} - 1} \rd x$ However, as $n$ is a strictly positive integer, we have $\dfrac n 2 - 1 > -1$.  So, the integral diverges in this case by Integral to Infinity of Reciprocal of Power of x, and hence the expectation also fails to exist in this case. For the integral to converge, we must have $\dfrac 1 2 - t > 0$. That is, when $t < \dfrac 1 2$.  In this case:  {{begin-eqn}} {{eqn	| l = \frac 1 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty x^{\paren {n / 2} - 1} e^{-\paren {\paren {1 / 2} - t} x } \rd x 	| r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} \paren {\frac 1 2 - t} } \int_0^\infty \paren {\frac u {\frac 1 2 - t} }^{n / 2 - 1} e^{- u} \rd u 	| c = substituting $u = \paren {\frac 1 2 - t} x$ }} {{eqn	| r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} \paren {\frac 1 2 - t}^{n / 2} } \int_0^\infty u^{n / 2 - 1} e^{- u} \rd u }} {{eqn	| r = \frac {\map \Gamma {n / 2} } {\map \Gamma {n / 2} } \paren {1 - 2 t}^{- n / 2} 	| c = {{Defof|Gamma Function}} }} {{eqn	| r = \paren {1 - 2 t}^{- n / 2} }} {{end-eqn}} {{qed}} Category:Moment Generating Functions Category:Chi-Squared Distribution lkc7zvpbzrtqy8y3o0q7y8nafzu4dyo
{{begin-eqn}} {{eqn	| l = \map {M_X} t 	| r = \expect {\map \exp {t X} } 	| c = {{Defof|Moment Generating Function}} }} {{eqn	| r = \expect {\map \exp {t \sum_{i \mathop = 1}^n k_i X_i} } }} {{eqn	| r = \expect {\prod_{i \mathop = 1}^n \map \exp {t k_i X_i} } 	| c = Exponential of Sum }} {{eqn	| r = \prod_{i \mathop = 1}^n \expect {\map \exp {t k_i X_i} } 	| c = Condition for Independence from Product of Expectations }} {{eqn	| r = \prod_{i \mathop = 1}^n \map {M_{X_i} } {k_i t} 	| c = {{Defof|Moment Generating Function}} }} {{end-eqn}} {{qed}} Category:Moment Generating Functions dzd7cx2ujmvr3c9hk7s355q607sojgk
By Square of Standard Gaussian Random Variable has Chi-Squared Distribution, we have:  :$X^2_i \sim \chi^2_1$ for $1 \le i \le n$.  So, by Sum of Chi-Squared Random Variables, we have:  :$\displaystyle \sum_{i \mathop = 1}^n X^2_i \sim \chi^2_{1 + 1 + 1 \ldots} = \chi^2_n$  {{qed}} Category:Chi-Squared Distribution jux10bcen80nce4qjw26kcvc98bec09
In $3$-dimensional Euclidean space the length of the curve is: :$\ds \int_{x_0}^{x_1} \sqrt {1 + y'^2 + z'^2} \rd x$ The sphere satisfies the following equation: {{begin-eqn}} {{eqn | l = \map g {x, y, z}       | r = x^2 + y^2 + z^2 - a^2 }} {{eqn | r = 0 }} {{end-eqn}} Consider its partial derivatives {{WRT}} $y$ and $z$. :$\dfrac {\partial g} {\partial y} = 2y$ :$\dfrac {\partial g} {\partial z} = 2z$ $g_y$ and $g_z$ vanish for $y = 0$ and $z = 0$ respectively. Substitution of this into the sphere equation tells us that $x^2 = a^2$. Therefore, the following analysis should exclude points with $x = \pm a$. By Simplest Variational Problem with Subsidiary Conditions for Curve on Surface, the length functional is replaced by the following auxiliary functional: :$\ds \int_{x_0}^{x_1} \sqbrk {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } \rd x$ It follows that: :$\map {\dfrac {\partial} {\partial y'} } {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } = \dfrac {y'} {\sqrt {1 + y'^2 + z'^2} }$ :$\map {\dfrac {\partial} {\partial y} } {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } = 2 y \lambda$ Analogous relations hold for $\map z x$.  Then by Euler's Equations the following equations of motion hold: :$2 y \map \lambda x - \dfrac \d {\d x} \dfrac {y'} {\sqrt{1 + y'^2 + z'^2} } = 0$ :$2 z \map \lambda x - \dfrac \d {\d x} \dfrac {z'} {\sqrt{1 + y'^2 + z'^2} } = 0$ {{qed}}
