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Prove Theorem 4.7.4.

Short Answer

Expert verified

\(Var\left( Y \right) = E\left( {Var\left( {Y|X} \right)} \right) + Var\left( {E\left( {Y|X} \right)} \right).\)

Step by step solution

01

Given information

X and Y are arbitrary random variables for which the necessary expectation and variance exist.

02

Verifying \(Var\left( Y \right) = E\left( {Var\left( {Y|X} \right)} \right) + Var\left( {E\left( {Y|X} \right)} \right).\)

Let \(E\left( Y \right) = \mu Y.\) then

\(\begin{align}Var\left( Y \right) &= E\left( {{Y^2}} \right) - E{\left( Y \right)^2}\\Var\left( Y \right) &= E\left( {{{\left( {Y - \mu Y} \right)}^2}} \right)\\ &= E\left\{ {{{\left( {\left( {Y - E\left( {Y|X} \right)} \right) + \left( {E\left( {Y|X} \right) - \mu Y} \right)} \right)}^2}} \right\}\end{align}\)

\(\begin{align} &= E\left\{ {{{\left( { + \left( {E\left( {Y|X} \right) - \mu Y} \right)} \right)}^2}} \right\}\\ &= E{\left( {Y - E\left( {Y|X} \right)} \right)^2} + 2E\left\{ {\left( {Y - E\left( {Y|X} \right)} \right)\left( {E\left( {Y|X} \right) - \mu Y} \right)} \right\} + \\E\left\{ {{{\left( {E\left( {Y|X} \right) - \mu Y} \right)}^2}} \right\}\end{align}\)

The three expectations in the total sum.

\(\begin{align}E\left\{ {{{\left( {Y - E\left( {Y|X} \right)} \right)}^2}} \right\} &= E\left( {E\left\{ {\left( {Y - E{{\left( {Y|X} \right)}^2}|X} \right)} \right\}} \right)\\ &= E\left( {Var\left( {Y|X} \right)} \right)\end{align}\)

Next,

\(\begin{align}E\left\{ {\left( {Y - E\left( {Y|X} \right)} \right)\left( {E\left( {Y|X} \right) - \mu Y} \right)} \right\}\\ = E\left( {E\left\{ {\left( {Y - E\left( {Y|X} \right)} \right)\left( {E\left( {Y|X} \right) - \mu Y} \right)|X} \right\}} \right)\end{align}\)

\(\begin{align} &= E\left( {\left( {E\left( {Y|X} \right) - \mu Y} \right)E\left\{ {Y - E\left( {Y|X} \right)|X} \right\}} \right)\\ &= E\left( {\left( {E\left( {Y|X} \right) - \mu Y} \right) \times 0} \right)\\ &= 0\end{align}\)

Finally, since the mean \(E\left( {Y|X} \right)\) is \(E\left( {E\left( {Y|X} \right)} \right) = \mu Y\)

\(E\left\{ {{{\left( {E\left( {Y|X} \right) - \mu Y} \right)}^2}} \right\} = {\mathop{\rm var}} \left( {E\left( {Y|X} \right)} \right).\)

It now follows that

\(Var\left( Y \right) = E\left( {Var\left( {Y|X} \right)} \right) + Var\left( {E\left( {Y|X} \right)} \right).\)

Hence proved the theorem that if X and Y are arbitrary random variables for which the necessary expectation and variance exist, then

\(Var\left( Y \right) = E\left( {Var\left( {Y|X} \right)} \right) + Var\left( {E\left( {Y|X} \right)} \right).\)

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Most popular questions from this chapter

Consider a utility function U for which\({\bf{U}}\left( {\bf{0}} \right){\bf{ = 0}}\)and\({\bf{U}}\left( {{\bf{100}}} \right){\bf{ = 1}}\). Suppose that a person who has this utility function is indifferent to accepting a gamble from which his gain will be 0 dollars with probability\({\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/ {\vphantom {{\bf{1}} {\bf{3}}}}\right.\ } \!\lower0.7ex\hbox{\({\bf{3}}\)}}\)or 100 dollars with probability\({\raise0.7ex\hbox{\({\bf{2}}\)} \!\mathord{\left/ {\vphantom {{\bf{2}} {\bf{3}}}}\right.\ } \!\lower0.7ex\hbox{\({\bf{3}}\)}}\)or accepting 50 dollars as a sure thing. What is the value\({\bf{U}}\left( {{\bf{50}}} \right)\)?

Find the median of the Cauchy distribution defined in Example 4.1.8

For all numbers a and b such that \(a < b\), find the variance of the uniform distribution on the interval \(\left( {a,b} \right)\).

Construct an example of a distribution for which the mean is finite but the variance is infinite.

LetYbe a discrete random variable whose p.f. is the

functionfin Example 4.1.4. LetX= |Y|. Prove that the

distribution ofXhas the p.d.f. in Example 4.1.5

\({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{\bf{1}}}{{{\bf{2}}\left| {\bf{x}} \right|\left( {\left| {\bf{x}} \right|{\bf{ + 1}}} \right)}}{\bf{,x = \pm 1, \pm 2 \ldots ,}}\\{\bf{0,Otherwise}}\end{array} \right.\)

\({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{\bf{1}}}{{{\bf{x}}\left( {{\bf{x + 1}}} \right)}}{\bf{,x = 1,2,3 \ldots ,}}\\{\bf{0,Otherwise}}\end{array} \right.\)

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