Question

In: Advanced Math

Probabilities Compute P(X1 + 2X2 -2X3 > 7) , where X1, X2, X3 are iid random...

Probabilities

Compute P(X1 + 2X2 -2X3 > 7) , where X1, X2, X3 are iid random variables with common distribution N(1,4). You must give the numerical answer.

Specify and show any formulas used for E, P, and Var

for example Var( X1 + 2X2 -2X3) = 4 + 22(4) + (-2)2(4) = 35. Please specify how all the individual numbers where obtained and the formula used.

Solutions

Expert Solution

Consider the following formula:

(i) For Independent and identically distributed (i.i.d.) random variables x1, x2 and constant a & b,

                                      Var(aX1+bX2) = a2Var(X1) + b2Var(X2)

(ii) For random variables x1, x2 and constant a & b,

                                       E(aX1+bX2) = aE(X1) + bE(X2)

(iii) For any real value a and continuous random variable X,

                                     P(X>a) = 1-P(x<=a)

Solution:

X1, X2, X3 are i.i.d. random variables with N(1,4). So,

for each i = 1,2,3; E(xi) = 1 and Var(Xi) = 4.

Let Y = X1+2X2-2X3

Mean of Y is

E(Y) = E(X1+2X2-2X3) = E(X1) + 2E(X2) -2E(X3) (From (ii))

                                     = 1+2(1)-2(1) = 1

Var(Y) = Var(X1+2X2-2X3) = Var(X1) + 22Var(X2) + (-2)2Var(X3)    (from (i))

                                            = 4 + 22(4) + (-2)24 = 36

So deviation of Y = (Var(Y))1/2 = (36)1/2 = 6

So, P(X1+2X2-2X3>7) = P(Y>7) = 1 - P(Y<=7) = 1 - P((Y - E(Y))/(deviation of Y) <= (7 - E(Y))/(deviation of Y) )

                                                                            = 1 - P((Y - 1)/6 <= (7 - 1)/6 )

                                                                            = 1 - P(Z <= 1)    (Z = (Y - 1)/6 is a standard normal variable)

                                                                           = 1 - Phi(1)     

                                                                         = 1 - 0.8413     (From standard normal table )

                                                                           = 0.1587


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