{"id":697,"date":"2016-12-15T17:36:04","date_gmt":"2016-12-15T22:36:04","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/estg1003\/?p=697"},"modified":"2026-04-04T10:54:36","modified_gmt":"2026-04-04T15:54:36","slug":"pdf-bivariada-ejercicio","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/stp-u02eva\/pdf-bivariada-ejercicio\/","title":{"rendered":"pdf - Bivariada Ejercicios"},"content":{"rendered":"\n

1. pdf Bivariada Ejercicios<\/h2>\n\n\n\n

Referencia<\/strong>: Le\u00f3n Garc\u00eda Ejercicio 5.16 p.252<\/p>\n\n\n\n

Ejemplo<\/p>\n\n\n\n

Encuentre la constante c de normalizaci\u00f3n y las pdf marginales de la siguiente funci\u00f3n:<\/p>\n\n\n f_{X,Y}(x,y)= \\begin{cases} c e^{-x} e^{-y} &, 0\\leq y \\leq x < \\infty \\\\ 0 & ,\\text{otro caso} \\end{cases} <\/span>\n\n\n\n

Soluci\u00f3n<\/h3>\n\n\n\n

la funci\u00f3n es v\u00e1lida en la regi\u00f3n mostrada:

La constante c se encuentra cumpliendo la condici\u00f3n de normalizaci\u00f3n:<\/p>\n\n\n 1 = \\int_{0}^{\\infty} \\int_{0}^{x} c e^{-x} e^{-y} dy dx = <\/span>\n\n\n = \\int_{0}^{\\infty} c e^{-x} (1-e^{-x}) dx = \\frac{c}{2}<\/span>\n\n\n 1 = \\frac{c}{2} <\/span>\n\n\n c = 2 <\/span>\n\n\n\n

Determinar las marginales, conociendo que c=2:<\/p>\n\n\n f_X (x) = \\int_{0}^{\\infty} f_{X,Y}(x,y) dy = <\/span>\n\n\n = \\int_{0}^{x} 2 e^{-x} e^{-y} dy = 2 e^{-x} \\int_{0}^{x} e^{-y} dy = <\/span>\n\n\n = 2 e^{-x} \\left. \\left[ - e^{-y} \\right]\\right|_{0}^{x} = 2 e^{-x} \\left[ -e^{-x}-(-e^{0}) \\right] = <\/span>\n\n\n f_X (x) = 2 e^{-x} (1- e^{-x}) <\/span>\n\n\n 0\\leq x < \\infty <\/span>\n\n\n\n

\n\n\n f_Y (y) = \\int_{0}^{\\infty} f_{X,Y}(x,y) dx = <\/span>\n\n\n = \\int_{y}^{\\infty} 2 e^{-x} e^{-y} dx = 2 e^{-y} \\int_{y}^{\\infty} e^{-x} dx = <\/span>\n\n\n = 2 e^{-y} \\left. \\left[ - e^{-x} \\right]\\right|_{y}^{\\infty} = 2 e^{-y} \\left[-e^{-\\infty}-(-e^{-y})\\right] = <\/span>\n\n\n = 2 e^{-y} 2 e^{-y} <\/span>\n\n\n f_Y (y) = 2 e^{-2y} <\/span>\n\n\n 0\\leq y < \\infty <\/span>\n\n\n\n
\n\n\n\n

Tarea<\/strong>: Verificar que las funciones marginales cumplen que el integral es 1<\/p>\n\n\n

