Recall that, in the previous readings, we defined the conditional distribution function of \\(X\\), given that \\(\\text{Y}=\\text{y}\\) as:<\/p>\n
$$
\n\\text{g}(\\text{x} \\mid \\text{y})=\\frac{\\text{f}(\\text{x}, \\text{y})}{\\text{f}_{\\text{Y}}(\\text{y})}, \\quad \\text { provided that } \\text{f}_{\\text{Y}}(\\text{y})>0
\n$$<\/p>\n
Similarly, the conditional distribution function of \\(Y\\), given that \\(X=\\text{x}\\) is defined by:<\/p>\n
$$
\n\\text{h}(\\text{y} \\mid \\text{x})=\\frac{\\text{f(x, y)}}{\\text{f}_{\\text X}(\\text x)}, \\quad \\text { provided that } \\text{f}_{\\text X}(\\text x)>0
\n$$
\nNow, if we have the above conditional distributions, we can go ahead to calculate conditional variance as well as the conditional standard deviation.<\/p>\n
The conditional variance of \\(X\\), given that \\(Y=y\\), is defined by:<\/p>\n
$$
\n\\operatorname{Var}(\\text X \\mid \\text Y=\\text y)=\\text E\\left(\\text X^{2} \\mid \\text Y=\\text y\\right)-[\\text E(\\text X \\mid \\text Y=\\text y)]^{2}
\n$$
\nWhere
\n$$
\n\\text E\\left(\\text X^{2} \\mid \\text Y= \\text y\\right)=\\sum_{\\text x} \\text x^{2} \\cdot \\text g(\\text x \\mid \\text Y=\\text y)
\n$$
\nand
\n$$
\n\\text{E}(\\text{X} \\mid \\text{Y}=\\text{y})=\\sum_{\\text{x}} \\text{x} \\cdot \\text{g}(\\text{x} \\mid \\text{Y}=\\text{y})
\n$$
\nNote that this is analogous to the variance of a single random variable.<\/p>\n
We know that standard deviation is defined simply as the square root of variance.<\/p>\n
This implies that the conditional standard deviation of \\(\\text{X}\\), given that \\(\\text{Y}=\\text{y}\\) is defined by:
\n$$
\n\\sigma_{\\text{X} \\mid \\text{Y}=\\text{y}}=\\sqrt{\\text{Var}(\\text{X} \\mid \\text{Y}=\\text{y})}
\n$$
\nNow, the conditional variance of \\(\\text{Y}\\) given \\(\\text{X}=\\text{x}\\) is defined by:
\n$$
\n\\text{Var}(\\text{Y} \\mid \\text{X}=\\text{x})=\\text{E}\\left(\\text{Y}^{2} \\mid \\text{X}=\\text{x}\\right)-[\\text{E}(\\text{Y} \\mid \\text{X}=\\text{x})]^{2}
\n$$
\nWhere
\n$$
\n\\text{E}\\left(\\text {Y}^{2} \\mid \\text{X}=\\text{x}\\right)=\\sum_{\\text{y}} \\text{y}^{2} * \\text h(\\text{y}\\mid \\text {X}=\\text {x})
\n$$
\nand
\n$$
\n\\text{E}(\\text{Y} \\mid \\text{X}=\\text{x})=\\sum_{\\text{x}} \\text{y} \\cdot \\text{h}(\\text{y} \\mid \\text{X}=\\text{x})
\n$$<\/p>\n
