Question #71378

The random variables X and Y have a density of
fXY (x, y) = x + y for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, and fXY (x, y) = 0 elsewhere. Evaluate
E [X] and E [XY ].
1

Expert's answer

2017-12-05T13:26:06-0500

Answer for Question #71378


fX(x)=01(x+y)dy=x+12f _ {X} (x) = \int_ {0} ^ {1} (x + y) d y = x + \frac {1}{2}fY(y)=01(x+y)dx=y+12f _ {Y} (y) = \int_ {0} ^ {1} (x + y) d x = y + \frac {1}{2}E[X]=01x(x+12)dx=712E [ X ] = \int_ {0} ^ {1} x \left(x + \frac {1}{2}\right) d x = \frac {7}{12}E[Y]=01y(y+12)dy=712E [ Y ] = \int_ {0} ^ {1} y \left(y + \frac {1}{2}\right) d y = \frac {7}{12}E[XY]=E[X]E[Y]KXY,KXY=0101xy(x+y)dydx=13E \left[ X Y \right] = E \left[ X \right] E \left[ Y \right] - K _ {X Y}, \quad K _ {X Y} = \int_ {0} ^ {1} \int_ {0} ^ {1} x y (x + y) d y d x = \frac {1}{3}E[XY]=(712)213=4914448144=1144E [ X Y ] = \left(\frac {7}{12}\right) ^ {2} - \frac {1}{3} = \frac {49}{144} - \frac {48}{144} = \frac {1}{144}


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