hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn	| l = \expect X 	| r = \prod_{r \mathop = 0}^0 \frac {\alpha + r} {\alpha + \beta + r} 	| c = Raw Moment of Beta Distribution }} {{eqn	| r = \frac {\alpha + 0} {\alpha + \beta + 0} }} {{eqn	| r = \frac \alpha {\alpha + \beta} }} {{end-eqn}} {{qed}}
From the definition of the Beta distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac {x^{\alpha - 1} \paren {1 - x}^{\beta - 1} } {\map \Beta {\alpha, \beta} }$ From Variance as Expectation of Square minus Square of Expectation:  :$\displaystyle \var X = \int_0^1 x^2 \map {f_X} X \rd x - \paren {\expect X}^2$ So:  {{begin-eqn}} {{eqn	| l = \var X 	| r = \frac 1 {\map \Beta {\alpha, \beta} } \int_0^1 x^{\alpha + 1} \paren {1 - x}^{\beta - 1} \rd x - \frac {\alpha^2} {\paren {\alpha + \beta}^2} 	| c = Expectation of Beta Distribution }} {{eqn	| r = \frac {\map \Beta {\alpha + 2, \beta} } {\map \Beta {\alpha, \beta} } - \frac {\alpha^2} {\paren {\alpha + \beta}^2} 	| c = {{Defof|Beta Function|index = 1}} }} {{eqn	| r = \frac {\map \Gamma {\alpha + 2} \, \map \Gamma \beta} {\map \Gamma {\alpha + \beta + 2} } \cdot \frac {\map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta} - \frac {\alpha^2} {\paren {\alpha + \beta}^2} 	| c = {{Defof|Beta Function|index = 3}} }} {{eqn	| r = \frac {\alpha \paren {\alpha + 1} } {\paren {\alpha + \beta} \paren {\alpha + \beta + 1} } \cdot \frac {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} } - \frac {\alpha^2} {\paren {\alpha + \beta}^2} 	| c = Gamma Difference Equation }} {{eqn	| r = \frac {\paren {\alpha^2 + \alpha} \paren {\alpha + \beta} } {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } - \frac {\alpha^2 \paren {\alpha + \beta + 1} } {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } }} {{eqn	| r = \frac {\alpha ^3 + \alpha^2 \beta + \alpha^2 + \alpha \beta - \alpha^3 - \alpha^2 \beta - \alpha^2} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } }} {{eqn	| r = \frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } }} {{end-eqn}} {{qed}} Category:Variance Category:Beta Distribution t0rawvs64jf7qakf0oxwz2slz9x4gh2
From the definition of the Gamma distribution, $X$ has probability density function: :$\displaystyle f_X\left({x}\right) = \frac { \beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\Gamma \left({\alpha}\right)}$ From the definition of a moment generating function: :$\displaystyle M_X \left({t}\right) = \mathbb E \left[{ e^{t X} }\right] = \int_0^\infty e^{tx} f_X \left({x}\right) \rd x$ First take $t < \beta$.  Then: {{begin-eqn}} {{eqn	| l = M_X \left({t}\right) 	| r = \frac {\beta^\alpha} {\Gamma \left({\alpha}\right)} \int_0^\infty x^{\alpha - 1} e^{-\left({\beta - t}\right) x} \rd x }} {{eqn	| r = \frac {\beta^\alpha} {\Gamma \left({\alpha}\right)} \int_0^\infty \left({ \frac u {\beta - t} }\right)^{\alpha - 1} e^{-u} \frac {\rd u} {\beta - t} 	| c = substituting $\left({\beta - t}\right) x = u$ }} {{eqn	| r = \frac{\beta^\alpha} {\Gamma \left({\alpha}\right) \left({\beta - t}\right)^\alpha} \int_0^\infty u^{\alpha - 1} e^{-u}  \rd u }} {{eqn	| r = \frac {\beta^\alpha \Gamma \left({\alpha}\right)} {\Gamma \left({\alpha}\right) \left({\beta - t}\right)^\alpha} 	| c = {{Defof|Gamma Function}} }} {{eqn	| r = \left({ \frac \beta {\beta - t} }\right)^\alpha }} {{eqn	| r = \left({\frac {\beta - t} \beta}\right)^{-\alpha} }} {{eqn	| r = \left({1 - \frac t \beta}\right)^{-\alpha} }} {{end-eqn}} Now take $t = \beta$.  