{"id":687,"date":"2016-12-14T11:05:59","date_gmt":"2016-12-14T16:05:59","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/estg1003\/?p=687"},"modified":"2026-04-04T10:54:22","modified_gmt":"2026-04-04T15:54:22","slug":"pdf-univariada-ejercicio","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/stp-u02eva\/pdf-univariada-ejercicio\/","title":{"rendered":"pdf - univariada Ejercicio"},"content":{"rendered":"\n

Referencia<\/strong>: Ross 2.33 p89<\/p>\n\n\n\n

Ejercicio<\/h2>\n\n\n\n

Sea X una variable aleatoria con densidad de probabilidad:<\/p>\n\n\n f(x)= \\begin{cases} c(1-x^2)&, -1<x<1 \\\\ 0 &, \\text{en otro caso} \\end{cases} <\/span>\n\n\n\n

a) \u00bfCu\u00e1l es el valor de c? que permite hacer la funci\u00f3n una pdf.<\/p>\n\n\n\n

b) \u00bfCu\u00e1l es la funci\u00f3n de distribuci\u00f3n acumulada de X?<\/p>\n\n\n\n

Soluci\u00f3n<\/em><\/strong><\/p>\n\n\n\n

a)Solo es v\u00e1lido en el rango [-1,1], por lo que el integral es:<\/p>\n\n\n 1 = \\int_{-1}^{1} c(1-x^2) dx = c \\int_{-1}^{1} (1-x^2) dx <\/span>\n\n\n = c \\int_{-1}^{1}dx - c \\int_{-1}^{1}x^2 dx = c \\left. x \\right|_{-1}^{1} - c \\left. \\frac{x^3}{3} \\right|_{-1}^{1} <\/span>\n\n\n = c[1-(-1)] - c\\frac{1^3-(-1^3)}{3} = 2c -\\frac{2c}{3} <\/span>\n\n\n 1 = \\frac{4c}{3} <\/span>\n\n\n c=\\frac{3}{4} <\/span>\n\n\n\n

Soluci\u00f3n<\/em><\/strong><\/p>\n\n\n\n

b)La funci\u00f3n es la integral hasta x:<\/p>\n\n\n F(y) = \\frac{3}{4} \\int_{-1}^{y} (1-x^2) dx = \\frac{3}{4} \\left. \\left[ x - \\frac{x^3}{3} \\right] \\right|_{-1}^{y} = <\/span>\n\n\n = \\frac{3}{4}\\left[(y+1) - \\left( \\frac{y^3}{3} + \\frac{1}{3} \\right) \\right] = <\/span>\n\n\n F(y) = \\frac{3}{4} \\left[ y+\\frac{2}{3} - \\frac{y^3}{3} \\right], 1<y<1 <\/span>\n\n\n\n

\n\n\n\n

Instrucciones en Python<\/h2>\n\n\n\n

usando el resultado anterior:<\/p>\n\n\n

\nimport matplotlib.pyplot as plt\nimport numpy as np\n\ndef fxdensidad(X):\n    n = len(X)\n    Y = np.zeros(n,dtype=float)\n    \n    c = 3\/4\n    for i in range(0,n,1):\n        x = X[i]\n        if (x>=-1 and x<=1):\n            y = c*(1-x**2)\n            Y[i] = y\n    return(Y)\n\n# INGRESO\n# rango [a,b] y muestras\na = -1\nb = 1\nm = 100\n\n# PROCEDIMIENTO\ndeltax = (b-a)\/m\nx  = np.linspace(a,b,m)\nfx = fxdensidad(x)\n\n# Funci\u00f3n de distribuci\u00f3n acumulada\nFy = np.cumsum(fx)*deltax\n\n# SALIDA Gr\u00e1fico\nplt.plot(x,fx,label='pdf')\nplt.plot(x,Fy,label='cdf')\nplt.xlabel('x')\nplt.legend()\nplt.show()\n<\/pre><\/div>\n\n\n
\"pmf<\/figure>\n\n\n
\n# Verificando resultado del integral vs la suma acumulada\ndef Fxacumulada(X):\n    n = len(X)\n    Y = np.zeros(n,dtype=float)\n    c = 3\/4\n    for i in range(0,n,1):\n        x=X[i]\n        if (x>=-1 and x<=1):\n            y=c*(x+ 2\/3 -(x**3)\/3)\n            Y[i]=y\n    return(Y)\n\n# PROCEDIMIENTO\nFycalc = Fxacumulada(x)\n\n# SALIDA Gr\u00e1fico\nplt.plot(x,fx,label='pdf')\nplt.plot(x,Fy,label='cdf')\nplt.plot(x,Fycalc,label='calculada')\nplt.xlabel('x')\nplt.legend()\nplt.show()\n<\/pre><\/div>\n\n\n
\"pmf<\/figure>\n","protected":false},"excerpt":{"rendered":"
Referencia: Ross 2.33 p89 Ejercicio Sea X una variable aleatoria con densidad de probabilidad: a) \u00bfCu\u00e1l es el valor de c? que permite hacer la funci\u00f3n una pdf. b) \u00bfCu\u00e1l es la funci\u00f3n de distribuci\u00f3n acumulada de X? Soluci\u00f3n a)Solo es v\u00e1lido en el rango [-1,1], por lo que el integral es: Soluci\u00f3n b)La funci\u00f3n […]<\/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-687","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\/687","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=687"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/687\/revisions"}],"predecessor-version":[{"id":23274,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/687\/revisions\/23274"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=687"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=687"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=687"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}