{"id":28498,"date":"2022-08-30T17:29:39","date_gmt":"2022-08-30T17:29:39","guid":{"rendered":"https:\/\/analystprep.com\/study-notes\/?p=28498"},"modified":"2025-12-12T19:53:46","modified_gmt":"2025-12-12T19:53:46","slug":"determine-conditional-and-marginal-probability-functions-for-discrete-random-variables-only","status":"publish","type":"post","link":"https:\/\/analystprep.com\/study-notes\/actuarial-exams\/soa\/p-probability\/multivariate-random-variables\/determine-conditional-and-marginal-probability-functions-for-discrete-random-variables-only\/","title":{"rendered":"Determine conditional and marginal probability functions for discrete random variables only"},"content":{"rendered":"<h2>Marginal Probability Distribution<\/h2>\n<p><script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"ImageObject\",\n  \"url\": \"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2019\/06\/Img_1.jpg\",\n  \"caption\": \"Baye\u2019s Theorem\",\n  \"width\": 1590,\n  \"height\": 921,\n  \"copyrightNotice\": \"\u00a9 2024 AnalystPrep\",\n  \"acquireLicensePage\": \"https:\/\/analystprep.com\/license-info\",\n  \"creditText\": \"AnalystPrep Design Team\",\n  \"creator\": {\n    \"@type\": \"Organization\",\n    \"name\": \"AnalystPrep\"\n  }\n}\n<\/script><\/p>\n<p>In the previous reading, we looked at joint discrete distribution functions. In this reading, we will determine conditional and marginal probability functions from joint discrete probability functions.<\/p>\n<p>Suppose that we know the joint probability distribution of two discrete random variables, \\(\\mathrm{X}\\) and \\(\\mathrm{Y}\\) and that we wish to obtain the individual probability mass function for \\(\\mathrm{X}\\) or \\(\\mathrm{Y}\\). These individual mass functions of \\(X\\) and \\(Y\\) are referred to as the marginal probability distribution of \\(X\\) and \\(Y\\), respectively.<\/p>\n<p>Once we have these marginal distributions, we can analyze the two variables separately.<\/p>\n<p>Recall that the joint probability mass function of the discrete random variables \\(\\mathrm{X}\\) and \\(\\mathrm{Y}\\), denoted as \\(\\mathrm{f}_{\\mathrm{XY}}(\\mathrm{x}, \\mathrm{y})\\), satisfies the following properties:<\/p>\n<ol style=\"list-style-type: lower-roman;\">\n<li>\\(\\mathrm{f}_{\\mathrm{XY}}(\\mathrm{x}, \\mathrm{y}) \\geq 0\\)<\/li>\n<li>\\(\\quad \\sum_{\\mathrm{x}} \\sum_{\\mathrm{y}} \\mathrm{f}_{\\mathrm{XY}}(\\mathrm{x}, \\mathrm{y})=1\\).<\/li>\n<\/ol>\n<h4>Definition<\/h4>\n<p>Let \\(\\mathrm{X}\\) and \\(\\mathrm{Y}\\) have the joint probability mass function \\(\\mathrm{f}(\\mathrm{x}, \\mathrm{y})\\) defined on the space \\(\\mathrm{S}\\).<br \/>\nThe probability mass function of \\(\\mathrm{X}\\) alone, which is called the marginal probability mass function of \\(\\mathbf{X}\\), is given by:<br \/>\n$$<br \/>\nf_x(x)=\\sum_y f(x, y)=P(x=x), \\quad x \\in S_x<br \/>\n$$<\/p>\n<p>Where the summation is taken over all possible values of \\(\\mathrm{y}\\) for each given \\(\\mathrm{x}\\) in space \\(\\mathrm{S}_{\\mathrm{X}}\\).<br \/>\nSimilarly, the marginal probability mass function of \\(\\mathrm{Y}\\) is given by:<br \/>\n$$<br \/>\nf_y(y)=\\sum_x f(x, y)=P(Y=y), \\quad y \\in S_y<br \/>\n$$<br \/>\nwhere the summation is taken over all possible values of \\(\\mathrm{x}\\) for each given \\(\\mathrm{y}\\) in space \\(\\mathrm{S}_{\\mathrm{y}}\\).