Our integral becomes:  {{begin-eqn}} {{eqn 	| l = \frac {\beta^\alpha} {\Gamma \left({\alpha}\right)} \int_0^\infty x^{\alpha - 1} \rd x 	| r = \frac {\beta^\alpha} {\alpha \Gamma \left({\alpha}\right)} \left[{x^\alpha}\right]_0^\infty 	| c = Primitive of Power, Fundamental Theorem of Calculus }} {{eqn	| r = 0 + \frac {\beta^\alpha} {\alpha \Gamma \left({\alpha}\right)} \lim_{x \to \infty} x^\alpha }} {{eqn	| o = \to 	| r = \infty 	| c = for $\alpha > 0$, $x^\alpha$ increases without bound as $x \to \infty$ }} {{end-eqn}} So $\mathbb E \left[{ e^{\beta X} }\right]$ does not exist. Finally take $t > \beta$. Note that $-\left({\beta - t}\right)$ is therefore positive. As a consequence of Exponential Dominates Polynomial, we have $x^{\alpha - 1} < e^{-\left({\beta - t}\right) x}$ for sufficiently large $x$.  Therefore, in this case, the integrand increases without bound.  We conclude that the integral is divergent, hence $\mathbb E \left[{ e^{tX} }\right]$ does not exist for $t > \beta$.  {{qed}} Category:Moment Generating Functions Category:Gamma Distribution lpa0mtzce9kzn1wrwc76otqzuudn3ky
{{begin-eqn}} {{eqn | l = \nabla \times \paren {\mathbf f + \mathbf g}       | r = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ \dfrac \partial {\partial x} & \dfrac \partial {\partial y} & \dfrac \partial {\partial z} \\ f_x + g_x & f_y + g_y & f_z + g_x \end {vmatrix}       | c = {{Defof|Curl Operator}} }} {{eqn | r = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ \dfrac \partial {\partial x} & \dfrac \partial {\partial y} & \dfrac \partial {\partial z} \\ f_x & f_y & f_z \end {vmatrix} + \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ \dfrac \partial {\partial x} & \dfrac \partial {\partial y} & \dfrac \partial {\partial z} \\ g_x & g_y & g_x \end {vmatrix}       | c = Determinant as Sum of Determinants }} {{eqn | r = \nabla \times \mathbf f + \nabla \times \mathbf g       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \nabla \cdot \left({g \, \mathbf f}\right)       | r = \sum_{k \mathop = 1}^n \frac {\partial \left({g f_k}\right)} {\partial x_k}       | c = {{Defof|Divergence Operator}} }} {{eqn | r = \sum_{k \mathop = 1}^n \left({g \frac {\partial f_k} {\partial x_k} + \frac {\partial g} {\partial x_k} f_k}\right)       | c = Product Rule for Derivatives }} {{eqn | r = g \sum_{k \mathop = 1}^n \frac {\partial f_k} {\partial x_k} +  \sum_{k \mathop = 1}^n \frac {\partial g} {\partial x_k} f_k       | c =  }} {{eqn | r = g \left({\nabla \cdot \mathbf f}\right) + \sum_{k \mathop = 1}^n \frac {\partial g} {\partial x_k} f_k       | c = {{Defof|Divergence Operator}} }} {{eqn | r = g \left({\nabla \cdot \mathbf f}\right) + \left({\sum_{k \mathop = 1}^n \frac {\partial g} {\partial x_k} \mathbf e_k}\right) \cdot \left({\sum_{k \mathop = 1}^n f_k \mathbf e_k}\right)       | c = {{Defof|Dot Product}} }} {{eqn | r = g \left({\nabla \cdot \mathbf f}\right) + \left({\nabla g}\right) \cdot \mathbf f       | c = {{Defof|Gradient Operator}}, {{Defof|Vector (Linear