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

`Cov(X,Y) = -( 25)/(9)`

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) = (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) = (1)/(0.5 * 10 * 10)`

`f(X,Y) = (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 {(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 {(XY)/(50)} \, dY} \, dX`

Rewrite as:

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

Integrate Y

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

Expand

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

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

Rewrite as:

`E(XY) =(1)/(100)\int\limits^(10)_0 X(10 - X)^2\ dX`

Expand

`E(XY) =(1)/(100)\int\limits^(10)_0 X*(100 - 20X + X^2)\ dX`

`E(XY) =(1)/(100)\int\limits^(10)_0 100X - 20X^2 + X^3\ dX`

Integrate

`E(XY) =(1)/(100) [(100X^2)/(2) - (20X^3)/(3) + (X^4)/(4)]|\limits^(10)_0`

Expand

`E(XY) =(1)/(100) ([(100*10^2)/(2) - (20*10^3)/(3) + (10^4)/(4)] - [(100*0^2)/(2) - (20*0^3)/(3) + (0^4)/(4)])`

`E(XY) =(1)/(100) ([(10000)/(2) - (20000)/(3) + (10000)/(4)] - 0)`

`E(XY) =(1)/(100) ([5000 - (20000)/(3) + 2500])`

`E(XY) =50 - (200)/(3) + 25`

Take LCM

`E(XY) = (150-200+75)/(3)`

`E(XY) = (25)/(3)`

Calculate E(X)

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

Rewrite as:

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

Integrate Y

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

Expand

`E(X) =(1)/(50)\int\limits^(10)_0 ( [X*(10 - X)] - [X * 0])\ dX`

`E(X) =(1)/(50)\int\limits^(10)_0 ( [X*(10 - X)]\ dX`

`E(X) =(1)/(50)\int\limits^(10)_0 10X - X^2\ dX`

Integrate

`E(X) =(1)/(50)(5X^2 - (1)/(3)X^3)|\limits^(10)_0`

Expand

`E(X) =(1)/(50)[(5*10^2 - (1)/(3)*10^3)-(5*0^2 - (1)/(3)*0^3)]`

`E(X) =(1)/(50)[5*100 - (1)/(3)*10^3]`

`E(X) =(1)/(50)[500 - (1000)/(3)]`

`E(X) = 10- (20)/(3)`

Take LCM

`E(X) = (30-20)/(3)`

`E(X) = (10)/(3)`

Calculate E(Y)

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

Rewrite as:

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

Integrate Y

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

Expand

`E(Y) =(1)/(50)\int\limits^(10)_0 ( [((10 - X)^2)/(2)] - [((0)^2)/(2)])\ dX`

`E(Y) =(1)/(50)\int\limits^(10)_0 ( [((10 - X)^2)/(2)] )\ dX`

`E(Y) =(1)/(50)\int\limits^(10)_0 [(100 - 20X + X^2)/(2)] \ dX`

Rewrite as:

`E(Y) =(1)/(100)\int\limits^(10)_0 [100 - 20X + X^2] \ dX`

Integrate

`E(Y) =(1)/(100)( [100X - 10X^2 + (1)/(3)X^3]|\limits^(10)_0)`

Expand

`E(Y) =(1)/(100)( [100*10 - 10*10^2 + (1)/(3)*10^3] -[100*0 - 10*0^2 + (1)/(3)*0^3] )`

`E(Y) =(1)/(100)[100*10 - 10*10^2 + (1)/(3)*10^3]`

`E(Y) =10 - 10 + (1)/(3)*10`

`E(Y) =(10)/(3)`

Recall that:

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

`Cov(X,Y) = (25)/(3) - (10)/(3)*(10)/(3)`

`Cov(X,Y) = (25)/(3) - (100)/(9)`

Take LCM

`Cov(X,Y) = (75- 100)/(9)`

`Cov(X,Y) = -( 25)/(9)`