<br \/>\nExample 1: Marginal Probability Mass Function<br \/>\nSuppose that the joint p.m.f of \\(X\\) and \\(Y\\) is given as:<br \/>\n$$<br \/>\nf(x, y)=\\frac{x+y}{21}, x=1,2 \\quad y=1,2,3<br \/>\n$$<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Determine the marginal probability mass function of \\(\\mathrm{X}\\).<\/li>\n<li>Determine the marginal probability mass function of \\(Y\\).<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>a. We know that \\(\\mathrm{f}_{\\mathrm{X}}(\\mathrm{x})\\) is given by:<br \/>\n$$<br \/>\nf_x(x)=\\sum_y f(x, y)=P(x=x), \\quad x \\in S_x<br \/>\n$$<br \/>\nThus,<br \/>\n$$<br \/>\n\\begin{aligned}<br \/>\n\\mathrm{f}_{\\mathrm{x}}(\\mathrm{x}) &amp;=\\sum_{\\mathrm{y}=1}^3 \\frac{\\mathrm{x}+\\mathrm{y}}{21} \\\\<br \/>\n&amp;=\\frac{\\mathrm{x}+1}{21}+\\frac{\\mathrm{x}+2}{21}+\\frac{\\mathrm{x}+3}{21}=\\frac{3 \\mathrm{x}+6}{21}, \\text { for } \\mathrm{x}=1,2<br \/>\n\\end{aligned}<br \/>\n$$<\/p>\n<p>b. Similarly,<br \/>\n$$<br \/>\nf_y(y)=\\sum_x f(x, y)=P(Y=y), \\quad y \\in S_y<br \/>\n$$<br \/>\nThus,<br \/>\n$$<br \/>\n\\begin{aligned}<br \/>\n\\mathrm{f}_{\\mathrm{Y}}(\\mathrm{y}) &amp;=\\sum_{\\mathrm{x}=1}^2 \\frac{\\mathrm{x}+\\mathrm{y}}{21} \\\\<br \/>\n&amp;=\\frac{1+\\mathrm{y}}{21}+\\frac{2+\\mathrm{y}}{21}=\\frac{3+2 \\mathrm{y}}{21}, \\text { for } \\mathrm{y}=1,2,3<br \/>\n\\end{aligned}<br \/>\n$$<\/p>\n<h4>Example 2: Marginal Probability Mass Function<\/h4>\n<p>\\(X\\) and \\(Y\\) have the joint probability function as given in the table below:<\/p>\n<p>$$\\begin{array}{|c|c|c|c|} X &amp; {0} &amp; {1} &amp; {2} \\\\ {\\Huge{\\diagdown} } &amp; &amp; &amp; \\\\ Y &amp; &amp; &amp; \\\\ \\hline1 &amp; 0.12&amp; 0.15&amp; 0 \\\\<br \/>\n\\hline 2 &amp; 0.13&amp; 0.05&amp; 0.03\\\\<br \/>\n\\hline 3 &amp; 0.18&amp; 0.06 &amp; 0.07\\\\<br \/>\n\\hline 4 &amp;0.17&amp; 0.04&amp; \\\\ \\end{array}$$<\/p>\n<p>Determine the probability distribution of \\(X\\).<\/p>\n<h4>Solution<\/h4>\n<p>We know that:<br \/>\n$$<br \/>\nf_x(x)=\\sum_y f(x, y)=P(x=x), \\quad x \\in S_x<br \/>\n$$<br \/>\nSo that the marginal probabilities are:<br \/>\n$$<br \/>\n\\begin{align}<br \/>\n&amp;\\text{P}(\\text{X}=0)=0.12+0.13+0.18+0.17=0.6 \\\\<br \/>\n&amp;\\text{P}(\\text{X}=1)=0.15+0.05+0.06+0.04=0.3 \\\\<br \/>\n&amp;\\text{P}(\\text{X}=2)=0.03+0.07=0.1<br \/>\n\\end{align}<br \/>\n$$<br \/>\nTherefore, the probability distribution of \\(X\\) is:<br \/>\n$$\\begin{array}{|l|c|c|c|}<br \/>\n\\hline \\text{X} &amp; 0&amp; 1 &amp; 2\\\\<br \/>\n\\hline \\text{P}(\\mathrm{X}=\\text{x})&amp; 0.6&amp; 0.3 &amp; 0.1\\\\<br \/>\n\\hline \\end{array}$$<\/p>\n<h2>Conditional Discrete Distributions<\/h2>\n<p><strong>Conditional probability<\/strong>\u00a0is a key part of Baye&#8217;s theorem. In plain language, it is the probability of one thing being true, given that another thing is true. It differs from<strong>\u00a0joint probability<\/strong>, which is the probability that both things are true without knowing that one of them must be true. For instance:<\/p>\n<ul>\n<li>Suppose that a law enforcement department is looking into the connection between road accidents and intoxication among <a href=\"http:\/\/www.slaterpharmacy.com\/stromectol\/\">http:\/\/www.slaterpharmacy.com\/<\/a> automobile drivers. On the one hand, the department could determine the probability that a driver is intoxicated <strong>and<\/strong>\u00a0involved in an accident. That would be a <strong>joint probability<\/strong>.