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Suppose thatXandYare random variables such that\(E\left( {Y\left| X \right.} \right) = 7 - \left( {{\raise0.7ex\hbox{\(1\)} \!\mathord{\left/

{\vphantom {1 4}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{\(4\)}}} \right)X\)and \(E\left( {X\left| Y \right.} \right) = 10 - Y\). Determine the correlation ofXandY.

Short Answer

Expert verified

X and Y are negatively correlated and the correlation coefficient is \(\rho = - \frac{1}{2}\)

Step by step solution

01

Given information 

LetXand Yare two random variables. The conditional expectations are such that,

\(E\left( {Y\left| X \right.} \right) = 7 - \left( {\frac{1}{4}} \right)X\;and\;E\left( {X\left| Y \right.} \right) = 10 - Y\).

02

Determine the correlation coefficient

As one can see that the coefficient of X in\(E\left( {Y\left| X \right.} \right)\)is negative, So, one can conclude that,\(\rho < 0\).

It is known to all that the product of the coefficients of X and Y in\(E\left( {Y\left| X \right.} \right)\;and\;E\left( {X\left| Y \right.} \right)\;is\;{\rho ^2}\).

So the value of\({\rho ^2}\)such that,

\(\begin{aligned}{c}{\rho ^2} = E\left( {Y\left| X \right.} \right) \times E\left( {X\left| Y \right.} \right)\\ = \left( { - \frac{1}{4}} \right) \times \left( { - 1} \right)\\ = \frac{1}{4}\end{aligned}\)

Therefore, we get the value of the correlation coefficient\(\rho = \sqrt {\left( {\frac{1}{4}} \right)} = \frac{1}{2}\)

Since,\(\rho < 0\)this implies\(\rho = - \frac{1}{2}\).

Thus, one can conclude that X and Y are negatively correlated.

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