Algebra)|Vector}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \nabla \times \paren {g \, \mathbf f}       | r = \paren {\dfrac {\partial g f_z} {\partial y} - \dfrac {\partial g f_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial g f_x} {\partial z} - \dfrac {\partial g f_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial g f_y} {\partial x} - \dfrac {\partial g f_x} {\partial y} } \mathbf k       | c = {{Defof|Curl Operator}} }} {{eqn | r = \paren {g \dfrac {\partial f_z} {\partial y} + \dfrac {\partial g} {\partial y} f_z - g \dfrac {\partial f_y} {\partial z} - \dfrac {\partial g} {\partial z} f_y} \mathbf i       | c = Product Rule for Derivatives }} {{eqn | o =        | ro= +       | r = \paren {g \dfrac {\partial f_x} {\partial z} + \dfrac {\partial g} {\partial z} f_x - g \dfrac {\partial f_z} {\partial x} - \dfrac {\partial g} {\partial x} f_z} \mathbf j       | c =  }} {{eqn | o =        | ro= +       | r = \paren {g \dfrac {\partial f_y} {\partial x} + \dfrac {\partial g} {\partial x} f_y - g \dfrac {\partial f_x} {\partial y} - \dfrac {\partial g} {\partial y} f_x } \mathbf k       | c =  }} {{eqn | r = \map g {\paren {\dfrac {\partial f_z} {\partial y} - \dfrac {\partial f_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial f_x} {\partial z} - \dfrac {\partial f_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial f_y} {\partial x} - \dfrac {\partial f_x} {\partial y} } \mathbf k}       | c = rearrangement }} {{eqn | o =        | ro= +       | r = \paren {\dfrac {\partial g} {\partial y} f_z - \dfrac {\partial g} {\partial z} f_y} \mathbf i + \paren {\dfrac {\partial g} {\partial z} f_x - \dfrac {\partial g} {\partial x} f_z} \mathbf j + \paren {\dfrac {\partial g} {\partial x} f_y - \dfrac {\partial g} {\partial y} f_x} \mathbf k       | c =  }} {{eqn | r = \map g {\nabla \times \mathbf f} + \paren {\dfrac {\partial g} {\partial y} f_z - \dfrac {\partial g} {\partial z} f_y} \mathbf i + \paren {\dfrac {\partial g} {\partial z} f_x - \dfrac {\partial g} {\partial x} f_z} \mathbf j + \paren {\dfrac {\partial g} {\partial x} f_y - \dfrac {\partial g} {\partial y} f_x} \mathbf k       | c = {{Defof|Curl Operator}} }} {{eqn | r = \map g {\nabla \times \mathbf f} + \paren {\dfrac {\partial g} {\partial x} \mathbf i + \dfrac {\partial g} {\partial y} \mathbf j + \dfrac {\partial g} {\partial z} \mathbf k} \times \paren {f_x \mathbf i + f_y \mathbf j + f_z \mathbf k}       | c = {{Defof|Cross Product}} }} {{eqn | r = \map g {\nabla \times \mathbf f} + \paren {\nabla g} \times \mathbf f       | c = {{Defof|Gradient Operator}}, {{Defof|Vector (Linear Algebra)|Vector}} }} {{end-eqn}} {{qed}}
From the definition of the Beta distribution, $X$ has probability density function: :$\displaystyle f_X \left({x}\right) = \frac { x^{\alpha - 1} \left({1 - x}\right)^{\beta - 1} } {\Beta \left({\alpha, \beta}\right)}$ From the definition of a moment generating function: :$\displaystyle M_X \left({t}\right) = \mathbb E \left[{ e^{t X} }\right] = \int_0^1 e^{tx} f_X \left({x}\right) \rd x$ So: {{begin-eqn}} {{eqn	| l = M_X \left({t}\right) 	| r = \frac 1 {\Beta \left({\alpha, \beta}\right)} \int_0^1 e^{t x} x^{\alpha - 1} \left({1 - x}\right)^{\beta - 1} \rd x }} {{eqn	| r = \frac 1 {\Beta \left({\alpha, \beta}\right)} \int_0^1 \left({ \sum_{k \mathop = 0}^\infty \frac {\left({t x}\right)^k} {k!