\nimport numpy as np\nimport matplotlib.pyplot as plt\nfrom mpl_toolkits.mplot3d import Axes3D\n\n# Funci\u00f3n evaluada\ndef fxydensidad(X,Y):\n    n,m =np.shape(X)\n    Z=np.zeros(shape=(n,m),dtype=float)\n    \n    c=2\n    for i in range(0,n,1):\n        for j in range(0,m,1):\n            x=X[i,j]\n            y=Y[i,j]\n            \n            if (y>=0 and y<=x):\n                z=c*np.exp(-x)*np.exp(y)\n                Z[i,j]=z\n    return(Z)\n\n# PROGRAMA ---------\n\n# INGRESO\n# Rango de evaluaci\u00f3n\nxa = 0\nxb = 4\nya = 0\nyb = 4\n# muestras por eje\nnx = 500\nny = 500\n\n# PROCEDIMIENTO\n\n# Matriz de evaluaci\u00f3n\ny = np.linspace(ya,yb,ny)\nx = np.linspace(xa,xb,nx)\nX,Y = np.meshgrid(x,y)\n\n# Eval\u00faa la funci\u00f3n\nZ = fxydensidad(X,Y)\n\n# SALIDA\nfigura  = plt.figure(1)\ngrafica = figura.add_subplot(1, 1, 1, projection='3d')\ngrafica.plot_wireframe(X, Y, Z, rstride=10, cstride=10)\nplt.show()\n<\/pre><\/div>\n\n\n
\"\"<\/a><\/figure>\n\n\n
\n# Zona de integraci\u00f3n\ndef arealimite(X):\n    n = len(X)\n    yinferior = np.zeros(n,dtype=int)\n    ysuperior = X\n    return(yinferior, ysuperior)\n\n# PROCEDIMIENTO\nyinferior, ysuperior = arealimite(x)\n\n# SALIDA\nplt.plot(x, yinferior)\nplt.plot(x, ysuperior)\nplt.fill_between(x, yinferior, ysuperior, where=(ysuperior>=yinferior))\nplt.show()\n<\/pre><\/div>\n\n\n
\"bivariada<\/figure>\n\n\n
\n# Forma de las marginales\ndef marginalx(x):\n    fx = 2*np.exp(-x)*(1-np.exp(-x))\n    return(fx)\n\n# PROCEDIMIENTO\ndeltax = (xb-xa)\/nx\nfx = marginalx(x)\nFx = np.cumsum(fx*deltax)\nintegrax = np.sum(fx*deltax)\n\n# SALIDA\nplt.plot(x,fx, label='f(x)')\nplt.plot(x,Fx, label='F(x)')\nplt.xlabel('x')\nplt.legend()\nplt.show()\n<\/pre><\/div>\n\n\n
\"\"<\/a><\/figure>\n\n\n
\n# Forma de las marginales\ndef marginaly(y):\n    fy = 2*np.exp(-2*y)\n    return(fy)\n\n# PROCEDIMIENTO\ndeltay = (yb-ya)\/ny\nfy = marginaly(y)\nFy = np.cumsum(fy*deltay)\nintegray = np.sum(fy*deltay)\n\n# SALIDA\nprint(' El integral sobre el area es: ', integray)\n\nplt.plot(y,fy, label='f(y)')\nplt.plot(y,Fy, label='F(y)')\nplt.xlabel('y')\nplt.legend()\nplt.show()\n<\/pre><\/div>\n\n\n
\"bivariada<\/figure>\n\n\n\n
\n\n\n\n
2. pdf Bivariada Ejercicio02<\/h2>\n\n\n\n
Referencia<\/strong><\/em>: Le\u00f3n-Garc\u00eda 5.17 p 253<\/p>\n\n\n\n
Ejemplo<\/h2>\n\n\n\n
Encuentre P[ X + Y \u2264 1] de la funci\u00f3n en el ejemplo 5.16 mostrada a continuaci\u00f3n :<\/p>\n\n\n f_{X,Y}(x,y)= \\begin{cases} c e^{-x} e^{-y} & , 0\\leq y \\leq x < \\infty \\\\ 0 & ,\\text{otro caso} \\end{cases} <\/span>\n\n\n\n
Soluci\u00f3n<\/h3>\n\n\n\n
La la regi\u00f3n para integraci\u00f3n es [ X + Y \u2264 1] donde la pdf no es cero. Se obtiene la probabilidad del evento al a\u00f1adir(integrar) rect\u00e1ngulos infinitesimales de ancho dy como se indica en la figura:<\/p>\n\n\n\n
\"bivariada<\/figure>\n\n\n X + Y \\leq 1 <\/span>\n\n\n Y \\leq 1 - X <\/span>\n\n\n P[ X + Y \\leq 1] = \\int_{0}^{1\/2} \\int_{y}^{1-y} 2 e^{-x} e^{-y} dx dy <\/span>\n\n\n = \\int_{0}^{1\/2} 2 e^{-y} \\int_{y}^{1-y} e^{-x} dx dy = \\int_{0}^{1\/2} 2 e^{-y} \\left. [-e^{-x}] \\right|_{y}^{1-y} dy = <\/span>\n\n\n = \\int_{0}^{1\/2} 2 e^{-y} [-e^{-(1-y)}-(-e^{-y})] dy = \\int_{0}^{1\/2} [2 e^{-2y}- 2 e^{-y-(1-y)}] dy = <\/span>\n\n\n = \\int_{0}^{1\/2} [2 e^{-2y}- 2 e^{-1}] dy = \\left. \\left[ 2\\frac{e^{-2y}}{-2} - 2 e^{-1}y\\right] \\right|t_{0}^{1\/2} = <\/span>\n\n\n = [ - e^{-2 (1\/2)} - 2 e^{-1} (1\/2) ] - [ -e^{0}-0] = -e^{-1}-e^{-1} +1 <\/span>\n\n\n P[ X + Y \\leq 1] = 1- 2e^{-1} = 0.26424111765711533 <\/span>\n\n\n\n
