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For all random variables X, Y , and Z, let\({\bf{Cov}}\left( {{\bf{X,Y|z}}} \right)\)denote the covariance of X and Y in their conditional joint distribution given Z = z.

Prove that.\({\bf{Cov}}\left( {{\bf{X,Y}}} \right){\bf{ = E}}\left( {{\bf{Cov}}\left( {{\bf{X,Y|z}}} \right)} \right){\bf{ + Cov}}\left( {{\bf{E}}\left( {{\bf{X|Z}}} \right){\bf{,E}}\left( {{\bf{Y|Z}}} \right)} \right)\)

Short Answer

Expert verified

\(Cov\left( {X,Y} \right) = E\left( {Cov\left( {X,Y|z} \right)} \right) + Cov\left( {E\left( {X|Z} \right),E\left( {Y|Z} \right)} \right)\)

Step by step solution

01

Given information

For the random variables x,y ,z \(Cov\left( {X,Y|z} \right)\)is the covariance of x and yin their cdf given Z=z

02

Caculating covariance.

\(\begin{aligned}{c}Cov\left( {X,Y} \right) = E\left( {\left( {X - \mu x} \right)\left( {Y - \mu y} \right)} \right)\\ = E\left\{ {\left( {X - E\left( {X|Z} \right) + E\left( {X|Z} \right) - \mu x} \right) \times \left( {Y - E\left( {Y|Z} \right) + E\left( {Y|Z} \right) - \mu y} \right)} \right\}\\ = E\left\{ {\left( {X - E\left( {X|Z} \right)} \right)\left( {Y - E\left( {Y|Z} \right)} \right)} \right\} + E\left\{ {\left( {X - E\left( {X|Z} \right)} \right)\left( {E\left( {Y|Z} \right) - \mu y} \right)} \right\}\\ + E\left\{ {\left( {E\left( {X|Z} \right) - \mu x} \right)\left( {Y - E\left( {Y|Z} \right)} \right)} \right\} + E\left\{ {\left( {E\left( {X|Z} \right) - \mu x} \right)\left( {E\left( {Y|Z} \right) - \mu y} \right)} \right\}\end{aligned}\).......(1)

In the given equation taking the last 4 expectations.

In the 1st expectation:

if we 1st calculate the conditional expectation given Z and then take the expectation over Z, we get

\(E\left( {Cov\left( {X,Y|Z} \right)} \right)\)

In the 2nd and 3rd expectation:

We obtain value 0 when we take the conditional expectation given Z.

In the 4th expectation:

\(Cov\left( {E\left( {X|Z} \right),E\left( {Y|Z} \right)} \right)\)

Hence

\(Cov\left( {X,Y} \right) = E\left( {Cov\left( {X,Y|z} \right)} \right) + Cov\left( {E\left( {X|Z} \right),E\left( {Y|Z} \right)} \right)\)

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Most popular questions from this chapter

Show that two random variablesXandYcannot possibly have the following properties:\(E\left( X \right) = 3\),\(E\left( Y \right) = 2\),\(E\left( {{X^2}} \right) = 10\),\(E\left( {{Y^2}} \right) = 29\), and\(E\left( {XY} \right) = 0\).

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Suppose that this decision maker is trying to decide whether or not to buy a lottery ticket for \(1. The lottery ticket pays \)500 with a probability of 0.001, and it pays $0 with a probability of 0.999. What would the values of α have to be for this decision-maker to prefer buying the ticket to not buying it?

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