The \\(n\\)-th moment about the origin of a random variable is the expected value of its \\(n\\)-th power.<\/p>\n
In this reading, however, we will mostly look at moments about the mean, also called central moments. The \\(n\\)-th central moment of a random variable \\(X\\) is the expected value of the \\(n\\)-th power of the deviation of \\(X\\) from its mean.<\/p>\n
The first and the second central moments are the mean and variance, respectively.<\/p>\n
Let \\(\\text{X}\\) and \\(\\text{Y}\\) be two discrete random variables with joint probability function, \\(\\text{f}(\\text{x}, \\text{y})\\) defined on the space \\(\\text{S}\\). If there exists a function, \\(\\text{g}(\\text{x}, \\text{y})\\) of these two random variables, then the expected value(mean) of \\(g(x, y)\\) is given by:<\/p>\n
$$
\nE[g(x, y)]=\\sum_{(x, y) \\in S} g(x, y) f(x, y)
\n$$
\nThe second moment(variance) of \\(\\text{g}(\\text{x}, \\text{y})\\) is defined as:<\/p>\n
$$
\n\\text{Var}(\\text{X}, \\text{Y})=\\text{E}\\left(\\text{g}\\left(\\text{x}^{2}, \\text{y}^{2}\\right)\\right)-\\text{E}[\\text{g}(\\text{x}, \\text{y})]^{2}
\n$$<\/p>\n
An actuary determines that the annual number of claims from two policyholders, \\(\\text{A}\\) and \\(\\text{B}\\), is jointly distributed as:<\/p>\n
$$
\nf(x, y)=\\frac{x^{2}+3 y}{96} \\quad x=1,2,3,4 \\quad y=1,2
\n$$
\nLet \\(X\\) be the number of annual claims from policyholder \\(\\text{A}\\) and \\(Y\\) be the number of annual claims from policyholder \\(B\\).<\/p>\n
Find \\(\\text{E}(\\text{XY})\\)<\/p>\n
The possible values for this distribution are:
\n$$
\n(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(4,1),(4,2)
\n$$
\nNow,
\n$$
\n\\begin{align}
\n\\text{E}[\\text{XY}] &=\\sum_{(\\text{x}, \\text{y}) \\in \\text{S}} \\text{g}(\\text{x}, \\text{y}) \\text{f}(\\text{x}, \\text{y})=\\sum_{(\\text{x}, \\text{y}) \\in \\text{S}}(\\text{xy}) \\frac{\\text{x}^{2}+3 \\text{y}}{96} \\\\
\n&=(1) \\frac{1+3}{96}+(2) \\frac{1+6}{96}+(2) \\frac{4+3}{96}+(4) \\frac{4+6}{96}\\\\&+(3) \\frac{9+3}{96}+(6) \\frac{9+6}{96} +(4) \\frac{16+3}{96}+(8) \\frac{16+6}{96}\\\\ &=\\frac{4}{96}+\\frac{14}{96}+\\frac{14}{96}+\\frac{40}{96}+\\frac{36}{96}+\\frac{90}{96}+\\frac{76}{96}+\\frac{176}{96} \\\\
\n&=\\frac{75}{16}=4 \\frac{11}{16}
\n\\end{align}
\n$$<\/p>\n
In the previous reading, we defined the conditional distributions for two discrete random variables, \\(X\\) and \\(Y\\), with a joint pmf \\(f(x, y)\\) and marginal functions \\(f_{x}(x)\\) and \\(f_{y}(y)\\) as:<\/p>\n
$$
\n\\text{g}(\\text{x} \\mid \\text{y})=\\frac{\\text{f}(\\text{x}, \\text{y})}{\\text{f}_{\\text{Y}}(\\text{y})} \\quad \\text { provided that } \\text{f}_{\\text{Y}}(\\text{y})>0
\n$$
\nand,
\n$$
\n\\text{h}(\\text{y} \\mid \\text{x})=\\frac{\\text{f}(\\text{x}, \\text{y})}{\\text{f}_{\\text{X}}(\\text{x})} \\quad \\text { provided that } \\text{f}_{\\text{X}}(\\text{x})>0