que limita la figura que genera la funci\u00f3n a:<\/p>\n\n\n\n
... tarea ...<\/p>\n\n\n\n
\n\n\n\n
Instrucciones en Python<\/h2>\n\n\n
\nfrom mpl_toolkits.mplot3d import Axes3D\nimport matplotlib.pyplot as plt\nimport numpy as np\n\n# Funci\u00f3n evaluada\ndef fxydensidad(X,Y):\n    n,m = np.shape(X)\n    Z = np.zeros(shape=(n,m),dtype=float)\n    \n    c = 2\n    for i in range(0,n,1):\n        for j in range(0,m,1):\n            x = X[i,j]\n            y = Y[i,j]\n            if (y>=0 and y<=x and (x + y)<=1):\n                z = c*np.exp(-x)*np.exp(y)\n                Z[i,j] = z\n    return(Z)\n\n# PROGRAMA\n# INGRESO\n# Rango de evaluaci\u00f3n\nxa = 0\nxb = 1.5\nya = 0\nyb = 1.5\n#muestras por eje\nnx = 200\nny = 200\n\n# PROCEDIMIENTO\n# Matriz de evaluaci\u00f3n\ny = np.linspace(ya,yb,ny)\nx = np.linspace(xa,xb,nx)\nX,Y = np.meshgrid(x,y)\n# Evalua la funci\u00f3n\nZ = fxydensidad(X,Y)\n\n# Zona de integraci\u00f3n\ndef arealimite(X):\n    n = len(X)\n    yinferior = np.zeros(n,dtype=float)\n    ysuperior = np.zeros(n,dtype=float)\n    for i in range(0,n,1):\n        x = X[i]\n        if (x<0.5):\n            y = x\n        if (x>=0.5):\n            y = 1 - x\n        ysuperior[i] = y\n    \n    return(yinferior, ysuperior)\n\n# PROCEDIMIENTO\nyinferior , ysuperior = arealimite(x) \n\n# SALIDA GRAFICAS\nfigura1 = plt.figure(1)\nplt.plot(x,yinferior)\nplt.plot(x,ysuperior)\nplt.fill_between(x, yinferior, ysuperior,\n                 where= (ysuperior>=yinferior))\nplt.xlabel('x')\n<\/pre><\/div>\n\n\n
\"bivariada<\/figure>\n\n\n
\n# SALIDA\nfigura2 = plt.figure(2)\ngrafica2 = figura2.add_subplot(1, 1, 1, projection='3d')\ngrafica2.plot_wireframe(X, Y, Z, rstride=10, cstride=10)\nplt.show()\n<\/pre><\/div>\n\n\n
...tarea...<\/p>\n","protected":false},"excerpt":{"rendered":"
1. pdf Bivariada Ejercicios Referencia: Le\u00f3n Garc\u00eda Ejercicio 5.16 p.252 Ejemplo Encuentre la constante c de normalizaci\u00f3n y las pdf marginales de la siguiente funci\u00f3n: Soluci\u00f3n la funci\u00f3n es v\u00e1lida en la regi\u00f3n mostrada:La constante c se encuentra cumpliendo la condici\u00f3n de normalizaci\u00f3n: Determinar las marginales, conociendo que c=2: Tarea: Verificar que las funciones marginales […]<\/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-697","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\/697","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=697"}],"version-history":[{"count":7,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/697\/revisions"}],"predecessor-version":[{"id":23273,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/697\/revisions\/23273"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=697"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=697"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=697"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}