{"id":28487,"date":"2022-08-30T13:33:38","date_gmt":"2022-08-30T13:33:38","guid":{"rendered":"https:\/\/analystprep.com\/study-notes\/?p=28487"},"modified":"2022-09-14T11:42:00","modified_gmt":"2022-09-14T11:42:00","slug":"explain-and-perform-calculations-concerning-joint-probability-functions-and-cumulative-distribution-functions-for-discrete-random-variables-only","status":"publish","type":"post","link":"https:\/\/analystprep.com\/study-notes\/actuarial-exams\/soa\/p-probability\/multivariate-random-variables\/explain-and-perform-calculations-concerning-joint-probability-functions-and-cumulative-distribution-functions-for-discrete-random-variables-only\/","title":{"rendered":"Explain and perform calculations concerning joint probability functions and cumulative distribution functions for discrete random variables only"},"content":{"rendered":"<h2>Joint Discrete Probability Distributions<\/h2>\n<p><span style=\"font-weight: 400;\">We are often interested in experiments that involve the intersection of two or more events.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">For example:<\/span><\/p>\n<ol style=\"list-style-type: lower-roman;\">\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">An experimenter tossing a fair die is interested in the intersection of getting, say, a 5 and a 6.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The intersection of events of height and weight measures and so on.<\/span><\/li>\n<\/ol>\n<p><strong>Definition<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">Let \\(X\\) and \\(Y\\) be two discrete random variables defined on a two-dimensional discrete space, \\(S\\).\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The <\/span><b>joint probability mass function <\/b><span style=\"font-weight: 400;\">of \\(X\\) and \\(Y\\)<\/span><span style=\"font-weight: 400;\">\u00a0is defined as<\/span><\/p>\n<p><span style=\"font-weight: 400;\">$$f(x, y) = P(X = x, Y = y)$$<\/span><\/p>\n<p><b>Properties of a Joint Discrete Distribution<\/b><\/p>\n<ol>\n<li>\\(0 \\leq f(x, y) \\leq 1\\); It means that the probability of any possible outcome must not be less than 0 and greater than 1.<\/li>\n<li>\u00a0\\(\\sum_{\\mathrm{x}} \\sum_{\\mathrm{y}} \\mathrm{f}(\\mathrm{x}, \\mathrm{y})=1, \\forall(\\mathrm{x}, \\mathrm{y}) \\in \\mathrm{S}\\); The condition requires all the probabilities over the entire space, \\(S\\) to sum up to 1.<\/li>\n<li>\\(\\mathrm{P}[(\\mathrm{X}, \\mathrm{Y} \\in \\mathrm{A})]=\\sum_{\\mathrm{x}, \\mathrm{y} \\in \\mathrm{A}} \\sum \\mathrm{f}(\\mathrm{x}, \\mathrm{y})\\); where \\(\\mathrm{A}\\) is a subset of the space \\(\\mathrm{S}\\). It means that whenever we want to establish the probability of a given event \\(A\\), we do so by simply summing up the probabilities of the \\((x, y)\\) values in \\(A\\).<\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">We will now look at a few examples to shed more light on the above definition.<\/span><\/p>\n<h4>Example 1: Joint Discrete Distribution\u00a0<\/h4>\n<p><span style=\"font-weight: 400;\">Suppose a certain local bank had three deposit or withdrawal counters. Two investors arrive at the counters at different times when the counters are serving no other customers. Each investor chooses a counter at random, independently of the other.