Variance<\/strong><\/li>\n<\/ul>\nMoments about the mean describe the shape of the probability function of a random variable.<\/p>\n
Properties of Expectation<\/h3>\n
Recall that the expected value of a random variable \\(X\\) is defined by<\/p>\n
$$ E[X] = \\sum_{x} {xp(x)} $$<\/p>\n
where \\(X\\) is a discrete random variable with probability mass function \\(p(x)\\), and by<\/p>\n
$$ E[X] = \\int_{-\\infty}^{\\infty} xf(x)dx $$<\/p>\n
when \\(X\\) is a continuous random variable with probability density function \\(f(x)\\). Since \\(E[X]\\) is a weighted average of the possible values of \\(X\\), it follows that if \\(X\\) must lie between a and b, then so must its expected value, i.e,.<\/p>\n
If<\/p>\n
$$ P(a\\leq X \\leq b) = 1 $$<\/p>\n
Then<\/p>\n
$$ a \\leq E[X] \\leq b $$<\/p>\n
Expectations of a Sum<\/h3>\n
It follows that,<\/p>\n
\\(E [ag(X ) + bh(Y )] = aE [g(X )] + bE [h(Y )]\\)<\/p>\n
where a<\/em> and b<\/em> are constants.<\/p>\nWhat this equation tells us is that\u00a0handling the expected value of a linear combination of functions is no more difficult than handling the expected values of the individual functions.<\/p>\n
This handling also extends to situations where we have more than to variables. Expected values can easily be found from marginal distributions.<\/p>\n
Example: Expectation of a sum<\/h4>\n
You have been given the following joint pmf. Verify that \\(E[{ X }^{ 2 }+3Y]=E[{ X }^{ 2 }]+E[3Y]\\)<\/p>\n
$$ \\begin{array}{c|c|c|c|c} {\\begin{matrix} X \\\\ \\huge{\\diagdown} \\\\ Y \\end{matrix}} & {0} & {1} &{2} \\\\ \\hline {1} & {0.1} & {0.1} & {0} \\\\ \\hline {2} & {0.1} & {0.1} & {0.2} \\\\ \\hline {3} & {0.2} & {0.1} & {0.1} \\end{array} $$<\/p>\n<\/p>\n
Solution<\/strong><\/p>\nReading values from the table, we have:<\/p>\n
\\((E[{ X }^{ 2 }+3Y]=0.1({ 0 }^{ 2 }+3\\times 1)+0.1({ 0 }^{ 2 }+3\\times 2)+0.2({ 0 }^{ 2 }+3\\times 3)\\)<\/p>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \\(+0.1({ 1 }^{ 2 }+3\\times 1)+0.1({ 1 }^{ 2 }+3\\times 2)+0.1({ 1 }^{ 2 }+3\\times 3)\\)<\/p>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \\(+0({ 2 }^{ 2 }+3\\times 1)+0.2({ 2 }^{ 2 }+3\\times 2)+0.1({ 2 }^{ 2 }+3\\times 3)=8.1\\)<\/p>\n
Looking at the terms on the right side above,<\/p>\n
\\(E[{ X }^{ 2 }]={ 0 }^{ 2 }\\times 0.4+{ 1 }^{ 2 }\\times 0.3+{ 2 }^{ 2 }\\times 0.3=1.5\\)<\/p>\n
\\(E[3Y]=(3\\times 1)0.2+(3\\times 2)0.4+(3\\times 3)0.4=6.6\\)<\/p>\n
Thus, \\(E[{ X }^{ 2 }]+E[3Y]=8.1\\), and the reasult has been verified.<\/p>\n
