Given :-
Var(X)= 4,
Var(Y) = 1,
Cov(X,Y) = -1
Now we have to find Var(X-2Y+10)
Therefore,
Var(X-2Y+10) = Var (X) - Var(2Y) + Var(10)
For variance property, any variance of any constant value is equal to zero.
Here 10 is constant value. So Var(10)=0
Var(X-2Y+10) = Var(X) + 4Var(Y) - 0
= Var(X) + 4Var(Y) + 2Cov(X, -2Y) ----- using formula
= Var(X) + 4Var(Y) - 4Cov(X, Y)
= 4 + 4*1 - 4*(-1)
Var(X-2Y+10) = 12
9. Suppose Var(X] = 4, Var[Y-1, and Cov(X, Y] =-1. Calculate VarX-2Y + 101.
Suppose Var[X]=4, Var[Y]=1,and Cov [X,Y]= -1 . calculate Var [X-2Y+10]
X,Y, and Z are random variables.
Var(X) = 2, Var(Y) = 1, Var(Z) = 5, Cov(X,Y) = 3, Cov(X, Z) = -2, Cov(Y,Z) = 7. Determine Var(3X – 2Y - 2+10)
Given Var(X) = 4, Var(Y) = 1, and Var(X+2Y) = 10, What is Var(2X-Y-3)? I know the answer is 15, I'm particularly interested in the specific steps involved with finding the cov(X,Y) in this problem. Please explain in detail, step by step how you come to cov(X,Y) = 0.5 in this equation. Please include any formulas you would need to use to find the cov(X,Y) in this equation.
6 Suppose that X and Y are random variables such that Var(X) Var(Y)-2 and Cov(x,y)- 1. Find the value of Var(3.X-Y+2)
6. Suppose that X and Y are random variables such that Var(X)=Var(y)-2 and Cov(x,y)-1. the value of Var(ax-y-2). Find
6 Suppose that X and Y are random variables such that Var(X)-Var(Y)-2 and Cov(x,y)- 1. Find the value of Var(3.X-Y + 2)
10. Suppose Var|X] -2, VarlY] - 2, Var[Z]1, CovlX, Y10, CovlX, Z and Cov[Y, Z] =-1. Calculate Var(X + Y-2Z + 5.
Var(Y) =y Var(X)=x Cov(X,Y) =z What is Cov(XY,XY)
For random variables X, Y, and Z, Var(X) = 4, Var(Y) = 9, Var(Z) = 16, E[XY] = 6, E[XZ] = −8, E[Y Z] = 10, E[X] = 1, E[Y ] = 2 and E[Z] = 3. Calculate the followings: (b) Cov(−3Y , −4Z ). (d) Var(Y − 3Z). (e) Var(10X + 5Y − 5Z).
Suppose that EX-EY-0, var(X) = var(Y) = 1, and corr(X,Y) = 0.5. (i) Compute E3X -2Y]; and (ii) var(3X - 2Y) (ii) Compute E[X2]