{"id":182493,"date":"2025-01-14T06:24:23","date_gmt":"2025-01-14T06:24:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182493"},"modified":"2025-01-14T06:24:25","modified_gmt":"2025-01-14T06:24:25","slug":"moment-generating-function-for-binomial-distribution-let-x-have-the-binomial-distribution-with-probability-distribution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/14\/moment-generating-function-for-binomial-distribution-let-x-have-the-binomial-distribution-with-probability-distribution\/","title":{"rendered":"Moment generating function for binomial distribution Let X have the binomial distribution with probability distribution"},"content":{"rendered":"\n

Moment generating function for binomial distribution Let X have the binomial distribution with probability distribution<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Show that<\/p>\n\n\n\n

(a) M(t) = ( 1 \u2212 p + pet ) n for all t<\/p>\n\n\n\n

(b) E(X) = np and Var(X) = np ( 1 \u2212 p )<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Let ( X ) be a random variable following a binomial distribution, i.e., ( X \\sim \\text{Binomial}(n, p) ), where ( n ) is the number of trials and ( p ) is the probability of success in each trial. The probability mass function (PMF) for ( X ) is given by:<\/p>\n\n\n\n

[
P(X = k) = \\binom{n}{k} p^k (1-p)^{n-k}, \\quad k = 0, 1, 2, \\dots, n
]<\/p>\n\n\n\n

Part (a): Moment Generating Function (MGF)<\/h3>\n\n\n\n

The moment generating function (MGF) of a random variable ( X ) is defined as:<\/p>\n\n\n\n

[
M_X(t) = E(e^{tX}) = \\sum_{k=0}^{n} e^{tk} \\cdot P(X=k)
]<\/p>\n\n\n\n

Substituting the PMF of ( X ), we get:<\/p>\n\n\n\n

[
M_X(t) = \\sum_{k=0}^{n} e^{tk} \\binom{n}{k} p^k (1-p)^{n-k}
]<\/p>\n\n\n\n

Factor out terms that do not depend on ( k ):<\/p>\n\n\n\n

[
M_X(t) = \\left( \\sum_{k=0}^{n} \\binom{n}{k} p^k (1-p)^{n-k} e^{tk} \\right)
]<\/p>\n\n\n\n

Recognize that the expression inside the summation is a binomial expansion of ( (p + (1-p)e^t)^n ):<\/p>\n\n\n\n

[
M_X(t) = (p + (1-p)e^t)^n
]<\/p>\n\n\n\n

Thus, the moment generating function of ( X ) is:<\/p>\n\n\n\n

[
M_X(t) = (1 – p + pe^t)^n
]<\/p>\n\n\n\n

This confirms part (a).<\/p>\n\n\n\n

Part (b): Mean and Variance of ( X )<\/h3>\n\n\n\n

To compute the mean and variance of ( X ), we use the properties of the MGF. First, recall that:<\/p>\n\n\n\n

    \n
  • The mean is given by ( E(X) = M_X'(0) ).<\/li>\n\n\n\n
  • The variance is given by ( \\text{Var}(X) = M_X”(0) – (M_X'(0))^2 ).<\/li>\n<\/ul>\n\n\n\n
      \n
    1. Mean ( E(X) ):<\/strong><\/li>\n<\/ol>\n\n\n\n

      Differentiate ( M_X(t) ) with respect to ( t ):<\/p>\n\n\n\n

      [
      M_X'(t) = n(1 – p + pe^t)^{n-1} \\cdot p e^t
      ]<\/p>\n\n\n\n

      Evaluate at ( t = 0 ):<\/p>\n\n\n\n

      [
      M_X'(0) = n(1 – p + p)^{n-1} \\cdot p = np
      ]<\/p>\n\n\n\n

      Thus, ( E(X) = np ).<\/p>\n\n\n\n

        \n
      1. Variance ( \\text{Var}(X) ):<\/strong><\/li>\n<\/ol>\n\n\n\n

        Differentiate ( M_X'(t) ) to obtain the second derivative:<\/p>\n\n\n\n

        [
        M_X”(t) = n(n-1)(1 – p + pe^t)^{n-2} \\cdot p^2 e^{2t} + n(1 – p + pe^t)^{n-1} \\cdot p e^t
        ]<\/p>\n\n\n\n

        Evaluate at ( t = 0 ):<\/p>\n\n\n\n

        [
        M_X”(0) = n(n-1)(1 – p + p)^{n-2} \\cdot p^2 + n(1 – p + p)^{n-1} \\cdot p
        ]<\/p>\n\n\n\n

        This simplifies to:<\/p>\n\n\n\n

        [
        M_X”(0) = n(n-1) p^2 + np
        ]<\/p>\n\n\n\n

        Now, compute the variance:<\/p>\n\n\n\n

        [
        \\text{Var}(X) = M_X”(0) – (M_X'(0))^2 = n(n-1) p^2 + np – (np)^2 = np(1 – p)
        ]<\/p>\n\n\n\n

        Thus, the variance of ( X ) is ( \\text{Var}(X) = np(1 – p) ).<\/p>\n\n\n\n

        Conclusion<\/h3>\n\n\n\n

        We have shown that for a binomial random variable ( X ) with parameters ( n ) and ( p ):<\/p>\n\n\n\n

          \n
        • The moment generating function is ( M_X(t) = (1 – p + pe^t)^n ).<\/li>\n\n\n\n
        • The mean is ( E(X) = np ), and the variance is ( \\text{Var}(X) = np(1 – p) ).<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

          Moment generating function for binomial distribution Let X have the binomial distribution with probability distribution Show that (a) M(t) = ( 1 \u2212 p + pet ) n for all t (b) E(X) = np and Var(X) = np ( 1 \u2212 p ) The Correct Answer and Explanation is : Let ( X ) […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-182493","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182493","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=182493"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182493\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182493"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182493"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182493"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}