Moments of Joint Random Variables<\/h2>\nDiscrete Case<\/strong><\/h3>\nLet \\(X,Y\\) be a pair of joint random variables with a joint probability function \\(f(x,y)\\) on the space \\(S\\). If there exists a function of these two namely \\(g(X,Y)\\) defined:<\/p>\n
$$ E[g(X,Y)] = \\sum_{(x,y) \\in S} g(x,y)f(x,y) $$<\/p>\n
Then this function is called the mathematical expectation (or expected value)<\/strong> of \\(g(X,Y)\\).<\/p>\nThis mathematical expectation is known as the first moment of joint random variables, or mean.<\/p>\n
The second moment is a derivative of the first moment and it is equal to:<\/p>\n
$$ E[g(X,Y)]= E(g(X^2,Y^2)) – (E[g(X,Y)])^2 = Var(X,Y) $$<\/p>\n
Example 1: Moments of Joint Random Variables<\/strong><\/h4>\nLet \\(X\\) and \\(Y\\) have the following pmf:<\/p>\n
$$ f(x,y) = \\frac{x^2 + 3y}{60} \\qquad x = 1,2,3,4\\quad y=1,2. $$<\/p>\n
Find the expected value with \\(g(X,Y) = XY\\)
\nSolution<\/strong>
\nThe possible values for this distribution are:<\/p>\n$$(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(4,1),(4,2)$$<\/p>\n
Then we proceed to calculate,<\/p>\n
$$\\begin{align} E[XY] & = \\sum_{(x,y) \\in S} g(x,y) f(x,y)\\\\ & = \\sum_{(x,y) \\in S} (xy) \\frac{x^2 + 3y}{60}\\\\ & = (1)\\frac{1+3}{60} + (2)\\frac{1+6}{60} + (2)\\frac{4+3}{60} + (4)\\frac{4+6}{60} + \\\\ & (3)\\frac{9+3}{60} + (6)\\frac{9+6}{60} + (4)\\frac{16+3}{60} + (8)\\frac{16+6}{60}\\\\ & = \\frac{4}{60} + \\frac{14}{60} + \\frac{14}{60} + \\frac{40}{60} + \\frac{36}{60} + \\frac{90}{60} + \\frac{76}{60} +\\frac{176}{60}\\\\ & = \\frac{15}{2}=7.5 \\end{align}$$<\/p>\n
This process is quite similar to calculating the mean of any mass function, univariate or multivariate.<\/p>\n
If \\(X\\) and \\(Y\\) are independent, then;<\/p>\n
\n- The expectation of the product of X and Y is the product of the individual expectations: \\(E(XY ) = E(X)E(Y )\\). This product formula holds for any expectation of a function \\(X\\) times a function of \\(Y\\)<\/li>\n
- The product formula holds for probabilities of the form P (some condition on X, some condition on Y) (where the comma denotes \u201cand\u201d): For instance, \\(P(X\\le 3, Y \\le 4)=P(X\\le 3)P(Y\\le 4)\\)<\/li>\n
- The variance of the sum of \\(X\\) and \\(Y\\) is the sum of the individual variances: \\(Var(X + Y ) = Var(X) + Var(Y )\\)<\/li>\n<\/ul>\n
Example 2:<\/strong> Moments of Joint Random Variables<\/strong><\/h4>\nLet X and Y have the following pmf:<\/p>\n
$$ f\\left( x, y\\right)=\\frac{ x^2+3 y}{96}\\ \\ \\ \\ \\ \\ \\ x=1,2,3,4\\ \\ \\ \\ y=1,2\\ $$<\/p>\n