} }\right) x^{\alpha - 1} \left({1 - x}\right)^{\beta - 1} \rd x 	| c = Power Series Expansion for Exponential Function }} {{eqn	| r = \frac 1 {\Beta \left({\alpha, \beta}\right)} \sum_{k \mathop = 0}^\infty \frac{t^k} {k!} \int_0^1 x^{\alpha + k - 1} \left({1 - x}\right)^{\beta - 1} \rd x 	| c = Power Series is Termwise Integrable within Radius of Convergence }} {{eqn	| r = \sum_{k \mathop = 0}^\infty \frac{t^k} {k!} \left({ \frac {\Beta \left({\alpha + k, \beta}\right)} {\Beta \left({\alpha, \beta}\right)} }\right) 	| c = {{Defof|Beta Function/Definition 1|Beta Function}} }} {{eqn	| r = \frac{\Beta \left({\alpha, \beta}\right)} {\Beta \left({\alpha, \beta}\right)} \frac {t^0} {0!} + \sum_{k \mathop = 1}^\infty \frac{t^k} {k!} \left({ \frac {\Beta \left({\alpha + k, \beta}\right)} {\Beta \left({\alpha, \beta}\right)} }\right) }} {{eqn	| r = 1 + \sum_{k \mathop = 1}^\infty \left({ \frac{\Gamma \left({\alpha + k}\right) \Gamma \left({\beta}\right)} {\Gamma \left({\alpha + \beta + k}\right)} \cdot \frac {\Gamma \left({\alpha+\beta}\right)} {\Gamma \left({\alpha}\right) \Gamma \left({\beta}\right)} }\right) \frac{t^k} {k!} 	| c = {{Defof|Beta Function/Definition 3|Beta Function}} }} {{eqn	| r = 1 + \sum_{k \mathop = 1}^\infty \left({ \frac{\Gamma \left({\alpha + k}\right)} {\Gamma \left({\alpha}\right)} \cdot \frac {\Gamma \left({\alpha+\beta}\right)} {\Gamma \left({\alpha + \beta + k}\right)} }\right) \frac{t^k} {k!} }} {{eqn	| r = 1 + \sum_{k \mathop = 1}^\infty \left({\frac {\Gamma \left({\alpha}\right) \prod_{r \mathop = 0}^k \left({\alpha + r}\right)} {\Gamma \left({\alpha}\right)} \cdot \frac {\Gamma \left({\alpha + \beta}\right)} {\Gamma \left({\alpha + \beta}\right) \prod_{r \mathop = 0}^k \left({\alpha + \beta + r}\right)} }\right) \frac{t^k} {k!} 	| c = Gamma Difference Equation }} {{eqn	| r = 1 + \sum_{k \mathop = 1}^\infty \left({ \prod_{r \mathop = 0}^{k - 1} \frac {\alpha + r} {\alpha + \beta + r} }\right) \frac{t^k} {k!} 	| c = Product of Products }} {{end-eqn}} {{qed}} Category:Moment Generating Functions Category:Beta Distribution 8l9c20slqhlw3uq6fv3mvjok1t3ye8l
By Moment Generating Function of Beta Distribution, the moment generating function, $M_X$, of $X$ is given by:  :$\displaystyle \map {M_X} t = \expect {e^{t X} } = 1 + \sum_{n \mathop = 1}^\infty \paren {\prod_{r \mathop = 0}^{n - 1} \frac {\alpha + r} {\alpha + \beta + r} } \frac{t^n} {n!}$ We also have:  {{begin-eqn}} {{eqn	| l = \expect {e^{t X} } 	| r = \expect {\sum_{n \mathop = 0}^\infty \frac{t^n X^n} {n!} } 	| c = Power Series Expansion for Exponential Function }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \expect {\frac {t^n X^n} {n!} } 	| c = Linearity of Expectation Function }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \expect {X^n} \frac{t^n} {n!} 	| c = Linearity of Expectation Function }} {{eqn	| r = \frac {t^0} {0!} \expect {X^0} + \sum_{n \mathop = 1}^\infty \expect {X^n} \frac{t^n} {n!} }} {{eqn	| r = 1 + \sum_{n \mathop = 1}^\infty \expect {X^n} \frac{t^n} {n!} 	| c = Expectation of Constant }} {{end-eqn}} Comparing coefficients yields the result.  {{qed}} Category:Beta Distribution Category:Raw Moments 6l4q5nxxlyv0pa3n0lcuiqnvph0sa3u
