Question

a) The maximum daily water level at an embankment with height 6m can be described with mean 2m and varians 1m².

What is the upper bound that the embankment will be flooded the given day?

b) For the stochastic variable  Z = 4X − 5Y − 5, it is given that E(X) = 5, E(Y ) = 3, Var(X) = 16 and Var(Y ) = 9 while Cov(X, Y ) = 8.

Find Var(Z)

Answer 1

(a) Let X be the height of daily water level. Then E(X) = 2 and . The embankment will be flooded the given day, if the height of daily water level is greater than the height of maximum daily water level i.e 6m. Hence the embankment will be flood if X > 6.

Using Chebyshev's inequality Pr (| X-E(X)| k) 1/k² . So if we put k = 4 and values of E(X) and  , we get

Pr (| X-2| 4*1) 1/4² => Pr (X-2 > 4 or X-2 < -4) => Pr( X > 6 or X < -2) 1/16.( We have used the result |X| > Y => X >Y or X< -Y)

Since X is height of daily water level , hence it cannot be negative and cannot be less than -2.

Hence Pr (X > 6) 0.0625. Hence the upper bound that the embankment will be flood the given day is 0.0625.

(b) We will use the result that Var (aX + bY + c) = a² *Var(X) + b² * Var (Y) +2 * a* b *Cov (X,Y)

Hence Var (Z) = Var (4X - 5Y- 5) = 16* Var (X) + 25* Var(Y) -2*4*5* Cov(X,Y) = 16*16+ 25*9 - 40*8 = 161.

So Var (Z) = 161.