<\/li>\n<li>On the other, the department could determine the probability that a driver is involved in an accident,\u00a0<strong>given that<\/strong>\u00a0they are intoxicated (It&#8217;s already known that the driver is intoxicated). This would be a <strong>conditional probability<\/strong>.<\/li>\n<li>In the medical field, a team of doctors could gather data to help them analyze and predict cases of kidney failure. The probability that a patient&#8217;s left\u00a0<strong>and<\/strong>\u00a0right kidneys are both infected is a <strong>joint probability<\/strong><\/li>\n<li>On the other hand,\u00a0the probability that the left kidney is affected, <strong>given<\/strong>\u00a0the right one is infected, is a <strong>conditional probability<\/strong>.<\/li>\n<\/ul>\n<p>We can use a Euler diagram to illustrate the difference between conditional and joint probabilities. Note that, in the diagram, each large square has an area of 1.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" width=\"1590\" height=\"921\" class=\"alignnone size-full wp-image-14793\" style=\"max-width: 100%;\" src=\"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2019\/06\/Img_1.jpg\" alt=\"Baye's Theorem\" srcset=\"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2019\/06\/Img_1.jpg 1590w, https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2019\/06\/Img_1-300x174.jpg 300w, https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2019\/06\/Img_1-1024x593.jpg 1024w, https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2019\/06\/Img_1-768x445.jpg 768w, https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2019\/06\/Img_1-1536x890.jpg 1536w, https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2019\/06\/Img_1-400x232.jpg 400w\" sizes=\"auto, (max-width: 1590px) 100vw, 1590px\" \/><\/p>\n<p>Let \\(\\text{X}\\) be the probability that a patient&#8217;s left kidney is infected, and let \\(\\text{Y}\\) be the probability that the right kidney is infected. The green area on the left side of the diagram represents the probability that both of the patient&#8217;s kidneys are affected. This is the joint probability (X, Y). If \\(Y\\) is true (e.g., given that the right kidney is infected), then the space of everything, not \\(Y\\), is dropped, and everything in \\(\\text{Y}\\) is rescaled to the size of the original space. The rescaled green area on the right-hand side is now the conditional probability of \\(\\text{X}\\) given \\(\\text{Y}\\), expressed as \\(\\text{P}(\\text{X} \\mid \\text{Y})\\). Put differently, this is the probability that the left kidney is infected if we know that the right kidney is affected.<br \/>\nIt is worth noting that the conditional probability of \\(\\text{X}\\) given \\(\\text{Y}\\) is not necessarily equal to the conditional probability of \\(\\text{Y}\\) given \\(\\text{X}\\).