Follows directly from Expectation of Function of Joint Probability Mass Distribution, thus: {{begin-eqn}} {{eqn | l = \expect {\alpha X + \beta Y}       | r = \sum_x \sum_y \paren {\alpha x + \beta y} \, \map \Pr {X = x, Y = y}       | c = Expectation of Function of Joint Probability Mass Distribution }} {{eqn | r = \alpha \sum_x x \sum_y \map \Pr {X = x, Y = y}       | c =  }} {{eqn | o = +       | r = \beta \sum_y y \sum_x \map \Pr {X = x, Y = y}       | c =  }} {{eqn | r = \alpha \sum_x x \, \map \Pr {X = x} + \beta \sum_y y \, \map \Pr {Y = y}       | c = {{Defof|Marginal Probability Mass Function}} }} {{eqn | r = \alpha \, \expect X + \beta \, \expect Y       | c = {{Defof|Expectation}} }} {{end-eqn}} {{qed}}
Let $\map {\operatorname {supp} } X$ and $\map {\operatorname {supp} } Y$ be the supports of $X$ and $Y$ respectively.  Let $f_{X, Y} : \map {\operatorname {supp} } X \times \map {\operatorname {supp} } Y \to \R$ be the joint probability density function of $X$ and $Y$.  Let $f_X$ and $f_Y$ be the marginal probability density functions of $X$ and $Y$.  Then: {{begin-eqn}} {{eqn	| l = \expect {\alpha X + \beta Y} 	| r = \int_{y \in \map {\operatorname {supp} } Y} \int_{x \in \map {\operatorname {supp} } X} \paren {\alpha x + \beta y} \map {f_{X, Y} } {x, y} \rd x \rd y }} {{eqn	| r = \alpha \int_{y \in \map {\operatorname {supp} } Y} \int_{x \in \map {\operatorname {supp} } X} x \map {f_{X, Y} } {x, y} \rd x \rd y + \beta \int_{y \in \map {\operatorname {supp} } Y} \int_{x \in \map {\operatorname {supp} } X} y \map {f_{X, Y} } {x, y} \rd x \rd y 	| c = Linear Combination of Definite Integrals }} {{eqn	| r = \alpha \int_{x \in \map {\operatorname {supp} } X} x \paren {\int_{y \in \map {\operatorname {supp} } Y} \map {f_{X , Y} } {x, y} \rd y} \rd x + \beta \int_{y \in \map {\operatorname {supp} } Y} y \paren {\int_{x \in \map {\operatorname {supp} } X} \map {f_{X, Y} } {x, y} \rd x} \rd y 	| c = rewriting }} {{eqn	| r = \alpha \int_{x \in \map {\operatorname {supp} } X} x \map {f_X} x \rd x + \beta \int_{y \in \map {\operatorname {supp} } Y} y \map {f_Y} y \rd y 	| c = {{Defof|Marginal Probability Density Function}} }} {{eqn	| r = \alpha \expect X + \beta \expect Y 	| c = {{Defof|Expectation of Continuous Random Variable}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {c \map f x}       | r = c \map {\dfrac \d {\d x} } {\map f x} + \map f x \map {\dfrac \d {\d x} } c       | c = Product Rule for Derivatives }} {{eqn | r = c \map {\dfrac \d {\d x} } {\map f x} + 0       | c = Derivative of Constant }} {{eqn | r = c \map {\dfrac \d {\d x} } {\map f x}       | c = }} {{end-eqn}} {{qed}}
From the Euler Form of the Gamma Function: {{begin-eqn}} {{eqn | l = \size {\map \Gamma {s + i t} }       | r = \lim_{M \mathop \to \infty} \size {\dfrac 1 {s + i t} \prod_{n \mathop = 1}^M \dfrac {\paren {1 + \frac 1 n}^{s + i t} } {1 + \frac {s + i t} n} } }} {{eqn | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \size {\dfrac {\paren {1 + \frac 1 n}^{s + i t} } {1 + \frac {s + i t} n} } }} {{eqn | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^{s + i t} } } {\size {1 + \frac {s + i t} n} } }} {{eqn | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^s} } {\size {1 + \frac {s + i t} n} }       | c = Modulus of Exponential of Imaginary Number is One }} {{end-eqn}} Because $\size b \le \size t$, we have that: {{begin-eqn}} {{eqn | l = b^2       | o = \le       | r = t^2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\dfrac b n}^2       | o = \le       | r = \paren {\dfrac t n}^2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {1 + \dfrac s n}^2 + \paren {\dfrac b n}^2       | o = \le       | r = \paren {1 + \dfrac s n}^2 + \paren {\dfrac t n}^2       | c =  }} {{eqn | ll= \leadsto       | l = \size {1 + \frac {s + i b} n}        | o = \le       | r = \size {1 + \frac {s + i t} n}       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadsto       | l = \dfrac 1 {\size {1 + \frac {s + i b} n} }       | o = \ge        | r = \dfrac 1 {\size {1 + \frac {s + i t} n} }       | c =  }} {{end-eqn}} Using this we obtain: {{begin-eqn}} {{eqn | l = \size {\map \Gamma {s + i t} }       | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^s} } {\size {1 + \frac {s + i t} n} } }} {{eqn | o = \le       | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^s} } {\size {1 + \frac {s + i b} n} } }} {{eqn | r = \dfrac {\size {s + i b} } {\size {s + i t} } \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i b} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^s} } {\size {1 + \frac{s + i b} n} } }} {{eqn | r = \dfrac {\size {s + i b} } {\size {s + i t} } \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i b} } \size {\prod_{n \mathop = 1}^M \dfrac {\paren {1 + \frac 1 n}^{s + i b} } {1 + \frac{s + i b} n} } }} {{eqn | r = \dfrac {\size {s + i b} } {\size {s + i t} } \size {\map \Gamma {s + i b} } }} {{end-eqn}} {{qed}} Category:Gamma Function pw70f4doua9xqrbrqlvo66jrr07gtth
Let $\map \det {\mathbf A}$ be invertible in $R$. From Matrix Product with Adjugate Matrix: {{begin-eqn}} {{eqn | l = \mathbf A \cdot \adj {\mathbf A}       | r = \map \det {\mathbf A} \cdot \mathbf I_n }} {{eqn | l = \adj {\mathbf A} \cdot \mathbf A       | r = \map \det {\mathbf A} \cdot \mathbf I_n }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \mathbf A \cdot \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} }       | r = \mathbf I_n }} {{eqn | l = \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} } \cdot \mathbf A       | r = \mathbf I_n }} {{end-eqn}} Thus $\mathbf A$ is invertible, and: :$\mathbf A^{-1} = \map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A}$ {{qed}}
Note that as $k > 0$ and $\sigma > 0$, we have $k \sigma > 0$. We therefore have: {{begin-eqn}} {{eqn	| l = \map \Pr {\size {X - \mu} \ge k \sigma} 	| r = \map \Pr {\paren {X - \mu}^2 \ge \paren {k \sigma}^2} }} {{eqn	| o = \le 	| r = \frac {\expect {\paren {X - \mu}^2} } {\paren {k \sigma}^2} 	| c = as $k \sigma > 0$, we can apply Markov's Inequality }} {{eqn	| r = \frac {\sigma^2} {k^2 \sigma^2} 	| c = {{Defof|Variance}} }} {{eqn	| r = \frac 1 {k^2} }} {{end-eqn}} {{qed}}
