Chapter 7: Q.7.4 (page 355)

The joint density function of $X$ and $Y$ is given by
$f (x, y) = \frac{1}{y} e^{- (y + x / y)}, x > 0, y > 0$
Find $E [X], E [Y]$ and show that $C o v (X, Y) = 1$

Short Answer

Expert verified

The value of $E [X] = 1$

The value of $E [Y] = 1$

The value of $Cov (X, Y) = 1$

Step by step solution

Given Information

The density of $X$ and $Y$ is $f (x, y) = \frac{1}{y} e^{- (y + x / y)}, x > 0, y > 0$

$E [X] = ?$

$E [Y] = ?$

$Cov (X, Y) = ?$

Explanation

Calculate the value of $E [X],$

$E [X] = \int_{0}^{\infty} \int_{0}^{\infty} x \cdot \frac{1}{y} e^{- (y + \frac{x}{y})} d y d x$

$= \int_{0}^{\infty} e^{- y} [\int_{0}^{\infty} \frac{x}{y} e^{- \frac{x}{y}} d x] d y$

$= \int_{0}^{\infty} y e^{- y} d y$

$= {y \frac{e^{- y}}{- 1}|}_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{- y}}{- 1} d y$

$= 0 + 1$

$= 1$

Explanation

Calculate the value of $E [Y],$

$E [Y] = \int_{0}^{\infty} \int_{0}^{\infty} y \cdot \frac{1}{y} e^{- (y + \frac{x}{y})} d x d y$

$= \int_{0}^{\infty} \int_{0}^{\infty} e^{- y} e^{- \frac{x}{y}} d x d y$

$= \int_{0}^{\infty} e^{- y} [\int_{0}^{\infty} e^{- \frac{x}{y}} d x] d y$

$= \int_{0}^{\infty} e^{- y} y d y$

$= 1$

Explanation

Calculate the value of $Cov (X, Y),$

$E (X Y) = \int_{0}^{\infty} \int_{0}^{\infty} x y \cdot \frac{1}{y} e^{- (y + \frac{x}{y})} d x d y$

$= \int_{0}^{\infty} e^{- y} [\int_{0}^{\infty} x e^{\frac{- x}{y}} d x] d y$

$= \int_{0}^{\infty} e^{- y} [{\frac{x e^{\frac{- x}{y}}}{\frac{- 1}{y}}|}_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{\frac{- x}{y}}}{\frac{- 1}{y}} d x] d y$

$= \int_{0}^{\infty} e^{- y} [0 + y^{2}] d y$

$= \int_{0}^{\infty} y^{2} e^{- y} d y$

$= - {y^{2} e^{- y}|}_{0}^{\infty} + \int_{0}^{\infty} 2 y e^{- y} d y$

$= 0 + 2 = 2$

Final Answer

Hence $Cov (X, Y) = E (X Y) - E (X) E (Y) = 2 - 1.1$

$= 2 - 1$

$= 1$

So, the value of $C o v (X, Y) = 1$

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The joint density function of $X$ and $Y$ is given by
$f (x, y) = \frac{1}{y} e^{- (y + x / y)}, x > 0, y > 0$
Find $E [X], E [Y]$ and show that $C o v (X, Y) = 1$

Short Answer

Step by step solution

Given Information

Explanation

Explanation

Explanation

Final Answer

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The joint density function ofXandYis given byf(x,y)=1ye-(y+x/y),x>0,y>0Find E[X],E[Y]and show that Cov(X,Y)=1

Short Answer

Step by step solution

Given Information

Explanation

Explanation

Explanation

Final Answer

One App. One Place for Learning.

Most popular questions from this chapter

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Theoretical and Mathematical Physics

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The joint density function of $X$ and $Y$ is given by
$f (x, y) = \frac{1}{y} e^{- (y + x / y)}, x > 0, y > 0$
Find $E [X], E [Y]$ and show that $C o v (X, Y) = 1$