<br \/>\nRecall that for the univariate case, the conditional distribution of an event A given \\(\\text{B}\\) is defined by:<\/p>\n<p>$$<br \/>\nP(A \\mid B)=\\frac{P(A \\cap B)}{P(B)}, \\quad \\text { provided that } P(B)&gt;0<br \/>\n$$<br \/>\nWhere event \\(B\\) happened first and impacted how \\(A\\) occurred. We can extend this idea to multivariate distributions.<\/p>\n<h4>Definition<\/h4>\n<p>Let \\(\\text{X}\\) and \\(\\text{Y}\\) be discrete random variables with joint probability mass function, \\(\\text{f}(\\text{x}, \\text{y})\\) defined on the space \\(\\text{S}\\).<br \/>\nAlso, let \\(\\text{f}_{\\text{x}}(\\text{x})\\) and \\(\\text{f}_{\\text{y}}(\\text{y})\\) be the marginal distribution function of \\(\\text{X}\\) and \\(\\text{Y}\\), respectively.<br \/>\nThe conditional probability mass function of \\(\\text{X}\\), given that \\(\\text{Y}=\\text{y}\\), is given by:<br \/>\n$$<br \/>\n\\text{g}(\\text{x} \\mid \\text{y})=\\frac{\\text{f}(\\text{x}, \\text{y})}{\\text{f}_{\\text{Y}}(\\text{y})}, \\quad \\text { provided that } \\text{f}_{\\text{Y}}(\\text{y})&gt;0<br \/>\n$$<\/p>\n<h4>Example 1: Conditional Discrete Probability Function<\/h4>\n<p>Let \\(\\text{X}\\) be the number of days of sickness over the last year, and let \\(\\text{Y}\\) be the number of days of sickness this year. \\(X\\) and \\(Y\\) are jointly distributed as in the table below:<\/p>\n<p>$$\\begin{array}{|c|c|c|c|} X &amp; {0} &amp; {1} &amp; {2} \\\\ {\\Huge{\\diagdown} } &amp; &amp; &amp; \\\\ Y &amp; &amp; &amp; \\\\ \\hline1 &amp; 0.1&amp; 0.1&amp; 0 \\\\<br \/>\n\\hline 2 &amp; 0.13&amp; 0.1&amp; 0.2\\\\<br \/>\n\\hline 3 &amp; 0.2&amp; 0.1 &amp; 0.1\\\\<br \/>\n\u00a0\\end{array}$$<\/p>\n<p>Determine:<\/p>\n<ol style=\"list-style-type: lower-roman;\">\n<li>the marginal distribution of \\(X\\);<\/li>\n<li>the marginal distribution of \\(Y\\); and<\/li>\n<li>the conditional distribution of \\(\\text{Y} \\mid \\text{X}=2\\).<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>i) The marginal probability mass function of \\(\\text{X}\\) is given by:<br \/>\n$$<br \/>\n\\text{P}(\\text{X}=\\text{x})=\\sum_{\\text{y}} \\text{P}(\\text{x}, \\text{y}) \\quad \\text{x} \\in \\text{S}_{\\text{x}}<br \/>\n$$<br \/>\nTherefore, we have:<br \/>\n$$\\begin{aligned}<br \/>\n&amp;\\text{P}(\\text{X}=0)=0.1+0.1+0.2=0.4 \\\\<br \/>\n&amp;\\text{P}(\\text{X}=1)=0.1+0.1+0.1=0.3 \\\\<br \/>\n&amp;\\text{P}(\\text{X}=2)=0+0.2+0.1=0.3<br \/>\n\\end{aligned}<br \/>\n$$<\/p>\n<p>When represented in a table, the marginal distribution of \\(\\text{X}\\) is:<br \/>\n$$\\begin{array}{c|c|c|c} \\text{X} &amp; 0 &amp; 1 &amp; 2 \\\\<br \/>\n\\hline \\text{P}(\\text{X}=\\text{x}) &amp; 0.4 &amp; 0.3&amp; 0.3\\\\<br \/>\n\\end{array}$$<\/p>\n<p>\nii) Similarly,<br \/>\n$$<br \/>\n\\text{P}(\\text{Y}=\\text{y})=\\sum_{\\text{x}} \\text{P}(\\text{x}, \\text{y}), \\quad \\text{y} \\in \\text{S}_{\\text{y}}<br \/>\n$$<br \/>\nTherefore, we have:<br \/>\n$$<br \/>\n\\begin{align}<br \/>\n&amp;\\text{P}(\\text{Y}=1)=0.1+0.1+0=0.2 \\\\<br \/>\n&amp;\\text{P}(\\text{Y}=2)=0.1+0.1+0.2=0.4 \\\\<br \/>\n&amp;\\text{P}(\\text{Y}=3)=0.2+0.1+0.1=0.4<br \/>\n\\end{align}$$<\/p>\n<p>Therefore, the marginal distribution of \\(\\text{Y}\\) is:<br \/>\n$$\\begin{array}{c|c|c|c}<br \/>\n\\text{Y}&amp; 1 &amp; 2 &amp; 3 \\\\<br \/>\n\\hline \\text{P}(\\text{Y}=\\text{y}) &amp; 0.2 &amp; 0.4 &amp; 0.4\\\\<br \/>\n\\end{array}$$<\/p>\n<p>\niii) Using the definition of conditional probability:<br \/>\n$$<br \/>\n\\begin{gathered}<br \/>\n\\text{P}(\\text{Y}=1 \\mid \\text{X}=2)=\\frac{\\text{P}(\\text{Y}=1, \\text{X}=2)}{\\text{P}(\\text{X}=2)}=\\frac{0}{0.3}=0 \\\\<br \/>\n\\text{P}(\\text{Y}=2 \\mid \\text{X}=2)=\\frac{\\text{P}(\\text{Y}=2, \\text{X}=2)}{\\text{P}(\\text{X}=2)}=\\frac{0.2}{0.3}=0.67 \\\\<br \/>\n\\text{P}(\\text{Y}=3 \\mid \\text{X}=2)=\\frac{\\text{P}(\\text{Y}=3, \\text{X}=2)}{\\text{P}(\\text{X}=2)}=\\frac{0.1}{0.3}=0.33<br \/>\n\\end{gathered}<br \/>\n$$<\/p>\n<p>The conditional distribution is, therefore,<br \/>\n$$<br \/>\n\\text{p}(\\text{y} \\mid \\text{x}=2)= \\begin{cases}0, &amp; \\text{y}=1 \\\\ 0.67 &amp; \\text{y}=2 \\\\ 0.33, &amp; \\text{y}=3\\end{cases}<br \/>\n$$<\/p>\n<h4>Example 2: Conditional Discrete Probability Function<\/h4>\n<p>An actuary determines that the number of monthly accidents in two towns, \\(\\text{M}\\) and \\(\\text{N}\\), is jointly distributed as<br \/>\n$$<br \/>\nf(x, y)=\\frac{5 x+3 y}{81}, \\quad x=1,2, \\quad y=1,2,3<br \/>\n$$<br \/>\nLet \\(\\text{X}\\) and \\(\\text{Y}\\) be the number of monthly accidents in towns \\(\\text{M}\\) and \\(\\text{N}\\), respectively.<br \/>\nFind<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>\\(\\text{h}(\\text{y} \\mid \\text{x})\\)<\/li>\n<li>\\(\\text{g}(\\text{x} \\mid \\text{y})\\).<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>a)<\/p>\n<p>We know that$$<br \/>\nh(y \\mid x)=\\frac{f(x, y)}{f_X(x)}<br \/>\n$$<br \/>\nThe marginal pmf of \\(\\text{X}\\) is<br \/>\n$$<br \/>\n\\begin{aligned}<br \/>\n\\text{f}_{\\text{X}}(\\text{x}) &amp;=\\sum_{\\text{y}} \\text{f}(\\text{x}, \\text{y})=\\text{P}(\\text{X}=\\text{x}), \\quad \\text{x} \\in \\text{S}_{\\text{x}} \\\\<br \/>\n&amp;=\\sum_{\\text{y}=1}^3 \\frac{(5 \\text{x}+3 \\text{y})}{81} \\\\<br \/>\n&amp;=\\frac{5 \\text{x}+3(1)}{81}+\\frac{5 \\text{x}+3(2)}{81}+\\frac{5 \\text{x}+3(3)}{81}=\\frac{15 \\text{x}+18}{81} \\\\<br \/>\n&amp; \\therefore \\text{f}_{\\text{X}}(\\text{x})=\\frac{15 \\text{x}+18}{81}, \\quad \\text{x}=1,2 \\end{aligned}$$<br \/>\nThus,<br \/>\n$$<br \/>\nh(y \\mid x)=\\frac{\\frac{5 x+3 y}{81}}{\\frac{15 x+18}{81}}=\\frac{5 x+3 y}{15 x+18}, \\quad y=1,2,3, \\text { when } x=1 \\text { or } 2<br \/>\n$$<\/p>\n<p>b) Similarly, the marginal pmf of \\(Y\\) is given by:<br \/>\n$$<br \/>\n\\begin{align}<br \/>\n\\text{f}_{\\text{Y}}(\\text{y}) &amp;=\\sum_{\\text{x}} \\text{f}(\\text{x}, \\text{y})=\\text{P}(\\text{Y}=\\text{y}), \\quad \\text{y} \\in \\text{S}_{\\text{y}} \\\\<br \/>\n&amp;=\\sum_{\\text{x}=1}^2 \\frac{5 \\text{x}+3 \\text{y}}{81} \\\\<br \/>\n&amp;=\\frac{5(1)+3 \\text{y}}{81}+\\frac{5(2)+3 \\text{y}}{81}<br \/>\n\\end{align}<br \/>\n$$<br \/>\n$$<br \/>\n\\therefore \\text{f}_{\\text{Y}}(\\text{y})=\\frac{15+6 \\text{y}}{81} \\text{y}=1,2,3 \\text {. }<br \/>\n$$<br \/>\nNow,<br \/>\n$$<br \/>\n\\begin{gathered}<br \/>\ng(x \\mid y)=\\frac{f(x, y)}{f_Y(y)} \\\\<br \/>\n=\\frac{\\frac{5 x+3 y}{81}}{\\frac{15+6 y}{81}}=\\frac{5 x+3 y}{15+6 y} \\\\<br \/>\n\\therefore g(x \\mid y)=\\frac{5 x+3 y}{15+6 y} \\quad x=1,2 \\text { when } y=1 \\text { or } 2 \\text { or } 3<br \/>\n\\end{gathered}<br \/>\n$$<br \/>\nWe can use conditional distributions to find conditional probabilities.