Find the expected value of X and Y<\/p>\n
Solution<\/strong><\/p>\nTo find the expected value of X, we need to find the marginal probability mass function of X which is given by;<\/p>\n
$$\\begin{align} f_{ x}\\left( x\\right) &=\\sum_{ y}{ f\\left( x, y\\right)= P\\left( X= x\\right),\\ x\\epsilon S_{ x}} \\\\\u00a0 f_{ X}\\left( x\\right) & =\\frac{ x^2+3\\left(1\\right)}{96}+\\frac{ x^2+3(2)}{96} \\\\ &=\\frac{ x^2+3}{96}+\\frac{ x^2+6}{96} \\\\ &=\\frac{ x^2+ x^2+3+6 }{96} \\\\ \\therefore\\ f_X( x) & =\\frac{2 x^2+9}{96} \\end{align}$$<\/p>\n
Therefore,<\/p>\n
$$ \\begin{align*} E\\left( x\\right) & =\\sum_{ x=1}^{ n}{ {xf}_{ X}\\left(x\\right)\\ \\ \\ \\ \\ \\ \\ } \\\\ & =\\sum_{ x=1\\ }^{4}{ {xf}_{ x}( x)} \\\\ & =\\sum_{ x=1}^{4}{ x\\frac{2 x^2+9}{96}\\ } \\\\ & =\\left(1\\right)\\frac{2\\left(1\\right)^2+9}{60}+\\left(2\\right)\\frac{2\\left(2\\right)^2+9}{96}+\\left(3\\right)\\frac{2\\left(3\\right)^2+9}{96}+\\left(4\\right)\\frac{2\\left(4\\right)^2+9}{96} \\\\ & =\\left(1\\right)\\frac{11}{96}+\\left(2\\right)\\frac{17}{96}+\\left(3\\right)\\frac{27}{96}+\\left(4\\right)\\frac{41}{96}=\\frac{145}{48}=3.02\\ \\end{align*} $$<\/p>\n
Similarly, to find the expected value of Y, we need to find finding the marginal probability mass function of Y which is given by;<\/p>\n
$$ \\begin{align*} f_{ y}\\left( y\\right) & =\\sum_{ x}{ f\\left( x, y\\right)= P\\left( Y= y\\right),\\ \\ \\ y\\epsilon S_{ y}} \\\\ & =\\ \\frac{1+3 y}{96}+\\frac{4+3 y}{96}+\\frac{9+3 y}{96}+\\frac{16+3 y}{96} \\\\ & =\\frac{12 y+30}{96} \\\\ \\end{align*} $$<\/p>\n
Therefore,<\/p>\n
$$ \\begin{align*} E\\left( y\\right) & =\\sum_{ y=1}^{ n}{ {yf}_{ Y}\\left( y\\right)\\ } \\\\ & =\\sum_{ y=1}^{2}{ {yf}_{ Y}\\left( y\\right)} \\\\ & =\\sum_{ y=1}^{2}{ y\\frac{12 y+30}{96}} \\\\ & =\\left(1\\right)\\frac{12\\left(1\\right)+30}{96}+\\left(2\\right)\\frac{12\\left(2\\right)+30}{96}\\ \\\\ &=\\left(1\\right)\\frac{42}{96}+\\left(2\\right)\\frac{54}{96}=\\frac{25}{16}\\ \\end{align*} $$<\/p>\n
We can also proceed to find the variance for the corresponding variables: For X, we know that:<\/p>\n
$$ \\text{Variance}= V\\left( X\\right)= E\\left( X^2\\right)-\\left[ E\\left( X\\right)\\right]^2 $$<\/p>\n
Therefore, you need to find \\( E( X^2)\\) and \\( E( X)\\).<\/p>\n
Continuing with the example above,<\/p>\n