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Let \\(X\\) <\/span><span style=\"font-weight: 400;\">be the number of investors who select counter 1, and let\u00a0\\(Y\\)<\/span><span style=\"font-weight: 400;\">\u00a0<\/span><span style=\"font-weight: 400;\">be the number of investors who select counter 2.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Determine the joint probability function of \\(X\\) and \\(Y\\).<\/span><\/p>\n<h4><b>Solution<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">First, we have to consider the sample space associated with the experiment.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Let the pair <\/span><span style=\"font-weight: 400;\">{\\(i,j\\)}<\/span><span style=\"font-weight: 400;\"> represent the simple event that the first investor selects counter \\(i\\) <\/span><span style=\"font-weight: 400;\">and the second investor choose counter \\(j\\)<\/span><span style=\"font-weight: 400;\">, where \\(i, j=1, 2, 3.\\)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">By using the \\(mn\\)<\/span><span style=\"font-weight: 400;\"> rule, the sample space consists of \\(3\\times 3=9\\)<\/span><span style=\"font-weight: 400;\">\u00a0sample points.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, each sample point is equal and has a probability of \\(\\frac{1}{9}\\)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Thus, the sample space for the experiment is given as:<\/span><\/p>\n<p>$$S=[\\{1,1\\},\\{1,2\\},\\{1,3\\},\\{2,1\\},\\{2,2\\},\\{2,3\\},\\{3,1\\},\\{3,2\\},\\{3,3\\}]$$<\/p>\n<p><span style=\"font-weight: 400;\">We know that:<\/span><\/p>\n<p>$$ \\text{f} \\left( { \\text{x} },{ \\text{y} } \\right) = \\text{P} \\left( \\text{X} = \\text{x},\\text{Y} = \\text{y} \\right) $$<\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the joint probability of \\(X\\) and \\(Y\\) is given as follows: <\/span><\/p>\n<p>$$ \\begin{array}{c|c|c|c} {\\quad \\text X }&amp; {0} &amp; {1} &amp; {2} \\\\ {\\Huge \\diagdown } &amp; &amp; &amp; \\\\ {\\text Y \\quad} &amp; &amp; &amp; \\\\ \\hline {0} &amp; {\\cfrac{1}{9}} &amp; {\\cfrac{2}{9}} &amp; {\\cfrac{1}{9}} \\\\ \\hline {1} &amp; {\\cfrac{2}{9}} &amp; {\\cfrac{2}{9}} &amp; {0} \\\\ \\hline {2} &amp; {\\cfrac{1}{9}} &amp; {0} &amp; {0} \\end{array} $$<\/p>\n<p><span style=\"font-weight: 400;\">We can now calculate other associated probabilities from the joint distribution above.<\/span><\/p>\n<h4>Example 2: Joint Discrete Distribution\u00a0<\/h4>\n<p><span style=\"font-weight: 400;\">Using the results in Example 1 above, calculate:<\/span><\/p>\n<ol style=\"list-style-type: lower-roman;\">\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">\\(P(X=2, Y=0 \\quad or \\quad 1)\\)<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">\\(P(Y=2)\\)<\/span><\/li>\n<\/ol>\n<p><b>Solution<\/b><\/p>\n<p><span style=\"font-weight: 400;\">i. We know that,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">$$f(x, y) = P(X = x, Y = y)$$<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Thus,\u00a0<\/span><\/p>\n<p>$$\\begin{aligned}<br \/>\n\\mathrm{P}(\\mathrm{X}=2, \\mathrm{Y}=0 \\text { or } 1)&amp;= \\mathrm{P}(\\mathrm{X}=2, \\mathrm{Y}=0)+\\mathrm{P}(\\mathrm{X}=2, \\mathrm{Y}\\\\&amp; =\\frac{1}{9}+0=\\frac{1}{9} <br \/>\n\\end{aligned}$$<\/p>\n<p><span style=\"font-weight: 400;\">ii. We are required to find <\/span><span style=\"font-weight: 400;\">P(Y=2)<\/span><span style=\"font-weight: 400;\">, and since it does not depend on the value of \\(X\\), it is the same as finding <\/span><span style=\"font-weight: 400;\">P(Y=2, X=0,1,2)<\/span><span style=\"font-weight: 400;\">. That is, we are summing over all the possible values of \\(X\\).