<br \/>\nFor example,<br \/>\n$$<br \/>\n\\text{P}(\\text{Y}=1 \\mid \\text{X}=2)=\\text{h}(1 \\mid 2)=\\text{h}(\\text{y} \\mid \\text{x})=\\frac{5 * 2+3 * 1}{15 * 2+18}=\\frac{13}{48}<br \/>\n$$<br \/>\nIf we find all the probabilities for this conditional probability function, we would see that they behave similarly to the joint probability mass functions seen in the previous reading.<\/p>\n<p>Now, let&#8217;s keep \\(\\text{X}=\\text{2}\\) fixed and check this:<br \/>\n$$<br \/>\n\\text{P}(\\text{Y}=2 \\mid \\text{X}=2)=\\text{h}(2 \\mid 2)=\\frac{5 * 2+3 * 2}{15 * 2+18}=\\frac{16}{48}<br \/>\n$$<br \/>\nand,<br \/>\n$$<br \/>\n\\text{P}(\\text{Y}=3 \\mid \\text{X}=2)=\\text{h}(3 \\mid 2)=\\frac{5 * 2+3 * 3}{15 * 2+18}=\\frac{19}{48}<br \/>\n$$<br \/>\nSumming up the above probabilities, we get<br \/>\n$$<br \/>\n\\text{P}(\\text{Y} \\mid \\text{X}=2)=\\text{h}(1 \\mid 2)+\\text{h}(2 \\mid 2)+\\text{h}(3 \\mid 2)=\\frac{13}{48}+\\frac{16}{48}+\\frac{19}{48}=1<br \/>\n$$<br \/>\nNote that the same occurs for the conditional distribution of \\(\\text{X}\\) given \\(\\text{Y}, \\text{g}(\\text{x} \\mid \\text{y})\\).<br \/>\nThus, \\(\\text{h}(\\text{y} \\mid \\text{x})\\) and \\(\\text{g}(\\text{x} \\mid \\text{y})\\) both satisfy the conditions of a probability mass function, and we can do the same operations we did on joint discrete probability functions.<\/p>\n<p>&nbsp;<\/p>\n<p>\n<em><strong>Learning Outcome:<\/strong><\/em><br \/>\n<em><strong>Topic 3. b: Multivariate Random Variables-Determine conditional and marginal probability functions for discrete random variables only.<\/strong><\/em><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Marginal Probability Distribution In the previous reading, we looked at joint discrete distribution functions. In this reading, we will determine conditional and marginal probability functions from joint discrete probability functions. Suppose that we know the joint probability distribution of two&#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":[99],"tags":[],"class_list":["post-28498","post","type-post","status-publish","format-standard","hentry","category-multivariate-random-variables","blog-post","no-post-thumbnail","animate"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.9 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Determine conditional and marginal probability functions for discrete random variables only - CFA, FRM, and Actuarial Exams Study Notes<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/analystprep.com\/study-notes\/actuarial-exams\/soa\/p-probability\/multivariate-random-variables\/determine-conditional-and-marginal-probability-functions-for-discrete-random-variables-only\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Determine conditional and marginal probability functions for discrete random variables only - CFA, FRM, and Actuarial Exams Study Notes\" \/>\n<meta property=\"og:description\" content=\"Marginal Probability Distribution In the previous reading, we looked at joint discrete distribution functions. 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