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suppose X and Y are random variables such that

E(X)=1 (Y)=2 V(X)=1 V(Y)=2 Cov(X,Y)=1

by using above values following expression are evaluated

E(X+2Y)=p

EXY)=q

Vat(X-2Y+1)=r

find pq+r

 

 

 

asked in Probability by Veteran (32.4k points)  
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IF X,Y are independent than  Cov(X,Y) $=$ 0  . Here its $\neq$ 0 . So we they are dependent .

E(X+2Y) = E(X) + 2E(Y) = 5 = p

Cov(X,Y) = E(XY) - E(X)E(Y)  on solving E(XY) = 3 =q .

Var(X - 2Y + 1) = Var(X) + 4Var(Y) + Var(1) + 2Cov(X,-2Y) + 2Cov(X,1) + 2C(Y,1)    [Var[constant]= 0 ]

Cov(X,Y) = 0 if X , Y are independent . So Cov (X,1) = 0 , Cov(Y) =0 .

                         = Var(X) + 4Var(Y) -4Cov(X,Y)       [ Cov(X,-Y) = -Cov(X,Y) ]

On putting the values we get  r=5

Therefore  pq+r = 20
answered by Boss (7.1k points)  


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