{"id":223630,"date":"2025-06-02T06:41:33","date_gmt":"2025-06-02T06:41:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223630"},"modified":"2025-06-02T06:41:35","modified_gmt":"2025-06-02T06:41:35","slug":"calculate-the-covariance-of-the-following-random-variables","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/calculate-the-covariance-of-the-following-random-variables\/","title":{"rendered":"Calculate the covariance of the following random variables"},"content":{"rendered":"\n

Q1: Calculate the covariance of the following random variables: Part A: X = 1A and Y = 1B for two indicator variables over sets A and B. For the special case of indicator functions, what is the relationship between correlated\/uncorrelated and dependent\/independent? Part B: X and Y, for a point (X,Y) uniformly distributed on a triangle with endpoints at (0, 0), (1, 0), and (0, 1).<\/p>\n\n\n\n

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

Q1: Covariance of Random Variables<\/strong><\/h3>\n\n\n\n
\n\n\n\n

Part A: Indicator Variables<\/strong><\/h3>\n\n\n\n

Let X=1AX = \\mathbf{1}_A, Y=1BY = \\mathbf{1}_B be indicator random variables.
The covariance<\/strong> between XX and YY is: Cov(X,Y)=E[XY]\u2212E[X]E[Y]\\text{Cov}(X, Y) = \\mathbb{E}[XY] – \\mathbb{E}[X]\\mathbb{E}[Y]<\/p>\n\n\n\n

Since X=1AX = \\mathbf{1}_A, Y=1BY = \\mathbf{1}_B, and XY=1A\u2229BXY = \\mathbf{1}_{A \\cap B}, we get: E[X]=P(A),E[Y]=P(B),E[XY]=P(A\u2229B)\\mathbb{E}[X] = \\mathbb{P}(A), \\quad \\mathbb{E}[Y] = \\mathbb{P}(B), \\quad \\mathbb{E}[XY] = \\mathbb{P}(A \\cap B) \u21d2Cov(X,Y)=P(A\u2229B)\u2212P(A)P(B)\\Rightarrow \\text{Cov}(X, Y) = \\mathbb{P}(A \\cap B) – \\mathbb{P}(A)\\mathbb{P}(B)<\/p>\n\n\n\n

\n\n\n\n

Special Case: Dependence vs. Correlation for Indicator Variables<\/strong><\/h4>\n\n\n\n
    \n
  • If XX and YY are independent<\/strong>, then P(A\u2229B)=P(A)P(B)\u21d2Cov(X,Y)=0\\mathbb{P}(A \\cap B) = \\mathbb{P}(A)\\mathbb{P}(B) \\Rightarrow \\text{Cov}(X,Y) = 0: Uncorrelated<\/em>.<\/li>\n\n\n\n
  • If Cov(X,Y)=0\\text{Cov}(X, Y) = 0<\/strong>, this does not<\/strong> imply independence in general. However, for indicator variables, if AA and BB are independent events<\/strong>, then the corresponding indicators are uncorrelated.<\/li>\n<\/ul>\n\n\n\n

    So, for indicator functions:<\/p>\n\n\n\n

      \n
    • Independence \u21d2 Uncorrelated<\/strong><\/li>\n\n\n\n
    • Uncorrelated \u21cf Independence<\/strong> (unless under special distributional assumptions)<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Part B: Uniform Distribution on a Triangle<\/strong><\/h3>\n\n\n\n

      Let (X,Y)(X, Y) be uniformly distributed on the triangle with vertices at (0,0),(1,0),(0,1)(0,0), (1,0), (0,1).<\/p>\n\n\n\n

      The support is the region: T={(x,y)\u2223x\u22650,y\u22650,x+y\u22641}T = \\{(x, y) \\mid x \\ge 0, y \\ge 0, x + y \\le 1\\}<\/p>\n\n\n\n

      Step 1: Joint PDF<\/strong><\/h4>\n\n\n\n

      Area of triangle = 12\\frac{1}{2}. So uniform PDF: f(x,y)=2for (x,y)\u2208Tf(x, y) = 2 \\quad \\text{for } (x, y) \\in T<\/p>\n\n\n\n

      Step 2: Compute Covariance<\/strong><\/h4>\n\n\n\n

      We use: Cov(X,Y)=E[XY]\u2212E[X]E[Y]\\text{Cov}(X,Y) = \\mathbb{E}[XY] – \\mathbb{E}[X] \\mathbb{E}[Y]<\/p>\n\n\n\n

