Question

write the joint density f(x, y). (b) find e(ax by c). (c) find ex2 . (hint: use the variance identity.)

Answer 1

i) Joint density function  is given as

               f(x,y) =  1/2Πσ1σ2  exp[(-1)/2 {((x-µ1)^2)/(σ1^2)+ ((y-µ2))/(σ2^2)}]

ii) E(aX + bY + c) = aµx + bµy + c

iii) E(X^2) = σx^2 + µx^2

We have been given that, X ~ N(µx , σx^2) and Y~ N(µy , σy^2)

and X and Y are independent that is

                                 ρ = 0

Note that the joint density function of two normal distribution is known as bivariate normal distribution which is given as

    (X , Y) ~ BVN(µ1 , µ2 , σ1^2 , σ2^2 , ρ)

And whose joint density function is          

f(x,y)  =   1/(2Πσ1σ2√(1-ρ^2))  exp[(-1)/(2(1-ρ^2)) {((x-µ1)^2)/(σ1^2)-2ρ (x-µ1)(y-µ2)/σ1σ2+ ((y-µ2))/(σ2^2)}]

  Where ρ = correlation coefficient of (x,y)

i) Joint density function of given X and Y which follows

(X , Y) ~ BVN(µx , µy , σx^2 , σy^2 , 0) is given as

f(x,y) =  1/2Πσ1σ2  exp[(-1)/2 {((x-µ1)^2)/(σ1^2)+ ((y-µ2))/(σ2^2)}]

ii) E(aX + bY + c) = aE(X) + bE(Y) + c

                        = aµx + bµy + c

iii) E(X^2) = V(X) + [E(X)]^2      {V(X) = E(X^2) - [E(X)]^2}

            = σx^2 + µx^2

Learn more about normal distribution here : brainly.com/question/4079902

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