From the definition of the chi distribution, $X$ has probability density function: :$\displaystyle \map {f_X} x = \dfrac 1 {2^{\paren {n / 2} - 1} \map \Gamma {n / 2} } x^{n - 1} e^{- x^2 / 2}$ From the definition of the expected value of a continuous random variable: :$\displaystyle \expect X = \int_0^\infty x \map {f_X} x \rd x$ So: {{begin-eqn}} {{eqn	| l = \expect X 	| r = \frac 1 {2^{\paren {n / 2} - 1} \map \Gamma {n / 2} } \int_0^\infty x^n e^{- x^2 / 2} \rd x }} {{eqn	| r = \frac 1 {2^{\paren {n / 2} - 1 + \paren {1 / 2} } \map \Gamma {n / 2} } \int_0^\infty \frac 1 {\sqrt t} \paren {\sqrt {2 t} }^n e^{- t} \rd t 	| c = substituting $x = \sqrt {2 t}$ }} {{eqn	| r = \frac {2^{\paren {n / 2} } } {2^{\paren {n / 2} - \paren {1 / 2} } \map \Gamma {n / 2} } \int_0^\infty t^{\paren {n - 1} / 2} e^{- t} \rd t  }} {{eqn	| r = \sqrt 2 \dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} } 	| c = {{Defof|Gamma Function}} }} {{end-eqn}} {{qed}} Category:Expectation Category:Chi Distribution evlrudbecnpad6hpuit0k7x9zmc1kjn
By Sum of Chi-Squared Random Variables, we have:  :$\displaystyle \sum_{i \mathop = 1}^k X_i \sim \chi^2_{n k}$ By Multiple of Chi-Squared Random Variable has Gamma Distribution, we then have:  :$\displaystyle \frac 1 k \sum_{i \mathop = 1}^k X_i \sim \map \Gamma {\frac {n k} 2, \frac 1 {\frac 2 k} } = \map \Gamma {\frac {n k} 2, \frac k 2}$ {{qed}} Category:Chi-Squared Distribution Category:Gamma Distribution dohpoyhfpps11w5re8r9gaub3yof2v6
=== Definition 1 implies Definition 2 === {{:Equivalence of Definitions of Separated Sets/Definition 1 implies Definition 2}}{{qed|lemma}} === Definition 2 implies Definition 1 === {{:Equivalence of Definitions of Separated Sets/Definition 2 implies Definition 1}}{{qed}} Category:Separated Sets 6zt9vvs4i7xuf52q7od413ug0xhpdpc
From Topological Closure is Closed, $B^-$ is closed in $T$. Let $U =  S \setminus B^-$ be the relative complement of $B^-$. By the definition of a closed set, $U$ is open in $T$. From Empty Intersection iff Subset of Relative Complement: :$A \subseteq S \setminus B^- = U$ From Relative Complement of Relative Complement: :$S \setminus U = B^-$ By the definition of the closure of a subset: :$B \subseteq B^- = S \setminus U$ From Empty Intersection iff Subset of Relative Complement: :$U \cap B = \O$ Similarly, let $V =  S \setminus A^-$ then $V \in \tau$ with: :$B \subset V$ and :$V \cap A = \O$
Let $U \in \tau$ be an arbitrary open set of $T$ such that $A \subseteq U$ and $U \cap B = \O$. From Empty Intersection iff Subset of Relative Complement: :$B \subseteq S \setminus U$ By the definition of a closed set, the relative complement $S \setminus U$ of $U$ is closed in $T$. From Set Closure is Smallest Closed Set in Topological Space: :$B^- \subseteq S \setminus U$ From Empty Intersection iff Subset of Relative Complement: :$B^- \cap U = \O$ Then {{begin-eqn}} {{eqn | l = B^- \cap A        | r = B^- \cap \paren {U \cap A}       | c = Intersection with Subset is Subset }} {{eqn | r = \paren {B^- \cap U} \cap A       | c = Intersection is Associative }} {{eqn | r = \O \cap A       | c = as $B^- \cap U = \O$ }} {{eqn | r = \O       | c = Intersection with Empty Set }} {{end-eqn}} Similarly: :$A^- \cap B = \O$