\n$$
\nWe may wish to find the mean and variance of the above conditional distributions.<\/p>\n
The conditional mean of \\(\\text{X}\\), given that \\(\\text{Y}=\\text{y}\\) is given by:<\/p>\n
$$
\n\\mu_{\\text{x} \\mid \\text{y}}=\\text{E}[\\text{X} \\mid \\text{Y}]=\\sum_{\\text{x}} \\text{x} * \\text{g}(\\text{x} \\mid \\text{y})
\n$$
\nAnd the conditional variance of \\(X\\), given that \\(Y=y\\) is given by:<\/p>\n
$$
\n\\sigma_{\\text{X} \\mid \\text{Y}}=\\text{E}(\\text{X}-\\text{E}[\\text{X} \\mid \\text{Y}])^{2}=\\sum_{\\text{X}}(\\text{X}-\\text{E}[\\text{X} \\mid \\text{y}])^{2} * \\text{g}(\\text{x} \\mid \\text{y})
\n$$
\nThis is simplified to:<\/p>\n
$$
\n\\sigma_{\\text{X} \\mid \\text{Y}}=\\text{E}\\left[\\text{X}^{2} \\mid \\text{y}\\right]-(\\text{E}[\\text{X} \\mid \\text{y}])^{2}
\n$$
\nSimilarly, the conditional mean of \\(Y\\), given that \\(X=x\\) is given by:<\/p>\n
$$
\n\\mu_{\\text{Y} \\mid \\text{X}}=\\text{E}[\\text{Y} \\mid \\text{x}]=\\sum_{\\text{y}} \\text{y} * \\text{h}(\\text{y} \\mid \\text{x})
\n$$
\nAnd the conditional variance of \\(Y\\), given that \\(X=x\\) is given by:<\/p>\n
$$
\n\\sigma_{\\text{Y} \\mid \\text{x}}=\\text{E}(\\text{Y}-\\text{E}[\\text{Y} \\mid \\text{x}])^{2}=\\sum_{\\text{y}}(\\text{Y}-\\text{E}[\\text{Y} \\mid \\text{x}])^{2} * \\text{h}(\\text{y} \\mid \\text{x})
\n$$
\nThis is simplified to:<\/p>\n
$$
\n\\sigma_{\\text{Y} \\mid \\text{x}}=\\text{E}\\left[\\text{Y}^{2} \\mid \\text{x}\\right]-(\\text{E}[\\text{Y} \\mid \\text{X}])^{2}
\n$$<\/p>\n
Let \\(X\\) be the number of annual tornadoes in country \\(M\\) and let \\(Y\\) be the number of annual tornadoes in country \\(N\\). \\(X\\) and \\(Y\\) have the following probability mass function:<\/p>\n
$$
\n\\text{f}(\\text{x, y})=\\frac{x^{2}+3 y}{96} \\quad \\text{x}=1,2,3,4, \\quad \\text{y}=1,2
\n$$
\nCalculate<\/p>\n
From the joint function, we can get the following marginal distribution functions:
\n$$
\nf_{X}(x)=\\frac{2 x^{2}+9}{96}, x=1,2,3,4
\n$$
\nand,
\n$$
\n\\text{f}_{\\text{Y}}(\\text{y})=\\frac{12 \\text y+30}{96},\\text{y}=1,2
\n$$
\nWe can also find conditional probability mass functions:
\n$$\\begin{align}
\n&\\text{g}(\\text{x}\\mid \\text{y})=\\frac{x^{2}+3 y}{12y+30};\\quad \\text{and}\\\\\\\\ &h(y \\mid x)=\\frac{x^{2}+3 y}{2 x^{2}+9} \\end{align}
\n$$
\nSo,
\na)
\n$$
\n\\begin{align}
\n\\text{E}(\\text{X} \\mid \\text{Y}=1) &=\\text{E}(\\text{g}(\\text{x} \\mid \\text{y}=1))=\\sum_{\\text{x}=1}^{4} \\text{x} * \\text{g}(\\text{x} \\mid \\text{y}=1)=\\sum_{\\text{x}=1}^{4} \\text{x} * \\frac{\\text{x}^{2}+3(1)}{12(1)+30} \\\\
\n&=(1) \\frac{1^{2}+3(1)}{12(1)+30}+(2) \\frac{2^{2}+3(1)}{12(1)+30}+(3) \\frac{3^{2}+3(1)}{12(1)+30}+(4) \\frac{4^{2}+3(1)}{12(1)+30} \\\\