$$ \\begin{align*} {Var}\\left( X\\right) & =\\sum_{ x=1}^{4}{ x^2 f_{ x}\\left( x\\right)-\\left[ E\\left( X\\right)\\right]^2}\\\\ &=\\sum_{ x=1}^{4}{ x^2\\frac{2 x^2+9}{96}-\\left(\\frac{145}{48}\\right)^2} \\\\ & =\\left(1\\right)^2\\frac{11}{96}+\\left(2\\right)^2\\frac{17}{96}+\\left(3\\right)^2\\frac{27}{96}+\\left(4\\right)^2\\frac{41}{96}-\\left(\\frac{145}{96}\\right)^2=\\frac{163}{16}-\\left(\\frac{145}{48}\\right)^2=1.062\\ \\\\ \\text{ Similarly, for Y:}\\\\ {Var}\\left( Y\\right) & =\\sum_{ y=1}^{2}{ y^2{ f}_{ y}\\left( y\\right)-\\left[ E\\left( Y\\right)\\right]^2} \\\\ & =\\sum_{ y=1}^{2}{ y^2\\frac{12 y+30}{96}-\\left(\\frac{25}{16}\\right)^2} \\\\ & =\\left(1\\right)^2\\frac{42}{96}+\\left(2\\right)^2\\frac{54}{96}-\\frac{625}{256}=\\frac{43}{16}-\\frac{625}{256}=\\frac{63}{256}\\ \\end{align*} $$<\/p>\n
Continuous Case<\/strong><\/h3>\nExample 3: M<\/strong>oments of Joint Random Variables<\/strong><\/h5>\nLet \\(X\\) and \\(Y\\) have the following pmf:<\/p>\n
$$ f(x,y) = \\begin{cases}\u00a03(x+y), & 0 < x \\le y \\le 1\\\\ 0, &\\text{otherwise}\\\\ \\end{cases}\u00a0$$<\/p>\n
Find the expected value of \\(X\\) and \\(Y\\)<\/p>\n
Solution<\/strong><\/p>\nTo find the mean\/expectation of X, we need to find the marginal distribution of X. We know that:<\/p>\n
$$f_x\\left(x\\right)=\\int_{-\\infty}^{\\infty}{f\\left(x,y\\right)dy, \\ \\ x\\epsilon S_x}$$<\/p>\n
Then<\/p>\n
$$\\begin{align} f_X\\left(x\\right) &=\\int_{x}^{1}{\\left(3x+3y\\right)\\ dy}=\\left|3xy+\\frac{3y^2}{2}\\right|_x^1\\\\ &=\\left(3x+\\frac{3}{2}\\right)-\\left(3x^2+\\frac{3x^2}{2}\\right) =3x-\\frac{9x^2}{2}+\\frac{3}{2}\\end{align}$$<\/p>\n
Hence we know that:<\/p>\n
$$E\\left[X\\right]=\\int_{-\\infty}^{\\infty}{xf_X\\left(x\\right)\\ dx}$$<\/p>\n
Then,<\/p>\n
$$\\begin{align} E\\left(X\\right)&=\\int_{0}^{1}{x\\left(3x-\\frac{9x^2}{2}+\\frac{3}{2}\\right)\\\u00a0 dx}\\\\ &=\\left|\\frac{3x^3}{3}-\\frac{9x^4}{8}+\\frac{3x^2}{4}\\right|_0^1\\\\ &=1-\\frac{9}{8}+\\frac{3}{4}=\\frac{5}{8}\\\\ &=1-\\frac{9}{8}+\\frac{3}{4}=\\frac{5}{8}\u00a0\\end{align}$$<\/p>\n
$$\\therefore E\\left(x\\right)=\\frac{5}{8} $$<\/p>\n
For variance of X, we know that,<\/p>\n
$$Var(X)=E\\left(X^2\\right)-\\left[E\\left(X\\right)\\right]^2$$<\/p>\n
Now,<\/p>\n
$$E\\left(X^2\\right)=\\int_{0}^{1}{x^2\\left(3x-\\frac{9x^2}{2}+\\frac{3}{2}\\right)\\ dx}\u00a0=\\frac{7}{10}$$<\/p>\n
Thus,<\/p>\n
$$Var(X) =\u00a0E\\left(X^2\\right)-\\left[E\\left(X\\right)\\right]^2=\\frac{7}{10}-\\left(\\frac{5}{8}\\right)^2=\\frac{99}{320}$$<\/p>\n<\/p>\n
Moments for Conditional Random Variables<\/h2>\n
Let \\(X\\) and \\(Y\\) be random variables with a joint probability function \\(f(x,y)\\) and marginal functions \\(f_x(x)\\) and \\(f_y(y)\\)<\/p>\n