The number of days of hospitalization for two individuals, \\(\\text{P}\\), and \\(Q\\), is jointly distributed as:<\/p>\n $$ Since we wish to find \\(\\operatorname{Var}(\\text{X} \\mid \\text{Y}=1)\\), we first need to determine the conditional distribution of \\(\\mathrm{X}\\) given \\(\\mathrm{Y}=1\\), namely, $$ $$ Using the output from Example 1 above, find \\(\\text{SD}(\\text{X} \\mid \\text{Y}=1)\\)<\/p>\n From Example 1, we have Recall that if \\(\\text{X}\\) and \\(\\text{Y}\\) are discrete random variables with joint probability mass function \\(\\text{f}(\\text{x}, \\text{y})\\) defined on the space \\(S\\), then the marginal distribution functions of \\(X\\) and \\(Y\\) are given by:<\/p>\n $$ The variance for the random variable \\(\\text{X}\\) is given by:<\/p>\n $$ $$ Let \\(\\text{X}\\) and \\(\\text{Y}\\) be the number of days of sickness for two individuals, \\(\\text{A}\\) and \\(\\text{B}\\). We know that, $$ Therefore, A laptop dealer specializes in two brands of laptops, HP and Lenovo. Let \\({X}\\) be the number of HP laptops sold in a day, and let \\(Y\\) be the number of Lenovo laptops sold in a day. The dealer has determined that the number of mobile phones sold in a day is jointly distributed as in the table below: Calculate the variance of \\(\\text{X}\\)<\/p>\n We need to find the marginal distribution function of \\(X\\) first:<\/p>\n We know that, We can calculate the variance and the standard deviation of \\(\\text{Y}\\) in a similar manner.<\/p>\n Let \\(\\alpha\\) and \\(\\beta\\) be non-zero constants. Then, it can be proven that: Learning Outcome<\/strong><\/em><\/p>\n Topic 3. d: Multivariate Random Variables – Calculate variance and standard deviation for conditional and marginal http:\/\/www.gulfportpharmacy.com\/<\/a> probability distributions for discrete random variables only.<\/strong><\/em><\/p>\n
\n Want to practice growth theory questions with full CFA-style solutions? Try AnalystPrep\u2019s free trial.
\n<\/a><\/p>\nExample 1: Conditional Variance for Discrete Random Variables<\/h4>\n
\n\\text{f(x, y)}=\\frac{\\text{x}+\\text{y}}{21}, \\text {x=1,2,3} \\quad \\text{y=1,2}
\n$$
\nFind \\(\\operatorname{Var}(\\text{X} \\mid \\text{Y}=1)\\)<\/p>\nSolution<\/h4>\n
\n$$
\n\\text{g}(\\text{x} \\mid \\text{y})=\\frac{\\text{f(x, y)}}{\\text{f}_{\\text{Y}}(\\text{y})}
\n$$
\nWhere,
\n$$
\n\\begin{align}
\n\\text{f}_{\\text{Y}}(\\text{y})=& \\sum_{\\text {all } x} \\text{P}(\\text{x}, \\text{y})=\\text{P}(\\text{Y}=\\text{y}), \\quad \\text{y in S}_{\\text{y}} \\\\
