September 29, 2010
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 | \documentclass [12pt]{article} \usepackage [papersize={6in, 9.1in}, text={5.5in, 8.8in}]{geometry} \usepackage {amsmath} \usepackage {mathspec} \usepackage {fontspec} \defaultfontfeatures {Scale=MatchLowercase} \setmainfont [Mapping=tex-text]{Minion Pro} %\setmainfont[Mapping=tex-text]{Hoefler Text} %\setmainfont[Mapping=tex-text]{Garamond} \setsansfont [Mapping=tex-text]{Candara} %\setsansfont[Mapping=tex-text]{Myriad Pro} %\setsansfont[Mapping=tex-text]{Comic Sans MS} %\setmonofont{Courier} \setmonofont {Monaco} \usepackage {paralist} \newcommand { \fpd }[2]{ \ensuremath { \frac { \partial {#1}}{ \partial {#2}}}} \begin {document} \thispagestyle {empty} The Cauchy distribution is a symmetric distribution on $(- \infty , \infty )$ with pdf \begin {equation*} f_X(x; \theta , \gamma ) = \frac1 \pi \cdot \frac { \gamma }{(x- \theta )^2+ \gamma ^2} \end {equation*} In this paper, we only deal with the case $ \theta =0$. Consider two independent Gaussian random variables $X,Y \sim N(0,1)$. We will prove that the ratio $X/Y$ is a Cauchy distribution by \begin {inparaenum}[(1)] \item defining the transformation $U=X/Y$ and $V=|Y|$, \item finding the joint pdf $F_{U,V}(u,v)$, and \item integrating out $V$ to obtain the marginal pdf of $U$. \end {inparaenum} Unfortunately, the mapping $U=X/Y$ and $V=|Y|$ is not one-to-one: the two points $(x,y)$ and $(-x,-y)$ map to the same $(u,v)$ We need to partition $(X,Y)$ into $A_0,A_1,A_2$ such that the mapping from $A_i$ to $(U,V)$ is one-to-one. \begin {enumerate} \item $A_0= \ {(X,Y):Y=0 \ }$: This exceptional case does not happen because $ \Pr [Y=0]=0$ when $Y \sim N(0,1)$. \item $A_1= \ {(X,Y):Y& gt ;0 \ }$: The mapping $U=X/Y$, $V=|Y|$ is one-to-one, and the inverse mappings are $h_{11}(u,v)=uv$, $h_{21}=v$. \item $A_2= \ {(X,Y):Y& lt ;0 \ }$: The mapping $U=X/Y$, $V=|Y|$ is one-to-one, and the inverse mappings are $h_{12}(u,v)=-uv$, $h_{22}=-v$. \end {enumerate} \begin {equation*} \begin {array}{lllll} J_1 & = \begin {vmatrix} \fpd {h_{11}}{u} & \fpd {h_{11}}{v} \ \ \fpd {h_{21}}{u} & \fpd {h_{21}}{v} \end {vmatrix} & = \begin {vmatrix} \fpd {uv}{u} & \fpd {uv}{v} \ \ \fpd {v}{u} & \fpd {v}{v} \end {vmatrix} &= \begin {vmatrix} v & u \ \ 0 & 1 \end {vmatrix} &= v \ \ J_2 &= \begin {vmatrix} \fpd {h_{12}}{u} & \fpd {h_{12}}{v} \ \ \fpd {h_{22}}{u} & \fpd {h_{22}}{v} \end {vmatrix} &= \begin {vmatrix} \fpd {(-uv)}{u} & \fpd {(-uv)}{v} \ \ \fpd {(-v)}{u} & \fpd {(-v)}{v} \end {vmatrix} &= \begin {vmatrix} -v & -u \ \ 0 & -1 \end {vmatrix} &= v \end {array} \end {equation*} %%% \begin {equation*} f_{X,Y}(x,y) = \frac {1}{ \sqrt {2 \pi }} \exp (-x^2/2) \frac {1}{ \sqrt {2 \pi }} \exp (-y^2/2) = \frac {1}{2 \pi } \exp \left (- \frac {x^2+y^2}{2} \right ) \end {equation*} %%% \begin {align*} f_{UV}(u,v) &= f_{XY}(h_{11}(u,v),h_{21}(u,v))|J_1| + f_{XY}(h_{12}(u,v),h_{22}(u,v))|J_2| \ \ &= \frac {1}{2 \pi } \exp \left (- \frac {(uv)^2 + v^2}{2} \right )|v| + \frac {1}{2 \pi } \exp \left (- \frac {(-uv)^2 + (-v)^2}{2} \right )|v| \ \ &= \frac {v}{ \pi } \exp \left (- \frac {v^2(u^2+1)}{2} \right ), \qquad - \infty & lt ; u & lt ; \infty , \quad0 & lt ;v& lt ; \infty \ \ \end {align*} %%% \begin {alignat*}{2} f_U(u) &= \int_0 ^ \infty \frac {v}{ \pi } \exp \left (- \frac {v^2(u^2+1)}{2} \right ) dv & \text {integrating out $V$} \ \ &= \int_0 ^ \infty \frac {1}{2 \pi } \exp \left (- \frac {u^2+1}{2}z \right ) dz & \qquad \text {Let $z=v^2$ and $dz=2vdv$} \ \ &= \frac {1}{2 \pi } \cdot \frac {2}{u^2+1} & \int_0 ^ \infty \exp (- \alpha z) dz = \frac1 \alpha \ \ &= \frac {1}{ \pi } \cdot \frac {1}{u^2+1}, \qquad - \infty & lt ; u& lt ; \infty \end {alignat*} \end {document} |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 | \documentclass [10pt]{article} \pagestyle {empty} \usepackage {amsmath,amssymb,amsthm} \usepackage [papersize={140mm, 210mm}, text={120mm, 200mm}]{geometry} % For fancy fonts \usepackage [T1]{fontenc} \usepackage {ccfonts,eulervm} \usepackage {tikz} \newcommand { \fpd }[2]{ \ensuremath { \frac { \partial {#1}}{ \partial {#2}}}} \thispagestyle {empty} \begin {document} The ratio of two Gaussian random variables $X,Y \sim N(0,1)$ is a Cauchy distribution. Let $U=X/Y$ and $V=|Y|$. If $Y=0$, $U$ and $V$ can be any value because $ \Pr (Y=0)=0$. If we restrict consideration to either positive or negative value of $Y$, then the transform from $ \mathcal {A}=(X,Y)$ to $ \mathcal {B}=(U,V)$ is one-to-one. \begin {equation*} A_1 = \ {(x,y): y & gt ; 0 \ }, \qquad A_2 = \ {(x,y): y & lt ; 0 \ }, \qquad A_3 = \ {(x,y): y = 0 \ } \end {equation*} These three partition $ \mathcal {A}= \mathbb {R}^2$ and $ \Pr [(X,Y) \in A_0]= \Pr [Y=0]=0$. \begin {figure}[h] \centering \begin {tikzpicture} [line/.style ={draw, thick, -latex, shorten & gt ;=2pt},] \def \myradius {1.5cm} % left circle \shadedraw [left color=blue!10, right color=blue!60, draw=blue!50!black] (0, 0) -- ( \myradius , 0mm) arc (0:150: \myradius ) -- cycle; \shadedraw [left color=red!10, right color=red!60, draw=red!50!black] (0, 0) -- ( \myradius , 0mm) arc (0:-150: \myradius ) -- cycle; \draw (0,0) circle ( \myradius ); \node at (5mm, 8mm) {$A_1$}; \node at (5mm, -8mm) {$A_2$}; \node at (-9mm, 0mm) {$A_0$}; \node at (0cm, \myradius +5mm) {$ \mathcal {A}= \mathbb {R}^2=(X,Y)$}; % right circle \shade [left color=black!10, right color=black!40, draw=black!50!black] (6cm, 0) circle ( \myradius ); \node at ( \myradius *4, \myradius +5mm) {$ \mathcal {B}= \ {(U,V):V \geq 0 \ }$}; % arrows \path ( \myradius *4-3mm, 8mm) edge [line, bend angle=10, bend right] node [above, midway] {$h_{11},h_{21}$} (8mm, 8mm); \path (8mm, 6mm) edge [line, bend angle=10, bend right] ( \myradius *4-3mm, 6mm); \path ( \myradius *4-3mm, -8mm) edge [line, bend angle=10, bend left] node [below, midway] {$h_{12},h_{22}$} (8mm, -8mm); \path (8mm, -6mm) edge [line, bend angle=10, bend left] ( \myradius *4-3mm, -6mm); \end {tikzpicture} \end {figure} \begin {alignat*}{3} \mathcal {B} \to \mathcal {A}_1:& x=h_{11}(u,v)=uv, & \quad &y=h_{21}(u,v)=v \ \ \mathcal {B} \to \mathcal {A}_2:& x=h_{12}(u,v)=-uv, & \quad &y=h_{22}(u,v)=-v \end {alignat*} \begin {equation*} J_1 = \left | \begin {array}{cc} \fpd {uv}{u} & \fpd {uv}{v} \ \ \fpd {v}{u} & \fpd {v}{v} \end {array} \right | = \begin {vmatrix} v & u \ \ 0 & 1 \ \ \end {vmatrix} = v, \quad J_2 = \left | \begin {array}{cc} \fpd {(-uv)}{u} & \fpd {(-uv)}{v} \ \ \fpd {(-v)}{u} & \fpd {(-v)}{v} \end {array} \right | = \begin {vmatrix} -v & -u \ \ 0 & -1 \ \ \end {vmatrix} = v \end {equation*} \begin {equation*} f_{XY}(x,y) = \frac {1}{ \sqrt {2 \pi }} \exp \left (- \frac {x^2}{2} \right ) \cdot \frac {1}{ \sqrt {2 \pi }} \exp \left (- \frac {y^2}{2} \right ) = \frac {1}{2 \pi } \exp \left (- \frac {x^2+y^2}{2} \right ) \end {equation*} \begin {align*} f_{UV}(u,v) &= \sum_ {i=0}^2 f_{XY}(h_{1i}(u,v),h_{2i}(u,v))|J_i| \ \ &= \frac {1}{2 \pi } \exp \left (- \frac {(uv)^2 + v^2}{2} \right )|v| + \frac {1}{2 \pi } \exp \left (- \frac {(-uv)^2 + (-v)^2}{2} \right )|v| \ \ &= \frac {v}{ \pi } \exp \left (- \frac {v^2(u^2+1)}{2} \right ) \end {align*} \begin {alignat*}{2} f_U(u) &= \int_0 ^ \infty \frac {v}{ \pi } \exp \left (- \frac {v^2(u^2+1)}{2} \right ) dv & \qquad ( \text {change of variable: $z=v^2$}) \ \ &= \int_0 ^ \infty \frac {1}{2 \pi } \exp \left (- \frac {u^2+1}{2}z \right ) dz & \left ( \int_0 ^ \infty \exp (- \alpha z) dz = 1/ \alpha \right ) \ \ &= \frac {1}{ \pi } \cdot \frac {1}{u^2+1} \end {alignat*} \end {document} |
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