<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Thus,\u00a0<\/span><\/p>\n<p>$$<br \/>\n{P}({Y}=2)=\\frac{1}{9}+0+0=\\frac{1}{9}<br \/>\n$$<span style=\"font-weight: 400;\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/span><\/p>\n<h4>Example 3: Joint Discrete Distribution\u00a0<\/h4>\n<p>Let \\(X\\) be the number of claims from males and \\(Y\\) be the number of claims from females. \\(X\\) and \\(Y\\) have the following joint probability distribution:<\/p>\n<p>$$f(x, y)=\\frac{y}{9 x}, \\quad \\text { for } x=1,2 ; y=1,2,3$$<\/p>\n<p>Calculate \\(\\text{P}\\left(\\text{X}+\\frac{\\text{Y}}{2}=2\\right)\\).<\/p>\n<h4>Solution<\/h4>\n<p>We first determine the pairs \\((x, y)\\) which satisfy the condition that \\(x+\\frac{y}{2}=2\\).<\/p>\n<p>\\(x+\\frac{y}{2}=2\\) only for the pair \\((1,2)\\).<\/p>\n<p>Now, we can proceed to calculate the required probability:<\/p>\n<p>$$P\\left(X+\\frac{Y}{2}=2\\right)=\\frac{2}{9 \\times 1}=\\frac{2}{9}.$$<\/p>\n<h4>Example 4: Joint Discrete Distribution<\/h4>\n<p>An analyst is concerned about the annual number of tsunamis in two countries, \\(M\\) and \\(N\\). Let \\(X\\) and \\(Y\\) be the annual number of tsunamis in countries $M$ and $N$, respectively. The analyst determines that \\(X\\) and \\(Y\\) are jointly distributed as below:<\/p>\n<p>$$f(x, y)=\\frac{x y}{10}, \\text { for } x=0,1 ; y=0,1,2,3,4$$<\/p>\n<p>Calculate \\({P}(\\text{X}+\\text{Y}&lt;3)\\).<\/p>\n<h4>Solution<\/h4>\n<p>\\(x+y&lt;3\\) for the pairs, \\((0,0) ;(0,1) ;(0,2) ;(1,0) \\quad \\text{and} \\quad (1,1)\\).<\/p>\n<p>Therefore,<\/p>\n<p>$$\\begin{aligned}<br \/>\nP(X+Y&lt;3) &amp;=\\frac{0}{10}+\\frac{0}{10}+\\frac{0}{10}+\\frac{0}{10}+\\frac{1}{10} \\\\&amp;=\\frac{1}{10} \\end{aligned}<br \/>\n$$<\/p>\n<h2>Joint Discrete Cumulative Distribution Functions<\/h2>\n<h4>Definition<\/h4>\n<p>The joint cumulative distribution function, \\(F_{XY}(x,y)\\) of two discrete random variables,\\(X\\) and \\(Y\\), is defined as the probability <a href=\"https:\/\/canadianpharmacy365.net\/product-category\/erectile-dysfunction\/\">https:\/\/canadianpharmacy365.net\/<\/a> that the random variable \\(X\\) is a specified value of \\(x\\) and that the random variable \\(Y\\) is a specified value of \\(y\\), namely, \\(F_{XY}(x,y)=P(X\u2264x, Y\u2264y)\\).<\/p>\n<p>Now, consider an experiment of a sample of size n, i.e., \\(X_1, X_2,\u2026, X_n\\). The cumulative distribution function of \\(X_1, X_2,\u2026, X_n\\) is given by: $$F\\left(X_{1}, X_{2}, \\ldots, X_{n}\\right)=\\sum_{w_{1} \\leq x_{1}} \\sum_{w_{2} \\leq x_{2}} \\ldots \\sum_{w_{n} \\leq x_{n}} f\\left(w_{1}, w_{2}, \\ldots, w_{n}\\right).$$<\/p>\n<p>The following result hold for two random variables, \\(X\\) and \\(Y\\):<\/p>\n<p>$$P\\left(x_1&lt;X\u2264x_2, y_1&lt; Y\u2264y_2\\right) = F\\left(x_2,y_2\\right) + F\\left(x_1, y_1\\right) &#8211; F\\left(x_1, y_2\\right)- F\\left(x_2,y_1\\right).$$<\/p>\n<p>The above result holds if and only if \\(x_1&lt; x_2\\) and \\(y_1&lt; y_2\\).