\n\\Rightarrow & \\text{f}_{\\text{Y}}(\\text{y})=\\frac{(1)+\\text{y}}{21}+\\frac{(2)+\\text{y}}{21}+\\frac{(3)+\\text{y}}{21}=\\frac{6+3 \\text{y}}{21}\\end{align}
\n$$
\nTherefore, we have,
\n$$
\n\\text {g}(\\text{x} \\mid\\text{y} )=\\frac{\\text{f}\\text{(x, y)}}{\\text{f}_{\\text{Y}}(\\text{y})}=\\frac{\\frac{\\text{x}+\\text{y}}{21}}{\\frac{6+3 y}{21}}=\\frac{\\text{x}+\\text{y}}{\\text{3y}+6}
\n$$
\nWe can now go ahead to find the conditional variance:<\/p>\n
\n\\operatorname{Var}(\\text{X} \\mid \\text{Y}=\\text{y})=\\sigma_{\\text{X} \\mid \\text{Y}=\\text{y}}^{2}=\\text{E}\\left(\\text{X}^{2} \\mid \\text{Y}=\\text{y}\\right)-[\\text{E}(\\text{X} \\mid \\text{Y}=\\text{y})]^{2}
\n$$
\nWe need:<\/p>\n
\n\\operatorname{Var}(\\text{X} \\mid \\text{Y}=1)=\\text{E}\\left(\\text{X}^{2} \\mid \\text{Y}=1\\right)-[\\text{E}(\\text{X} \\mid \\text{Y}=1)]^{2}
\n$$
\nNow,
\n$$
\n\\begin{align}
\n\\text{E}(\\text{X} \\mid \\text{Y}&=1)=\\sum_{\\text{x}=1}^{3} \\text{x} \\cdot \\text{g}(\\text{x} \\mid \\text{y}=1) \\\\
\n&=\\sum_{\\text{x}=1}^{3} \\text{x} \\frac{(\\text{x}+(1))}{3(1)+6} \\\\
\n&=(1) \\frac{(1+(1))}{3(1)+6}+2 \\frac{(2+(1))}{3(1)+6}+3 \\frac{(3+(1))}{3(1)+6} \\\\ &=1\\left(\\frac{2}{9}\\right)+2\\left(\\frac{1}{3}\\right)+3\\left(\\frac{4}{9}\\right)=\\frac{20}{9} \\end{align}
\n$$
\nWe also need,
\n$$
\n\\begin{align}\\text{E}\\left(\\text{X}^{2} \\mid \\text{Y}=\\text{y}\\right)&=\\sum_{\\text{x}=1}^{3} \\text{x}^2*\\text{g}(\\text{x} \\mid \\text{y}=1) \\\\&=\\sum_{\\text{x}=1}^{3} \\text{x}^{2} \\frac{(\\text{x}+(1))}{3(1)+6} \\\\&
\n=\\left(1^{2}\\right) \\frac{(1+(1))}{3(1)+6}+2^{2} \\frac{(2+(1))}{3(1)+6}+3^{2} \\frac{(3+}{3(1)} \\\\&
\n=1\\left(\\frac{2}{9}\\right)+4\\left(\\frac{1}{3}\\right)+9\\left(\\frac{4}{9}\\right)=\\frac{50}{9} \\\\&
\n\\Rightarrow \\operatorname{Var}(\\text{X} \\mid \\text{Y}=1)=\\text{E}\\left(\\text{X}^{2} \\mid \\text{Y}=1\\right)-[ \\text{E}(\\text{X} \\mid \\text{Y}=1)]^{2} \\\\&
\n=\\frac{50}{9}-\\left(\\frac{20}{9}\\right)^{2}=\\frac{50}{81}
\n\\end{align}$$<\/p>\nExample 2: Conditional Standard Deviation for Discrete Random Variables<\/h4>\n
Solution<\/h4>\n
\n$$
\n\\text{V}(\\text{X} \\mid \\text{Y}=1)=\\frac{50}{81}
\n$$
\nThus,
\n$$
\n\\text{SD}(\\text{X} \\mid \\text{Y}=1)=\\sqrt{\\text{V}(\\text{X} \\mid \\text{Y}=1)}=\\sqrt{\\frac{50}{81}}=0.7856
\n$$<\/p>\nVariance and Standard Deviation for Marginal Discrete Distributions<\/h2>\n
