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Let X denote the size of a bodily injury claim and Y denote the size of the corresponding property damage claim. Let Z1 = X + Y. From prior experience we know Var(X) = 144, Var(Y) = 64 and Var(X + Y) = 308. It is expected that bodily injury claims will rise 10% next year and property damage will rise by a fixed amount of 5. Let Z2 be the new trial of bodily injury and property damage. Compute Cov(Z1, Z2 ).

Respuesta :

Answer:

[tex] Cov (Z_1, Z_2)= 1.1*144 +2.1*50 + 64 =213[/tex]

Step-by-step explanation:

Data given:

Var(X) = 144, Var(Y) = 64 and Var(X + Y) = 308

We have defined the following random variable:

[tex] Z_1 = X+Y[/tex]

And if we find the variance for Z1 we got:

[tex] Var(Z_1) = Var(X) +Var(Y) + 2 cov (X,Y)[/tex]

If we solve for the [tex] Cov(X,Y)[/tex] we got:

[tex]Cov(X,Y) =\frac{Var(X+Y)-Var(X)-Var(Y)}{2}[/tex]

And replacing the values that we have we got:

[tex]Cov(X,Y) =\frac{308-144-64}{2}=50[/tex]

Now we define the new random variable [tex] Z_2[/tex] like this:

[tex] Z_2 = 1.1 X + Y+5[/tex]

And we are interested on find the following covariance:

[tex] Cov (Z_1, Z_2) [/tex]

[tex] Cov (Z_1, Z_2)= Cox( X+Y, 1.1X +Y +5) [/tex]

And if we apply properties of covariance we have this:

[tex] Cov (Z_1, Z_2)= Cov (X,1.1X)+ Cov (X,Y) +Cov(X,5) +Cov(Y,1.1X) +Cov (Y,Y) + Cov (Y,5)[/tex]

We know that if a is a constant and X,Y random variables [tex] cov(a,X) =0[/tex], and [tex] Cov(X,X) = Var(X)[/tex] , [tex] Cov (X, aY) = aCov(X,y)[/tex], so then using these properties we got:

[tex] Cov (Z_1, Z_2)= 1.1 Var(X) +2.1 Cov (X,Y) + Var(Y)[/tex]

[tex] Cov (Z_1, Z_2)= 1.1*144 +2.1*50 + 64 =213[/tex]

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