MadeEasy Test Series: Probability - Expectation

Question

Suppose X and Y are random variables such that
E(X) = 1, Var (X) = 1, E(Y) = 2, Var (Y) = 2, Cov (X, Y) = 1
By using above values the following expressions are evaluated
(i) E(X + 2Y) = p

(ii) E(XY ) = q
(iii) Var(X – 2Y + 1) = r
The value of pq + r is ___________.

What is the formulae for point (iii) ?

Comments

Var(aX+bY)= a^2Var(X) + b^2Var(Y) + 2abCov(X,Y)

Var(Constant)= 0

So Var(aX+bY+C)=Var(aX+bY) +Var(X) = Var(aX+bY)

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It is - sign in between

Can you tell me the answer here?

I applied following formulae from Wikipedia but answer is not coming same

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If b is -ve then we put -b instead of b.

Did you put E(XY)=E(X)*E(Y) ?

If yes then that is where the mistake is done.

We apply that formula when X and Y are independent.

How to know whether X and Y are independent is when Cov(X,Y) =0.

Here Cov(X,Y) !=0 so we know that they are dependent.

So the formula becomes E(X,Y)=E(x)*E(Y)+ Cov(X,Y)

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The answer is 20. I have already taken this test :P

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My approach to (ii) point.From the above formula i mentioned i solved like

Var(X – 2Y + 1) = $ 1^2*V(X) + (-2)^2*V(Y) -2ab(COV(a,b)) $

=$ 1*1 + 4*2 -2*(1)(-2)(1) $

=$1+8+4$

=13

P.S:- Here i have taken b as -ve .Why is that incorrect as per image as shared?

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Got my mistake. Thank you!!!:)