{"id":711,"date":"2016-12-16T22:51:11","date_gmt":"2016-12-17T03:51:11","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/estg1003\/?p=711"},"modified":"2026-04-04T10:54:44","modified_gmt":"2026-04-04T15:54:44","slug":"covarianza-bivariada","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/stp-u02eva\/covarianza-bivariada\/","title":{"rendered":"Covarianza bivariada"},"content":{"rendered":"\n

Referencia<\/em><\/strong>: Ross 2.5.3 p50<\/p>\n\n\n\n

Covarianza<\/h3>\n\n\n\n

La covarianza y varianza de dos variables aleatorias X y Y se definen como:<\/p>\n\n\n Cov(X,Y) = E[(X-E[X])(Y- E[Y])] <\/span>\n\n\n = E[XY] - E[X]E[Y] <\/span>\n\n\n\n

Si X y Y son independientes, entonces Cov(X,Y) = 0<\/p>\n\n\n\n

Propiedades<\/h3>\n\n\n\n
    \n
  1. Cov(X,X) = Var(X)<\/li>\n\n\n\n
  2. Cov(X,Y) = Cov(Y,X)<\/li>\n\n\n\n
  3. Cov(c<\/em><\/strong>X,Y) = c<\/strong><\/em> Cov(X,Y)<\/li>\n\n\n\n
  4. Cov(X,Y+Z) = Cov(X,Y) + Cov(X,Z)<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Covarianza bivariada Ejemplo Ross 2_33<\/h2>\n\n\n\n

    Referencia: <\/strong>Ejemplo Ross 2.33 p51<\/p>\n\n\n\n

    La funci\u00f3n densidad conjunta de X,Y es<\/p>\n\n\n f(x,y) = \\frac{1}{y} e^{-(y+x\/y)} <\/span>\n\n\n donde: 0<x , y<\\infty <\/span>\n\n\n\n

    los intervalos tambi\u00e9n se expresan como: 0<x<\u221e  , 0<y<\u221e<\/p>\n\n\n\n

    a) Verifique que es una funci\u00f3n de densidad conjunta<\/p>\n\n\n\n

    b) Encuentre la Cov(X,Y)<\/p>\n\n\n\n

    \n\n\n\n

    Soluci\u00f3n propuesa<\/h2>\n\n\n\n

    Para mostrar que f(x,y) es una funci\u00f3n de densidad conjunta se debe mostar que es no negativa y que la integral doble es 1.<\/p>\n\n\n \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty}f(x,y) dy dx = <\/span>\n\n\n = \\int_{0}^{\\infty} \\int_{0}^{\\infty} \\frac{1}{y} e^{-(y+x\/y)} dy dx = <\/span>\n\n\n = \\int_{0}^{\\infty} e^{-y} \\int_{0}^{\\infty} \\frac{1}{y} e^{-x\/y} dx dy <\/span>\n\n\n\n

    Resolviendo parcialmente el integral<\/p>\n\n\n \\int_{0}^{\\infty} \\frac{1}{y} e^{-x\/y} dx = \\frac{1}{y} \\int_{0}^{\\infty} e^{-x\/y} dx = <\/span>\n\n\n = \\frac{1}{y} \\left. \\frac{e^{-x\/y}}{-\\frac{1}{y}}\\right|_{0}^{\\infty} = -[e^{-\\infty \/y} - e^{0} ] = -(0-1) = 1 <\/span>\n\n\n\n

    con el resultado se continua con el integral anterior:<\/p>\n\n\n = \\int_{0}^{\\infty} e^{-y} dy = \\left. \\frac{e^{-y}}{-1} \\right|_{0}^{\\infty} <\/span>\n\n\n = - (e^{-\\infty} - e^{0}) = <\/span>\n\n\n = -(0-1) = 1 <\/span>\n\n\n\n

    con lo que se comprueba que es una funci\u00f3n densidad de probabilidad conjunta.<\/p>\n\n\n\n

    \n\n\n\n

    Covarianza (X,Y)<\/h2>\n\n\n\n

    Se obtendr\u00e1n primero las funciones marginales para determinar el valor esperado de cada una de ellas:<\/p>\n\n\n f_Y(y) = \\int_{-\\infty}^{\\infty} f(x,y) dx = \\int_{0}^{\\infty} \\frac{1}{y} e^{-(y+x\/y)} dx = <\/span>\n\n\n = \\int_{0}^{\\infty} \\frac{1}{y} e^{-y} e^{-x\/y} dx = e^{-y} \\int_{0}^{\\infty} \\frac{1}{y}e^{-x\/y} dx = e^{-y} <\/span>\n\n\n\n

    usando parte del integral anterior, se conoce que es igual a 1<\/p>\n\n\n\n

    se requiere el valor esperado E[y]:<\/p>\n\n\n E[Y] = \\int_{-\\infty}^{\\infty} y f(y) dy = \\int_{0}^{\\infty} y e^{-y} dy = <\/span>\n\n\n\n