\n&=(1) \\frac{4}{42}+(2) \\frac{7}{42}+\\text { (3) } \\frac{12}{42}+(4) \\frac{19}{42}=\\frac{65}{21}=3.10
\n\\end{align}
\n$$
\nb)
\n$$
\n\\begin{align}
\n\\text{E}(\\text{Y} \\mid \\text{X}=3) &=\\text{E}(\\text{h}(\\text{y} \\mid \\text{x}=3))=\\sum_{\\text{y}=1}^{2} \\text{y} * \\text{h}(\\text{y} \\mid \\text{x}=3)=\\sum_{\\text{y}=1}^{2} \\text{y} * \\frac{3^{2}+3 \\text{y}}{2(3)^{2}+9} \\\\
\n&=(1) \\frac{3^{2}+3(1)}{2(3)^{2}+9}+(2) \\frac{3^{2}+3(2)}{2(3)^{2}+9} \\\\
\n&=\\frac{12}{27}+(2) \\frac{15}{27}=\\frac{14}{9}=1.56
\n\\end{align}
\n$$
\nc) Using the output from (a), we have:<\/p>\n
$$
\n\\begin{align} \\text{V}\\left(\\text{X} \\mid \\text{Y}=1\\right)&=\\text{E}\\left[\\text{X}-\\text{E}(\\text{X} \\mid \\text{Y}=1))^{2} \\mid \\text{Y}=1\\right]\\\\
\n&=\\sum_{\\text{x}=1}^{4}(\\text{x}-\\text{E}(\\text{X} \\mid \\text{Y}=1))^{2} \\text{g}(\\text{x} \\mid \\text{y}=1) \\\\
\n&=\\sum_{\\text{x}=1}^{4}\\left(\\text{x}-\\frac{65}{21}\\right)^{2} \\frac{\\text{x}^{2}+3(1)}{12(1)+30} \\\\
\n&=\\left(1-\\frac{65}{21}\\right)^{2} \\frac{4}{42}+\\left(2-\\frac{65}{21}\\right)^{2} \\frac{7}{42}+\\left(3-\\frac{65}{21}\\right)^{2} \\frac{12}{42}+\\left(4-\\frac{65}{21}\\right)^{2} \\frac{19}{42} \\\\
\n&=\\left(\\frac{1936}{441}\\right) \\frac{4}{42}+\\left(\\frac{529}{441}\\right) \\frac{7}{42}+\\left(\\frac{4}{441}\\right) \\frac{12}{42}+\\left(\\frac{361}{441}\\right) \\frac{19}{42}=0.9909
\n\\end{align}
\n$$<\/p>\n
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
$$
\n\\text{f}_{\\text{x}}(\\text{x})=\\sum_{\\text{y}} \\text f(\\text x, \\text y)=\\text P(\\text X=\\text x), \\quad \\text{x in S}_{\\text x}
\n$$
\nand,
\n$$
\n\\text{f}_{\\text{y}}(\\text{y})=\\sum_{\\text{x}} \\text{f}(\\text{x}, \\text{y})=\\text{P}(\\text{Y}=\\text{y}), \\quad \\text{y} \\in \\text{S}_{\\text{y}}
\n$$
\nOnce we have the marginal distribution functions of \\(X\\) and \\(Y\\), we can now go ahead to find the moments for \\(\\text{X}\\) and \\(\\text{Y}\\) separately.<\/p>\n
Let \\(\\text{X}\\) and \\(\\text{Y}\\) be the number of days of hospitalization for two patients, \\(P\\), and \\(Q\\), respectively.
\n$$
\n\\text f(\\text x, \\text y)=\\frac{\\text x^{2}+3 \\text y}{96} \\quad \\text x=1,2,3,4; \\quad \\text y=1,2
\n$$
\nCalculate:<\/p>\n
a) the expected number of days of hospitalization for patient \\(\\text P\\).<\/p>\n
b) the expected number of days of hospitalization for patient \\(\\text Q\\).<\/p>\n
a) To calculate the expected number of days of hospitalization for patient \\(\\text{P}\\), we need to find the marginal distribution function of \\(\\text{X}\\) first:
\nWe know that,
\n$$
\n\\text f_{\\text x}(\\text x)=\\sum_{\\text y} \\text{f(x, y)}=\\text P(\\text X=\\text x), \\quad \\text{x in S}_{\\text {x}}
\n$$
\n$$
\n\\begin{gather}