Discrete Case<\/h3>\n
The conditional pmf of \\(Y\\) given that \\(X=x\\) is defined by:<\/p>\n
$$h(y|x)=\\cfrac { f(x,y) }{ { f }_{ X }(x) }\\quad\\quad\\quad\\quad\\quad\\text{provided that}\\quad f_X(x)>0$$<\/p>\n
The conditional mean<\/strong> of \\(Y\\), given that \\(X=x\\) is defined:<\/p>\n$$ \\mu_{Y|x} = E[Y|x] = \\sum_{y} y h(y|x), $$<\/p>\n
and the conditional variance<\/strong> of \\(Y\\), given that \\(X=x\\) is defined:<\/p>\n$$ \\sigma_{Y|x} = E\\left\\{(Y – E[Y|x])^2|x \\right\\} = \\sum_{x}(y-E[Y|x])^2 h(y|x) $$<\/p>\n
This is simplified to:<\/p>\n
$$ \\sigma^2_{Y|x} = E[Y^2|x] – (E[Y|x])^2 $$<\/p>\n
Note also that:<\/p>\n
$$h\\left(y\\middle| x\\right)=\\frac{f\\left(x,y\\right)}{f_X\\left(x\\right)}\\ \\ provided\\ that\\ f_X\\left(x\\right)\u00a0 >0$$<\/p>\n
So that:<\/p>\n
$$\\mu_{x|y}=E[X|Y]=yxh(x|y)$$<\/p>\n
And<\/p>\n
$$\\sigma_{x|y}^2=E\\left[Y^2\\middle| X\\right]-E[X|Y]^2$$<\/p>\n
Example: Conditional Moments in the Discrete Case<\/h4>\n
The joint probability mass function of variables X and Y is given by:<\/p>\n
$$f(x,y) = \\frac{x^2 +3y}{60},\\ x=1,2,3,4;\\ y=1,2$$<\/p>\n
Calculate :<\/p>\n
a). E(X|Y=1)<\/p>\n
b). E(Y|X=3)<\/p>\n
c). V(X|Y=1)<\/p>\n
Solution\u00a0<\/strong><\/p>\nFrom the joint function, we can get the following marginal pmfs:<\/p>\n
$$f_X\\left(x\\right)=\\frac{2x^2+9}{60}\\ \\ \\text{and} \\ f_Y\\left(y\\right)=\\frac{12y+30}{60}$$<\/p>\n
We can also find conditional probability mass function:<\/p>\n
$$g\\left(x\\middle| y\\right)=\\frac{x^2+3y}{12y+30}\\ \\text{and}\\ h\\left(y\\middle| x\\right)=\\frac{x^2+3y}{2x^2+9}$$<\/p>\n
So,<\/p>\n
a). Finding \\(E(X|Y=1):<\/strong><\/p>\n$$\\begin{align} E\\left(X\\middle| Y=1\\right)&=E(g(x|y=1))\\\\ &=\\sum_{x=1}^{4}{xg\\left(x\\middle| y=1\\right)}\\\\&=\\sum_{x=1}^{4}{x\\frac{x^2+3\\left(1\\right)}{12\\left(1\\right)+30}} \\\\ &=\\frac{65}{21}=3.10\\end{align}$$<\/p>\n
b). Finding \\(E(Y|X=3)\\):<\/strong><\/p>\n$$\\begin{align} E\\left(Y\\middle| X=3\\right)&=E(h(y|x=3))\\\\ &=\\sum_{y=1}^{2}{yh(y|x=3)}\\\\ &=\\sum_{y=1}^{2}{y\\ \\frac{3^2+3y}{2\\left(3\\right)^2+9}}\\\\&=\\left(1\\right)\\frac{3^2+3\\left(1\\right)}{2\\left(3\\right)^2+9}+\\left(2\\right)\\frac{3^2+3\\left(2\\right)}{2\\left(3\\right)^2+9}\\\\&=\\frac{12}{27}+\\left(2\\right)\\frac{15}{27}=\\frac{14}{9}=1.56\\end{align}$$<\/p>\n
c). Finding \\(Var(X|Y=1)\\)<\/strong><\/p>\nUsing the output\u00a0 from (a), we have:<\/p>\n