\n\\text{f}_{\\text{X}}(\\text{x})=\\sum_{\\text{y}} \\text{f}(\\text{x}, \\text{y})=\\text{P}(\\text{X}=\\text{x}), \\quad \\text{x} \\in \\text{S}_{\\text{x}}
\n$$
\nand,
\n$$
\n\\text{f}_{\\text{Y}}(\\text{y})=\\sum_{\\text{x}} \\text{f(x, y)}=\\text{P}(\\text{Y}=\\text{y}), \\quad \\text{y in S}_{\\text{y}}
\n$$
\nOnce we have the marginal distribution functions of \\(X\\) and \\(Y\\), we can now go ahead to find individual variances and standard deviations for \\(X\\) and \\(Y\\).<\/p>\n
\n\\operatorname{Var}(\\text{X})=\\text{E}\\left(\\text{X}^{2}\\right)-[\\text{E}(\\text{X})]^{2}
\n$$
\nWhere
\n$$
\n\\text{E}\\left(\\text{X}^{2}\\right)=\\sum_{\\text{x}} \\text{x}^{2} * \\text{P}(\\text{X}=\\text{x})
\n$$
\nand
\n$$
\n\\text{E}(\\text{X})=\\sum_{\\text{X}} \\text{x} * \\text{P}(\\text{X}=\\text{x})
\n$$
\nSimilarly, the variance of the random variable \\(Y\\) is given by:<\/p>\n
\n\\operatorname{Var}(\\text{Y})=\\text{E}\\left(\\text{Y}^{2}\\right)-[\\text{E}(\\text{Y})]^{2}
\n$$
\nWhere
\n$$
\n\\text{E}\\left(\\text{Y}^{2}\\right)=\\sum_{\\text{x}} \\text{y}^{2} * \\text{P}(\\text{Y}=\\text{y})
\n$$
\nand
\n$$
\n\\text{E(Y)}=\\sum_{\\text{x}} \\text{y} * \\text{P(Y=y)}
\n$$
\nThe standard deviation for \\(\\text{X}\\) and \\(\\text{Y}\\) is the square root of their respective variances.
\n$$
\n\\text{SD}(\\text{X})=\\sqrt{\\text{E}\\left(\\text{X}^{2}\\right)-[\\text{E}(\\text{X})]^{2}}
\n$$
\nand,
\n$$
\n\\text{SD}(\\text{Y})=\\sqrt{\\text{E}\\left(\\text{Y}^{2}\\right)-[\\text{E}(\\text{Y})]^{2}}
\n$$<\/p>\nExample 1: Variance and Standard Deviation for Marginal Discrete Random Variables<\/h4>\n
\n$$
\n\\text{f(x, y)}=\\frac{\\text{x}+\\text{y}}{21}, \\text{x=1,2,3} \\quad \\text{y=1,2}
\n$$
\nCalculate the variance and the standard deviation of \\(X\\).<\/p>\nSolution<\/h4>\n
\n$$
\n\\operatorname{Var}(\\text{X})=\\text{E}\\left(\\text{X}^{2}\\right)-[\\text{E}(\\text{X})]^{2}
\n$$
\nFirst, we find the marginal probability mass function of \\(\\text{X}\\), which is given by:<\/p>\n
\n\\begin{align}
\n\\text{f}_{\\text{X}}(\\text{x}) &=\\sum_{\\text {all } \\text{y}} \\text{f}(\\text{x}, \\text{y}) \\text{x in S}_{\\text{x}} \\\\
\n&=\\frac{\\text{x}+(1)}{21}+\\frac{\\text{x}+(2)}{21}=\\frac{2 \\text{x}+3}{21}, \\quad \\text { for } \\text{x}=1,2,3
\n\\end{align}
\n$$
\nThen,
\n$$\\begin{align}
\n\\text{E(X)} &=\\sum_{\\text{x}=1}^{3}\\text{x} \\text{P}_{\\text{X}}(\\text{x})=\\sum_{\\text{x}=1}^{3} \\text{x} \\frac{2 x+3}{21} \\\\