By Expectation of Student's t-Distribution, we have that $\expect X$ exists {{iff}} $k > 2$.  Hence, take $k > 2$ from here on. By Expectation of Student's t-Distribution, we have that for $k > 2$:  :$\expect X = 0$ From Square of Random Variable with t-Distribution has F-Distribution, we have:  :$\expect {X^2} = \expect Y$ with $Y \sim F_{1, k}$, where $F_{1, k}$ is the $F$-distribution with $\tuple {1, k}$ degrees of freedom. Since $k > 2$, by Expectation of F-Distribution we have:  :$\expect {X^2} = \dfrac k {k - 2}$ We therefore have:  {{begin-eqn}} {{eqn	| l = \var X 	| r = \expect {X^2} - \paren {\expect X}^2 	| c = Variance as Expectation of Square minus Square of Expectation }} {{eqn	| r = \frac k {k - 2} - 0^2 }} {{eqn	| r = \frac k {k - 2} }} {{end-eqn}} {{qed}} Category:Variance Category:Student's t-Distribution hi8no3s0ipr6446akgnwy0saa5do08h
Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$. Let $\tau'_A$ be the subspace topology on $A$ induced by $\tau_B$. By the definition of a connected set, $A$ is connected in $T$ {{iff}} $\struct {A, \tau_A}$ is a connected topological space. Similarly, $A$ is connected in $T_B$ {{iff}} $\struct {A, \tau'_A}$ is a connected topological space. From Subspace of Subspace is Subspace, we have $\tau_A = \tau'_A$. Hence $A$ is connected in $T_B$ {{iff}} $A$ is connected in $T$. {{qed}} Category:Connected Spaces q9t9ucnz27q2wfju6e5shx4uvs914cm
{{begin-eqn}} {{eqn	| l = \expect X 	| r = \frac 1 {\lambda^1} \prod_{m \mathop = 0}^0 \paren {k + m} 	| c = Raw Moment of Erlang Distribution }} {{eqn	| r = \frac k \lambda }} {{end-eqn}} {{qed}} Category:Expectation Category:Erlang Distribution 2mvhbomtjhf2h95radm04h4kd0kzyb2
By Variance as Expectation of Square minus Square of Expectation, we have:  :$\var X = \expect {X^2} - \paren {\expect X}^2$ By Expectation of Erlang Distribution, we have:  :$\expect X = \dfrac k \lambda$ We also have:  {{begin-eqn}} {{eqn	| l = \expect {X^2} 	| r = \frac 1 {\lambda^2} \prod_{m \mathop = 0}^1 \paren {k + m} }} {{eqn	| r = \frac {k \paren {k + 1} } {\lambda^2}  }} {{eqn	| r = \frac {k^2 + k} {\lambda^2} }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = \var X 	| r = \frac {k^2 + k} {\lambda^2} - \paren {\frac k \lambda}^2 }} {{eqn	| r = \frac {k^2 + k - k^2} {\lambda^2} }} {{eqn	| r = \frac k {\lambda^2} }} {{end-eqn}} {{qed}} Category:Variance Category:Erlang Distribution hoxci9wul05aoxweky745hp408u3g4j
{{begin-eqn}} {{eqn	| l = \cov {X, Y} 	| r = \expect {\paren {X - \expect X} \paren {Y - \expect Y} } 	| c = {{Defof|Covariance}} }} {{eqn	| r = \expect {X Y - X \expect Y - Y \expect X + \expect X \expect Y} }} {{eqn	| r = \expect {X Y} - \expect Y \expect X - \expect Y \expect X + \expect X \expect Y 	| c = Linearity of Expectation Function, Expectation of Constant }} {{eqn	| r = \expect {X Y} - \expect X \expect Y }} {{end-eqn}} {{qed}} Category:Covariance dzcjh7la7k1pzk2uzicwmu51dxllf1a
{{begin-eqn}} {{eqn	| l = \cov {X, Y} 	| r = \expect {X Y} - \expect X \expect Y 	| c = Covariance as Expectation of Product minus Product of Expectations }} {{eqn	| r = \expect X \expect Y - \expect X \expect Y 	| c = Condition for Independence from Product of Expectations }} {{eqn	| r = 0 }} {{end-eqn}} {{qed}} Category:Covariance 1ar5z462t9jgr5eh2psrg6mps7hizg6
{{begin-eqn}} {{eqn	| l = \var {a X + b Y} 	| r = \expect {\paren {a X + b Y - \expect {a X + b Y} }^2} 	| c = {{Defof|Variance}} }} {{eqn	| r = \expect {\paren {a X + b Y - a \, \expect X - b \, \expect Y}^2} 	| c = Linearity of Expectation Function }} {{eqn	| r = \expect {\paren {a \paren {X - \expect X} + b \paren {Y - \expect Y} }^2} }} {{eqn	| r = \expect {a^2 \paren {X - \expect X}^2 + b^2 \paren {Y - \expect Y}^2 + 2 a b \paren {X - \expect X} \paren {Y - \expect Y} } }} {{eqn	| r = a^2 \, \expect {\paren {X - \expect X}^2} + b^2 \, \expect {\paren {Y - \expect Y}^2} + 2 a b \, \expect {\paren {X - \expect X} \paren {Y - \expect Y} } 	| c = Linearity of Expectation Function }} {{eqn	| r = a^2 \, \var X + b^2 \, \var Y + 2 a b \, \cov {X, Y} 	| c = {{Defof|Variance}}, {{Defof|Covariance}} }} {{end-eqn}} {{qed}} Category:Variance 8xmrbhcphrkjc31fvqzy05qqf3la33g