<\/p>\n<h4>Example 1: Joint Discrete Cumulative Distribution Function<\/h4>\n<p>Let \\(X\\) and \\(Y\\) be two discrete random variables whose joint pmf is given in the table below:<\/p>\n<p>$$ \\begin{array}{c|c|c|c} {\\quad \\text X }&amp; {0} &amp; {1} &amp; {2} \\\\ {\\Huge \\diagdown } &amp; &amp; &amp; \\\\ {\\text Y \\quad} &amp; &amp; &amp; \\\\ \\hline \\textbf{0} &amp; \\cfrac{1}{8} &amp; \\cfrac{1}{6} &amp; \\cfrac{1}{4} \\\\ \\hline \\textbf{1} &amp; \\cfrac{1}{6} &amp; \\cfrac{1}{8} &amp; \\cfrac{1}{6} \\end{array} $$<\/p>\n<p>Find \\({ \\text{F} }_{ \\text{XY} }\\) (0.5,1).<\/p>\n<h4>Solution<\/h4>\n<p>$$ \\text{P}\\left({ \\text{X} }\\le0.5,{ \\text{Y} }\\le1\\right) $$<\/p>\n<p>Thus,<\/p>\n<p>$$ \\begin{align*} { \\text{P} }_{ \\text{XY} } \\left(0,0 \\right)+ {\\text{P} }_{\\text{XY}} \\left(0,1\\right)&amp;=\\cfrac{1}{8}+\\cfrac{1}{6}=\\cfrac{7}{24} \\\\ \\therefore { \\text{F} }_{ \\text{XY} } \\left(0.5,1\\right) &amp; =\\cfrac{7}{24} \\end{align*} $$<\/p>\n<h4>Example 2: Joint Discrete Cumulative Distribution Function<\/h4>\n<p><span style=\"font-weight: 400;\">Let \\(X\\) and \\(Y\\) be the number of road accidents in two neighboring cities, \\(A\\) and \\(B\\), respectively, in June 2022.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\\(X\\) and \\(Y\\) have the following joint cumulative distribution function:<\/span><\/p>\n<p>$$F(x, y)=\\left(0.8^{\\text{x}}\\right)\\left(0.2^{\\text{y}}\\right), \\text { for } \\text{x}=0,1,2 \\ldots \\text; \\quad \\text{y}=0,1,2 \\ldots$$<\/p>\n<p>Find the probability that in June 2022, we will have exactly 3 claims from A and exactly 3 claims from B.<\/p>\n<h4>Solution.<\/h4>\n<p>We wish to find \\(P(\\text{X}=3, \\text{Y}=3)\\).<\/p>\n<p>We know that,<\/p>\n<p>$$F(x, y)=P(X \\leq x, Y \\leq y)$$<\/p>\n<p>We also know that,<\/p>\n<p>$$\\begin{aligned}{P}\\left({x}_{1}&lt;{X} \\leq \\mathrm{x}_{2}, \\mathrm{y}_{1}&lt;\\mathrm{Y} \\leq \\mathrm{y}_{2}\\right)&amp;=\\mathrm{F}\\left(\\mathrm{x}_{2}, \\mathrm{y}_{2}\\right)+\\mathrm{F}\\left(\\mathrm{x}_{1}, \\mathrm{y}_{1}\\right)-\\mathrm{F}\\left(\\mathrm{x}_{1}, \\mathrm{y}_{2}\\right)-\\mathrm{F}\\left(\\mathrm{x}_{2}, \\mathrm{y}_{1}\\right) \\\\<br \/>\n\\Rightarrow \\mathrm{P}(\\mathrm{X}=3, \\mathrm{Y}=3)&amp;=\\mathrm{F}(3,3)-\\mathrm{F}(2,3)-\\mathrm{F}(3,2)+\\mathrm{F}(2,2) \\\\<br \/>\n&amp;=\\left(0.8^{3}\\right)\\left(0.2^{3}\\right)+\\left(0.8^{2}\\right)\\left(0.2^{2}\\right)-\\left(0.8^{2}\\right)\\left(0.2^{3}\\right)-\\left(0.8^{3}\\right)\\left(0.2^{2}\\right) \\\\<br \/>\n&amp;=0.004096\\end{aligned}$$<\/p>\n<p>&nbsp;<\/p>\n<p><b><i>Learning Outcome:<\/i><\/b><\/p>\n<p><b><i>Topic 3. a: Explain and perform calculations concerning joint probability functions and cumulative distribution functions for discrete random variables only. <\/i><\/b><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Joint Discrete Probability Distributions We are often interested in experiments that involve the intersection of two or more events.\u00a0 For example: An experimenter tossing a fair die is interested in the intersection of getting, say, a 5 and a 6&#8230;.<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[89,99,90,92],"tags":[],"class_list":["post-28487","post","type-post","status-publish","format-standard","hentry","category-actuarial-exams","category-multivariate-random-variables","category-p-probability","category-soa","blog-post","no-post-thumbnail","animate"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.6 - 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