Let $f$ be the function: :$\map f x = \begin{cases} k^2 \sigma^2 & : \size {x - \mu} \ge k \sigma \\ 0 & : \text{otherwise} \end{cases}$ By construction, we see that: :$\map f x \le \size {x - \mu}^2 = \paren {x - \mu}^2$ for all $x$. This means that: :$\expect {\map f X} \le \expect {\paren {X - \mu}^2}$ By definition of variance: :$\expect {\paren {X - \mu}^2} = \var X = \sigma^2$ By definition of expectation of discrete random variable, we can show that: {{begin-eqn}} {{eqn | l = \expect {\map f X}       | r = k^2 \sigma^2 \map \Pr {\size {X - \mu} \ge k \sigma} + 0 \cdot \map \Pr {\size {X - \mu} \le k \sigma}       | c =  }} {{eqn | r = k^2 \sigma^2 \map \Pr {\size {X - \mu} \ge k \sigma}       | c =  }} {{end-eqn}} Putting this together, we have: {{begin-eqn}} {{eqn | l = \expect {\map f X}       | o = \le       | r = \expect {\paren {X - \mu}^2}       | c =  }} {{eqn | ll= \leadsto       | l = k^2 \sigma^2 \map \Pr {\size {X - \mu} \ge k \sigma}       | o = \le       | r = \sigma^2       | c =  }} {{end-eqn}} By dividing both sides by $k^2 \sigma^2$, we get:  :$\map \Pr {\size {X - \mu} \ge k \sigma} \le \dfrac 1 {k^2}$ {{qed}}
From Sign of Permutation on n Letters is Well-Defined, it is established that the sign each of $\pi_1$, $\pi_2$ and $\pi_1 \circ \pi_2$ is either $+1$ and $-1$. By Existence and Uniqueness of Cycle Decomposition, each of $\pi_1$ and $\pi_2$ has a unique cycle decomposition. Thus each of $\pi_1$ and $\pi_2$ can be expressed as the composite of $p_1$ and $p_2$ transpositions respectively. Thus $\pi_1 \circ \pi_2$ can be expressed as the composite of $p_1 + p_2$ transpositions. From Sum of Even Integers is Even, if $p_1$ and $p_2$ are both even then $p_1 + p_2$ is even. In this case: :$\map \sgn {\pi_1} = 1$ :$\map \sgn {\pi_2} = 1$ :$\map \sgn {\pi_1} \, \map \sgn {\pi_2} = 1 = 1 \times 1$ {{finish}} {{improve}} {{proofread}}
From Symmetric Group on n Letters is Isomorphic to Symmetric Group, it is sufficient to investigate the symmetric group on n letters. Let $\tau$ be a transposition. By definition of sign: :$\sgn \paren \rho = -1$ By definition of odd permutation: : $\tau \in R_o$ Moreover, also by definition of sign, for any $\rho \in R_e$ we have: :$\rho \circ \tau \in R_o$ Define the mapping $\phi$: :$\forall \rho \in R_e: \phi \paren \sigma = \rho \circ \tau$ We claim that $\phi$ is injective. Suppose: :$\phi \paren {\rho_1} = \phi \paren {\rho_2}$ Then: :$\rho_1 \circ \tau = \rho_2 \circ\tau$ Since each Transposition is Self-Inverse, it follows that: :$\rho_1 = \rho_1 \circ \tau \circ \tau = \rho_2 \circ \tau \circ \tau = \rho_2$ Thus $\phi$ is indeed injective. {{qed|lemma}} We claim that $\phi$ is also surjective. Let $\sigma \in R_o$.  We need to show that there exists an element $\rho\in R_e$ such that: :$\phi \paren \rho = \sigma$ By definition of sign: :$\sigma \circ \tau \in R_e$ Since each Transposition is Self-Inverse, it follows that: :$\phi \paren {\sigma \circ \tau} = \sigma \circ \tau \circ \tau = \sigma$ Thus $\phi$ is surjective. {{qed|lemma}} Since $\phi$ is injective and surjective, $\phi$ is the desired bijection. Hence the result by definition of set equivalence. {{qed}}
From Order of Symmetric Group: :$\order {S_n} = n!$ where: :$\order {S_n}$ denotes the order of $S_n$ :$n!$ denotes the factorial of $n$. By definition: :$\card {R_e} + \card {R_o} = order {S_n}$ From Odd and Even Permutations of Set are Equivalent: :$\card {R_e} = \card {R_o}$ The result follows. {{qed}}