Note that:  :$\map \Pr {Y^2 \ge 0} = 1$ so we have by Expectation of Non-Negative Random Variable is Non-Negative:  :$\expect {Y^2} \ge 0$ First, take $\expect {Y^2} > 0$.  Let $Z$ be a random variable with:  :$Z = X - Y \dfrac {\expect {X Y} } {\expect {Y^2} }$ Note that we have:  :$\map \Pr {Z^2 \ge 0} = 1$ so again applying Expectation of Non-Negative Random Variable is Non-Negative, we have:  :$\expect {Z^2} \ge 0$ That is:  {{begin-eqn}} {{eqn	| l = 0 	| o = \le 	| r = \expect {Z^2} }} {{eqn	| r = \expect {\paren {X - Y \frac {\expect {X Y} } {\expect {Y^2} } }^2} }} {{eqn	| r = \expect {X^2 - 2 X Y \frac {\expect {X Y} } {\expect {Y^2} } + Y^2 \frac {\paren {\expect {X Y} }^2} {\paren {\expect {Y^2} }^2} } 	| c = Square of Sum }} {{eqn	| r = \expect {X^2} - 2 \expect {X Y} \frac {\expect {X Y} } {\expect {Y^2} } + \expect {Y^2} \frac {\paren {\expect {X Y} }^2} {\paren {\expect {Y^2} }^2} 	| c = Linearity of Expectation Function }} {{eqn	| r = \expect {X^2} - \frac {\paren {\expect {X Y} }^2 } {\expect {Y^2} } }} {{end-eqn}} We therefore have:  :$\dfrac {\paren {\expect {X Y} }^2} {\expect {Y^2} } \le \expect {X^2}$ giving:  :$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$ as required.  It remains to address the case $\expect {Y^2} = 0$.  Note that since $\map \Pr {Y^2 \ge 0} = 1$, from Condition for Expectation of Non-Negative Random Variable to be Zero we necessarily have:  :$\map \Pr {Y^2 = 0} = 1$ That is:  :$\map \Pr {Y = 0} = 1$ This implies that the random variable $X Y$ has:  :$\map \Pr {X Y = 0} = 1$ From which, we have that:  :$\map \Pr {X Y \ge 0} = 1$ So, applying Expectation of Non-Negative Random Variable is Non-Negative again we have:  :$\expect {X Y} = 0$ With that, we have:  :$\paren {\expect {X Y} }^2 = 0$ and:  :$\expect {X^2} \expect {Y^2} = \expect {X^2} \times 0 = 0$ So the inequality:  :$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$ also holds in the case $\expect {Y^2} = 0$, completing the proof. {{qed}}
We have, by the definition of variance, that both:  :$\expect {\paren {X - \expect X}^2}$ and:  :$\expect {\paren {Y - \expect Y}^2}$ exist and are finite.  Therefore: {{begin-eqn}} {{eqn	| l = \paren {\cov {X, Y} }^2 	| r = \paren {\expect {\paren {X - \expect X} \paren {Y - \expect Y} } }^2 	| c = {{Defof|Covariance}} }} {{eqn	| o = \le  	| r = \expect {\paren {X - \expect X}^2} \expect {\paren {Y - \expect Y}^2} 	| c = Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares }} {{eqn	| r = \var X \, \var Y 	| c = {{Defof|Variance}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \paren {\map \rho {X, Y} }^2 	| r = \paren {\frac {\map {\operatorname {Cov} } {X, Y} } {\sqrt {\var X \var Y} } }^2 	| c = {{Defof|Pearson Correlation Coefficient}} }} {{eqn	| r = \frac {\paren {\map {\operatorname {Cov} } {X, Y} }^2} {\var X \var Y} }} {{eqn	| o = \le 	| r = 1 	| c = Square of Covariance is Less Than or Equal to Product of Variances }} {{end-eqn}} So: :$\size {\map \rho {X, Y} } \le 1$ {{qed}}
From Local Basis for Open Sets Implies Neighborhood Basis of Open Sets, it follows directly that: :each point of $T$ has a neighborhood basis  consisting entirely of connected sets in $T$.