Question

Annual windstorm losses, X and Y, in two different regions are independent, and each is uniformly distributed on the interval [0, 10]. Calculate the covariance of X and Y, given that X+ Y < 10.

Answer 1

Answer:

$Cov(X,Y) = -\frac{ 25}{9}$

Step-by-step explanation:

Given

$Interval =[0,10]$

$X + Y < 10$

Required

$Cov(X,Y)$

First, we calculate the joint distribution of X and Y

Plot $X + Y < 10$

So, the joint pdf is:

$f(X,Y) = \frac{1}{Area}$ --- i.e. the area of the shaded region

The shaded area is a triangle that has: height = 10; width = 10

So, we have:

$f(X,Y) = \frac{1}{0.5 * 10 * 10}$

$f(X,Y) = \frac{1}{50}$

$Cov(X,Y)$ is calculated as:

$Cov(X,Y) = E(XY) - E(X) \cdot E(Y)$

Calculate E(XY)

$E(XY) =\int\limits^X_0 {\int\limits^Y_0 {\frac{XY}{50}} \, dY} \, dX$

$X + Y < 10$

Make Y the subject

$Y < 10 - X$

So, we have:

$E(XY) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{XY}{50}} \, dY} \, dX$

Rewrite as:

$E(XY) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {XY}} \, dY} \, dX$

Integrate Y

$E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{XY^2}{2}}} }|\limits^{10 - X}_0 \, dX$

Expand

$E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2} - \frac{X(0)^2}{2}}} }\ dX$

$E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2}}} }\ dX$

Rewrite as:

$E(XY) =\frac{1}{100}\int\limits^{10}_0 X(10 - X)^2\ dX$

Expand

$E(XY) =\frac{1}{100}\int\limits^{10}_0 X*(100 - 20X + X^2)\ dX$

$E(XY) =\frac{1}{100}\int\limits^{10}_0 100X - 20X^2 + X^3\ dX$

Integrate

$E(XY) =\frac{1}{100} [\frac{100X^2}{2} - \frac{20X^3}{3} + \frac{X^4}{4}]|\limits^{10}_0$

Expand

$E(XY) =\frac{1}{100} ([\frac{100*10^2}{2} - \frac{20*10^3}{3} + \frac{10^4}{4}] - [\frac{100*0^2}{2} - \frac{20*0^3}{3} + \frac{0^4}{4}])$

$E(XY) =\frac{1}{100} ([\frac{10000}{2} - \frac{20000}{3} + \frac{10000}{4}] - 0)$

$E(XY) =\frac{1}{100} ([5000 - \frac{20000}{3} + 2500])$

$E(XY) =50 - \frac{200}{3} + 25$

Take LCM

$E(XY) = \frac{150-200+75}{3}$

$E(XY) = \frac{25}{3}$

Calculate E(X)

$E(X) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{X}{50}}} \, dY} \, dX$

Rewrite as:

$E(X) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {X}} \, dY} \, dX$

Integrate Y

$E(X) =\frac{1}{50}\int\limits^{10}_0 { (X*Y)|\limits^{10 - X}_0 \, dX$

Expand

$E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)] - [X * 0])\ dX$

$E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)]\ dX$

$E(X) =\frac{1}{50}\int\limits^{10}_0 10X - X^2\ dX$

Integrate

$E(X) =\frac{1}{50}(5X^2 - \frac{1}{3}X^3)|\limits^{10}_0$

Expand

$E(X) =\frac{1}{50}[(5*10^2 - \frac{1}{3}*10^3)-(5*0^2 - \frac{1}{3}*0^3)]$

$E(X) =\frac{1}{50}[5*100 - \frac{1}{3}*10^3]$

$E(X) =\frac{1}{50}[500 - \frac{1000}{3}]$

$E(X) = 10- \frac{20}{3}$

Take LCM

$E(X) = \frac{30-20}{3}$

$E(X) = \frac{10}{3}$

Calculate E(Y)

$E(Y) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{Y}{50}}} \, dY} \, dX$

Rewrite as:

$E(Y) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {Y}} \, dY} \, dX$

Integrate Y

$E(Y) =\frac{1}{50}\int\limits^{10}_0 { (\frac{Y^2}{2})|\limits^{10 - X}_0 \, dX$

Expand

$E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] - [\frac{(0)^2}{2}])\ dX$

$E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] )\ dX$

$E(Y) =\frac{1}{50}\int\limits^{10}_0 [\frac{100 - 20X + X^2}{2}] \ dX$

Rewrite as:

$E(Y) =\frac{1}{100}\int\limits^{10}_0 [100 - 20X + X^2] \ dX$

Integrate

$E(Y) =\frac{1}{100}( [100X - 10X^2 + \frac{1}{3}X^3]|\limits^{10}_0)$

Expand

$E(Y) =\frac{1}{100}( [100*10 - 10*10^2 + \frac{1}{3}*10^3] -[100*0 - 10*0^2 + \frac{1}{3}*0^3] )$

$E(Y) =\frac{1}{100}[100*10 - 10*10^2 + \frac{1}{3}*10^3]$

$E(Y) =10 - 10 + \frac{1}{3}*10$

$E(Y) =\frac{10}{3}$

Recall that:

$Cov(X,Y) = E(XY) - E(X) \cdot E(Y)$

$Cov(X,Y) = \frac{25}{3} - \frac{10}{3}*\frac{10}{3}$

$Cov(X,Y) = \frac{25}{3} - \frac{100}{9}$

Take LCM

$Cov(X,Y) = \frac{75- 100}{9}$

$Cov(X,Y) = -\frac{ 25}{9}$