$$\\begin{align}V\\left(X\\middle| Y=1\\right)&=E\\left[X-E\\left(X\\middle| Y=1\\right)\\right)^2|Y=1]\\\\ & =\\sum_{x=1}^{4}{\\left(x-E\\left(X\\middle| Y=1\\right)\\right)^2g(x|y=1)}\\\\ &=\\sum_{x=1}^{4}{\\left(x-\\frac{65}{21}\\right)^2\\frac{x^2+3\\left(1\\right)}{12\\left(1\\right)+30}}\\\\ &=0.99 \\end{align}$$<\/p>\n
Continuous Case:<\/h3>\n
When \\(X\\) and \\(Y\\) are continuous random variables, the conditional pdf, mean, and variance are given as follows:<\/p>\n
Conditional pdf:<\/p>\n
$$ g(x|y)=\\cfrac { f(x,y) }{ { f }_{ Y}(y) } \\quad\\quad\\quad\\quad\\quad\\text{provided that}\\quad f_Y(y)>0$$<\/p>\n
Conversely,<\/p>\n
$$h\\left(y\\middle| x\\right)=\\frac{f\\left(x,y\\right)}{f_X\\left(x\\right)}\\ \\text{provided\\ that}\\ f_X\\left(x\\right) > 0$$<\/p>\n
Conditional mean:<\/p>\n
$$E(Y|X)=\\int _{ -\\infty\u00a0 }^{ \\infty\u00a0 }{ yh(y|x)\\partial y } $$<\/p>\n
Also<\/p>\n
$$E\\left(Y\\middle| X\\right)=\\int_{-\\infty}^{\\infty}{yh\\left(y\\middle| x\\right)\\partial y}$$<\/p>\n
Conditional variance:<\/p>\n
$$\\text{Var}(Y|X)=E\\{ [Y-E(Y|x)]^{ 2 }|x\\} $$<\/p>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 $$=\\int _{ -\\infty\u00a0 }^{ \\infty\u00a0 }{ [y-E(Y|x)]^{ 2 }h(y|x)\\partial y }\u00a0$$<\/p>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 $$=E[{ Y }^{ 2 }|x]-{ [E(Y|x] }^{ 2 }$$<\/p>\n
Using the same logic,<\/p>\n
$$Var\\left(Y\\middle| X\\right)=E\\left[Y^2\\middle| X\\right]-\\left[E\\left(Y\\middle| X\\right)\\right]^2$$<\/p>\n
<\/h4>\nExample:\u00a0\u00a0Conditional Moments in the Continuous Case<\/strong><\/h4>\nLet<\/p>\n
$$ f(x,y) = \\frac{4}{3}(1-xy) \\qquad 0\\leq x\\leq1,\\quad 0\\leq y\\leq1 $$<\/p>\n
Find:<\/p>\n
a). g(x|y)<\/p>\n
b). E(x|y=1)<\/p>\n
Solution<\/strong><\/p>\na): Conditional function \\(g(x|y)\\)<\/strong><\/p>\nWe know that:<\/p>\n
$$\\begin{align} g\\left(x\\middle| y\\right)&=\\frac{f\\left(x,y\\right)}{f_Y\\left(y\\right)}\\\\ &=\\frac{\\frac{4}{3}\\left(1-xy\\right)}{\\int_{0}^{1}{\\frac{4}{3}\\left(1-xy\\right)dx}}\\\\&=\\frac{\\frac{4}{3}\\left(1-xy\\right)}{\\frac{4}{3}{\\left[x-\\frac{x^2y}{2}\\right]\\left(x-\\frac{x^2y}{2}\\right)}_{x=0}^{x=1}}\\\\&\\end{align}$$<\/p>\n
b): \u00a0Conditional mean of X given Y \\(E(X|Y=1)<\/strong><\/p>\nWe know that:<\/p>\n
$$\\begin{align} E\\left(x\\middle| y\\right)&=\\int_{0}^{1}{xg\\left(x\\middle| y\\right)dx}\\\\ &E\\left(x\\middle| y\\right)=\\int_{0}^{1}{x\\frac{1-xy}{1-\\frac{y}{2}}dx}\\\\ &=\\frac{1}{1-\\frac{y}{2}}{\\left[\\frac{x^2}{2}-\\frac{x^3y}{3}\\right]\\left(\\frac{x^2}{2}-\\frac{x^3y}{3}\\right)}_{x=0}^{x=1}\\\\ &=\\frac{1}{1-\\frac{y}{2}}\\left(\\frac{1}{2}-\\frac{y}{3}\\right)\\end{align}$$<\/p>\n