\n&=(1) \\frac{2(1)+3}{21}+(2) \\frac{2(2)+3}{21}+(3) \\frac{2(3)+3}{21}\\\\&=1\\left(\\frac{5}{21}\\right)+2\\left(\\frac{7}{21}\\right)+3\\left(\\frac{9}{21}\\right)=\\frac{46}{21}
\n\\end{align}
\n$$
\nand
\n$$
\n\\begin{align}
\n\\text{E}\\left(\\text{X}^{2}\\right) &=\\sum_{\\text {all }\\text{x}} \\text{x}^{2} \\text{P}_{\\text{X}}(\\text{x}) \\\\
\n&=(1)^{2}\\left(\\frac{5}{21}\\right)+(2)^{2}\\left(\\frac{7}{21}\\right)+(3)^{2}\\left(\\frac{9}{21}\\right)-\\left(\\frac{46}{21}\\right)^{2}=\\frac{38}{7}
\n\\end{align}
\n$$
\nThus,
\n$$
\n\\operatorname{Var(X)}=\\frac{38}{7}-\\left(\\frac{46}{21}\\right)^{2}=\\frac{278}{441} \\approx 0.6304
\n$$
\nWe know that the standard deviation of \\(\\text{X}\\) is the square root of its variance.<\/p>\n
\n$$
\n\\sigma_{\\text{X}}=\\sqrt{\\text{Var}(\\text{X})}=\\sqrt{0.6304}=0.7940
\n$$<\/p>\nExample 2: Variance and Standard Deviation for Marginal Discrete Random Variables<\/h4>\n
\n$$ \\begin{array}{c|c|c|c} {\\quad \\text X }& {1} & {2} & {3} \\\\ {\\Huge \\diagdown } & & & \\\\ {\\text Y \\quad} & & & \\\\ \\hline {0} & {\\cfrac{1}{6}} & {\\cfrac{1}{8}} & {\\cfrac{1}{6}} \\\\ \\hline {1} & {\\cfrac{1}{3}} & {\\cfrac{1}{12}} & {\\cfrac{1}{8}} \\\\ \\end{array} $$<\/p>\nSolution<\/h4>\n
\n$$
\n\\text{f}_{\\text{X}}(\\text{x})=\\sum_{\\text{y}} \\text{f(x, y)}=\\text{P(X=x)}, \\quad \\text{x in S}_{\\text{x}}
\n$$
\nNow,
\n$$
\n\\
\n$$
\n$$
\n\\begin{align}&\\text{P(X=1)}=\\frac{1}{6}+\\frac{1}{3}=\\frac{1}{2}\\\\ \\\\
\n&\\text{P}(\\text{X}=2)=\\frac{1}{8}+\\frac{1}{12}=\\frac{5}{24} \\\\ \\\\
\n&\\text{P}(\\text{X}=3)=\\frac{1}{6}+\\frac{1}{8}=\\frac{7}{24}
\n\\end{align}
\n$$
\nTherefore,
\n$$
\n\\text{E}(\\text{X})=\\sum_{\\text{x}=1}^{3} x P_{\\text{X}}(\\text{x})=1 * \\frac{1}{2}+2 * \\frac{5}{24}+3 * \\frac{7}{24}=\\frac{43}{24}
\n$$
\nand,
\n$$
\n\\text{E}\\left(\\text{X}^{2}\\right)=\\sum_{\\text{x}=1}^{3} \\text{x}^{2}(\\text{x})=1^{2} * \\frac{1}{2}+2^{2} * \\frac{5}{24}+3^{2} * \\frac{7}{24}=\\frac{95}{24}
\n$$
\nThus,
\n$$
\n\\begin{aligned}
\n\\operatorname{Var}(\\text{X}) &=\\text{E}\\left(\\text{X}^{2}\\right)-[\\text{E}(\\text{X})]^{2} \\\\
\n&=\\frac{95}{24}-\\left(\\frac{43}{95}\\right)^{2}=3.992
\n\\end{aligned}
\n$$
\nNote:<\/p>\nExam tips:<\/h4>\n
\ni) \\(\\operatorname{Var}(\\alpha)=0\\)
\nii) \\(\\operatorname{Var}(\\alpha \\text{X})=\\alpha^{2} * \\operatorname{Var}(\\text{X})\\)
\niii) \\(\\operatorname{Var}(\\alpha \\text{X}+\\beta)=\\alpha^{2} * \\operatorname{Var}(\\text{X})\\)<\/p>\n