      Compute expectations:<\/strong><\/p>\n\n\n\n

        \n
      1. E[X]=\u222b01\u222b01\u2212xx\u22c52\u2009dy\u2009dx=13\\mathbb{E}[X] = \\int_0^1 \\int_0^{1-x} x \\cdot 2 \\, dy\\, dx = \\frac{1}{3}<\/li>\n\n\n\n
      2. E[Y]=\u222b01\u222b01\u2212xy\u22c52\u2009dy\u2009dx=13\\mathbb{E}[Y] = \\int_0^1 \\int_0^{1-x} y \\cdot 2 \\, dy\\, dx = \\frac{1}{3}<\/li>\n\n\n\n
      3. E[XY]=\u222b01\u222b01\u2212xxy\u22c52\u2009dy\u2009dx=124\\mathbb{E}[XY] = \\int_0^1 \\int_0^{1-x} xy \\cdot 2 \\, dy\\, dx = \\frac{1}{24}<\/li>\n<\/ol>\n\n\n\n

        Cov(X,Y)=124\u221213\u22c513=124\u221219=\u2212572\\text{Cov}(X, Y) = \\frac{1}{24} – \\frac{1}{3} \\cdot \\frac{1}{3} = \\frac{1}{24} – \\frac{1}{9} = -\\frac{5}{72}<\/p>\n\n\n\n

        \n\n\n\n

        Final Answers<\/strong><\/h3>\n\n\n\n
          \n
        • Part A:<\/strong>
          Cov(X,Y)=P(A\u2229B)\u2212P(A)P(B)\\text{Cov}(X,Y) = \\mathbb{P}(A \\cap B) – \\mathbb{P}(A)\\mathbb{P}(B)
          Independence \u21d2 Uncorrelated, but not vice versa.<\/li>\n\n\n\n
        • Part B:<\/strong>
          Cov(X,Y)=\u2212572\\text{Cov}(X, Y) = -\\frac{5}{72}<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Explanation<\/strong><\/h3>\n\n\n\n

          In Part A, we examine the covariance of two indicator variables<\/strong>. These are binary random variables that take the value 1 if a condition or event occurs and 0 otherwise. The covariance formula simplifies nicely due to the binary nature of indicator variables: since the product XY=1XY = 1 only when both indicators are 1 (i.e., both events happen), the expected value E[XY]\\mathbb{E}[XY] becomes the probability of both events occurring simultaneously: P(A\u2229B)\\mathbb{P}(A \\cap B). The covariance is then the difference between this joint probability and the product of their individual probabilities: Cov(X,Y)=P(A\u2229B)\u2212P(A)P(B)\\text{Cov}(X, Y) = \\mathbb{P}(A \\cap B) – \\mathbb{P}(A)\\mathbb{P}(B)<\/p>\n\n\n\n

          This formula highlights the connection between covariance and independence: if two indicator variables are independent, their covariance is zero. However, the converse isn’t always true\u2014uncorrelated indicator variables aren’t necessarily independent unless additional assumptions hold.<\/p>\n\n\n\n

          In Part B, we consider a continuous bivariate distribution: a point uniformly distributed over a triangle. The triangle is bounded by the lines x=0x = 0, y=0y = 0, and x+y=1x + y = 1. The joint density function is constant over this region due to the uniformity. Calculating the covariance involves finding E[X]\\mathbb{E}[X], E[Y]\\mathbb{E}[Y], and E[XY]\\mathbb{E}[XY], which are computed via double integrals over the triangular region. The results show that the covariance is negative, indicating a weak inverse relationship between XX and YY: as one increases, the other tends to decrease slightly due to the constraint x+y\u22641x + y \\le 1. This makes intuitive sense, as increasing one coordinate in the triangle reduces the possible range for the other.<\/p>\n\n\n\n

          Thus, covariance captures the directional relationship between random variables, whether they are discrete indicators or continuous variables under geometric constraints.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          Q1: Calculate the covariance of the following random variables: Part A: X = 1A and Y = 1B for two indicator variables over sets A and B. For the special case of indicator functions, what is the relationship between correlated\/uncorrelated and dependent\/independent? Part B: X and Y, for a point (X,Y) uniformly distributed on a […]<\/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-223630","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223630","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=223630"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223630\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223630"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223630"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223630"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}