Therefore:<\/p>\n
$$E\\left(Xx\\middle| Y y=1\\right)=\\frac{1}{1-\\frac{1}{2}}\\left(\\frac{1}{2}-\\frac{1}{3}\\right)=\\frac{1}{3}$$<\/p>\n
To find variance in the continuous case, we would just integrate over the same region with \\(x^2\\) instead of x and then find the difference between this integral and \\(\\left[E\\left(x\\middle| y\\right)\\right]^2\\)<\/p>\n
<\/h4>\nConcept Reminders<\/h3>\n
We compute and define conditional expectations, variances, etc., as usual, but with conditional distributions in place of ordinary distributions:<\/p>\n
In the discrete case<\/strong>,<\/p>\n\\(E(X|Y)=E(X|Y=y)=\\sum _{ x }^{\u00a0 }{ xg(x|y) } \\)<\/p>\n
\\(E({ X }^{ 2 }|Y)=E({ X }^{ 2 }|Y=y)=\\sum _{ x }^{\u00a0 }{ { x }^{ 2 }g(x|y) } \\)<\/p>\n
\\(Var(X|Y)=Var(X|Y=y)=E({ X }^{ 2 }|y)-[E(X|y)]^{ 2 }\\)<\/p>\n
and in general,<\/p>\n
\\(E(X|\\text{condition})=\\sum _{ x }^{\u00a0 }{ xP(X=x|\\text{condition}) } \\)<\/p>\n
In the continuous case,<\/p>\n
\\(E(X|y)=E(X|Y=y)=\\int _{\u00a0 }^{\u00a0 }{ xg(x|y)\\partial x } \\)<\/p>\n
\\(E({ X }^{ 2 }|y)=E({ X }^{ 2 }|Y=y)=\\int _{\u00a0 }^{\u00a0 }{ { x }^{ 2 }g(x|y)\\partial x } \\)<\/p>\n
\\(Var(X|y)=Var(X|Y=y)=E({ X }^{ 2 }|y)-[E(X|y)]^{ 2 }\\)<\/p>\n
and in general,<\/p>\n
\\((E(X|{ \\text{condition} })=\\int _{ -\\infty\u00a0 }^{ \\infty\u00a0 }{ xP(X=x|\\text{condition}) } \\)<\/p>\n
The conditional density (pdf or pmf) of \\(X\\) given that \\(Y = y\\) is given by:<\/p>\n
\\(g(x|y)=\\cfrac { f(x,y) }{ { f }_{ Y }(y) } \\)<\/p>\n
The conditional density of \\(Y\\) given that \\(X=x\\) is given by:<\/p>\n
\\(h(y|x)=\\cfrac { f(x,y) }{ { f }_{ X }(x) } \\)<\/p>\n
Learning Outcome<\/strong><\/em><\/p>\nTopic 3.c:\u00a0Multivariate Random Variables – Calculate moments for joint, conditional, and marginal random variables.<\/strong><\/em><\/p>\n","protected":false},"excerpt":{"rendered":"Moments of a Probability Mass function The n-th moment about the origin of a random variable is the\u00a0expected value\u00a0of its\u00a0n-th power. Moments about the origin are \\(E(X),E({ X }^{ 2 }),E({ X }^{ 3 }),E({ X }^{ 4 }),….\\quad\\) For…<\/p>\n","protected":false},"author":1,"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-3007","post","type-post","status-publish","format-standard","hentry","category-multivariate-random-variables","blog-post","no-post-thumbnail","animate"],"acf":[],"yoast_head":"\n
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