\n=\\frac{\\text{x}^{2}+3(1)}{96}+\\frac{\\text{x}^{2}+3(2)}{96}=\\frac{\\text{x}^{2}+\\text{x}^{2}+3+6}{96} \\\\
\n\\therefore \\text{f}_{\\text{X}}(\\text{x})=\\frac{2 \\text{x}^{2}+9}{96}
\n\\end{gather}
\n$$
\nTherefore,
\n$$
\n\\begin{align}
\n\\text{E}(\\text{X}) &=\\sum_{\\text{x}=1}^{\\text{n}} \\text{xf}_ \\text{x}(\\text{x})=\\sum_{\\text{x}=1}^{4} \\text{xf} _{\\text{x}}(\\text{x})=\\sum_{\\text{x}=1}^{4} \\text{x} \\frac{2 \\text{x}^{2}+9}{96} \\\\
\n&=(1) \\frac{2(1)^{2}+9}{96}+(2) \\frac{2(2)^{2}+9}{96}+(3) \\frac{2(3)^{2}+9}{96}+(4) \\frac{2(4)^{2}+9}{96} \\\\
\n&=(1) \\frac{11}{96}+(2) \\frac{17}{96}+\\text { (3) } \\frac{27}{96}+(4) \\frac{41}{96}=\\frac{145}{48}=3.02
\n\\end{align}
\n$$
\nb) Similarly, to calculate the expected number of days of hospitalization for patient \\(Q\\), we need to find the marginal probability function of \\(Y\\) first:<\/p>\n
$$
\n\\begin{align}
\n\\text{f}_{\\text{y}}(\\text{y}) &=\\sum_{\\text{x}} \\text{f}(\\text{x}, \\text{y})=\\text{P}(\\text{Y}=\\text{y}), \\quad \\text{y} \\in \\text{S}_{\\text{y}} \\\\
\n&=\\frac{1+3 \\text{y}}{96}+\\frac{4+3 \\text{y}}{96}+\\frac{9+3 \\text{y}}{96}+\\frac{16+3 \\text{y}}{96} \\\\
\n&=\\frac{12 \\text{y}+30}{96}, \\text { for } \\text{y}=1,2
\n\\end{align}
\n$$
\nTherefore,
\n$$
\n\\begin{align}
\n\\text{E}(\\text{Y}) &=\\sum_{\\text{y}=1}^{\\text{n}} \\text{yf}(\\text{y})=\\sum_{\\text{y}=1}^{2} \\text{yf}(\\text{y})=\\sum_{\\text{y}=1}^{2} \\text{y} \\frac{12 \\text{y}+30}{96} \\\\
\n&=(1) \\frac{12(1)+30}{96}+(2) \\frac{12(2)+30}{96} \\\\
\n&=(1) \\frac{42}{96}+(2) \\frac{54}{96}=\\frac{25}{16}
\n\\end{align}
\n$$
\nWe can also proceed to find the variance for the corresponding variables:<\/p>\n
For \\(X\\), we know that:<\/p>\n
$$
\n\\text{V(X)}=\\text E\\left(\\text X^{2}\\right)-\\left[\\text{E(X)}\\right]^{2}
\n$$
\nTherefore, we need to find \\(\\text{E}\\left(\\text{X}^{2}\\right)\\) and \\(\\text{E}(\\text{X})\\). Continuing with the example above,<\/p>\n
$$
\n\\begin{align}
\n\\text{Var}(\\text{X}) &=\\sum_{\\text{x}=1}^{4} \\text{x}^{2} \\mathrm{f}_{\\text{x}}(\\text{x})-[\\text{E}(\\text{X})]^{2} \\\\
\n&=\\sum_{\\text{x}=1}^{4} \\text{x}^{2} \\frac{2 \\text{x}^{2}+9}{96}-\\left(\\frac{145}{48}\\right)^{2}\\\\& =(1)^{2} \\frac{11}{96}+(2)^{2} \\frac{17}{96}+(3)^{2} \\frac{27}{96}+(4)^{2} \\frac{41}{96}-\\left(\\frac{145}{48}\\right)^{2}=\\frac{163}{16}-\\left(\\frac{145}{48}\\right)^{2}=1.062
\n\\end{align}
\n$$
\nSimilarly, for \\(\\text{Y}\\) :
\n$$
\n\\begin{align}
\n\\text{Var}(\\text{Y}) &=\\sum_{\\text{y}=1}^{2} \\text{y}^{2} \\text{f}_{\\text{y}}(\\text{y})-[\\text{E}(\\text{Y})]^{2} \\\\
\n&=\\sum_{\\text{y}=1}^{2} \\text{y}^{2} \\frac{12 \\text{y}+30}{96}-\\left(\\frac{25}{16}\\right)^{2} \\\\
\n&=(1)^{2} \\frac{42}{96}+(2)^{2} \\frac{54}{96}-\\frac{625}{256}=\\frac{43}{16}-\\frac{625}{256}=\\frac{63}{256}
\n\\end{align}
\n$$<\/p>\n