    integraci\u00f3n por partes dv=e^{-y}dy <\/span>, v = - e^{-y} <\/span>, u=y, du=dy<\/p>\n\n\n \\int u dv = uv - \\int v du <\/span>\n\n\n E[Y] = \\left. - ye^{-y} \\right|_{0}^{\\infty} - \\int_{0}^{\\infty} (-e^{-y}) dy <\/span>\n\n\n = -(\\infty e^{-\\infty} - 0 e^{-0}) - \\left. e^{-y} \\right|_{0}^{\\infty} = <\/span>\n\n\n = 0 -[e^{-\\infty}- e^{0}] = -[0 -1] <\/span>\n\n\n E[Y] =1 <\/span>\n\n\n\n

    El valor esperado E[x] se obtiene como:<\/p>\n\n\n E[x] = \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} x f(x,y) dy dx = <\/span>\n\n\n = \\int_{0}^{\\infty} \\int_{0}^{\\infty} x \\frac{1}{y} e^{-(y+x\/y)} dy dx = <\/span>\n\n\n = \\int_{0}^{\\infty} e^{-y} \\int_{0}^{\\infty} \\frac{x}{y} e^{-x\/y} dx dy = <\/span>\n\n\n\n

    para el integral respecto a x:<\/p>\n\n\n \\int_{0}^{\\infty} \\frac{x}{y} e^{-x\/y} dx = <\/span>\n\n\n\n

    integraci\u00f3n por partes dv=\\frac{1}{y}e^{-x\/y} dx <\/span>, v = \\frac{\\frac{1}{y}}{-(1\/y)} e^{-x\/y}<\/span>, u=x, du=dx<\/p>\n\n\n \\int u dv = uv - \\int v du = <\/span>\n\n\n = \\left. x(-e^{-x\/y})\\right|_{0}^{\\infty} - \\int_{0}^{\\infty} - e^{-x\/y} dx = <\/span>\n\n\n = [- \\infty ye^{-\\infty\/y} - (-0e^{0\/y})] + \\int_{0}^{\\infty} e^{-x\/y} dx = <\/span>\n\n\n = [-0+0] + \\left. \\frac{e^{-x\/y}}{-(1\/y)}\\right|_{0}^{\\infty} = <\/span>\n\n\n =-y[e^{-\\infty\/y} - e^{-0\/y}] = y <\/span>\n\n\n\n

    reemplazando en E[x] y usando el resultado del integral de E[y]:<\/p>\n\n\n E[x] = \\int_{0}^{\\infty} y e^{-y} dy = 1 <\/span>\n\n\n\n

    Para calcular el valor de E[XY]:<\/p>\n\n\n E[XY] = <\/span>\n\n\n =\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} xy f(x,y) dy dx =<\/span>\n\n\n =\\int_{0}^{\\infty} \\int_{0}^{\\infty} xy \\frac{1}{y} e^{-(y+x\/y)} dy dx = <\/span>\n\n\n = \\int_{0}^{\\infty} y e^{-y} \\int_{0}^{\\infty} \\frac{x}{y} e^{-x\/y} dx dy = <\/span>\n\n\n = \\int_{0}^{\\infty} y e^{-y} (y) dy = \\int_{0}^{\\infty} y^2 e^{-y} dy = <\/span>\n\n\n\n

    nuevamente por partes: dv = e^{-y} dy, u=y^2 <\/span> se obtiene<\/p>\n\n\n E[XY] = \\int_{0}^{\\infty} y^2e^{-y} dy = <\/span>\n\n\n =\\left. -y^2 e^{-y} \\right|_{0}^{\\infty} + \\int_{0}^{\\infty} 2ye^{-y}dy = 2E[Y] =2 <\/span>\n\n\n\n

    En consecuencia:<\/p>\n\n\n Cov(X,Y) = E[XY] - E[X]E[Y] = 2-(1)(1) = 1 <\/span>\n","protected":false},"excerpt":{"rendered":"

    Referencia: Ross 2.5.3 p50 Covarianza La covarianza y varianza de dos variables aleatorias X y Y se definen como: Si X y Y son independientes, entonces Cov(X,Y) = 0 Propiedades Covarianza bivariada Ejemplo Ross 2_33 Referencia: Ejemplo Ross 2.33 p51 La funci\u00f3n densidad conjunta de X,Y es los intervalos tambi\u00e9n se expresan como: 0<x<\u221e  , […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-stp-unidades","format":"standard","meta":{"footnotes":""},"categories":[214],"tags":[],"class_list":["post-711","post","type-post","status-publish","format-standard","hentry","category-stp-u02eva"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/711","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=711"}],"version-history":[{"count":1,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/711\/revisions"}],"predecessor-version":[{"id":22152,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/711\/revisions\/22152"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=711"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=711"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=711"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}