Question

Let X₁, X_2, X₃ be uniform random variables on the interval (0, 1) with Cov(X_i , X_j ) = 1/24 for _i, j = 1, 2, 3, _i i != j. Calculate the variance of 2X₁ + X₂ − X_3.

Answer 1

Since $X_1, X_2, X_3$  are uniform random variables on the interval (0, 1), their probability distribution functions are,

 $f(X_1)=f(X_2)=f(X_3)= 1,\space 0\lt X_i\lt1, \space i=1,2,3$

$0, \space elsewhere$

$E(X_1)=\displaystyle\int^1_0 X_1dx_1={X_1^2\over 2}|^1_0={1\over2}$

$E(X_1^2)=\displaystyle\int^1_0 X_1^2dx_1={X_1^3\over3}|^1_0={1\over3}$

$Var(X_1)=E(X_1^2)-(E(X_1))^2={1\over3}-({1\over2})^2={1\over12}$

Since $X_2$ and $X_3$ have the same probability distribution function as $X_1$, they both will have the same variance as $X_1$, that is,

$Var(X_1)=Var(X_2)=Var(X_3)={1\over12}$ and

$Cov(X_i,X_j)={1\over24}\space for\space i=1,2,3,\space j=1,2,3, \space i\not=j$

Now,

$Var(2X_1 + X_2 − X_3)=E[(2X_1 + X_2 − X_3)-(2\mu_1+\mu_2-\mu_3)]^2$ , from definition

$=E[2(X_1-\mu_1)+(X_2+\mu_2)-(X_3-\mu_3)]^2$

$=4Var(X_1)+Var(X_2)+Var(X_3)+4Cov(X_1,X_2)-4Cov(X_1,X_3)-2Cov(X_2,X_3)$

$=4({1\over12})+{1\over12}+{1\over12}+4({1\over24})-4({1\over24})-2({1\over24})={6\over12}-{2\over24}={5\over12}$

Therefore,  the variance of 2X₁ + X₂ − X₃ is ${5\over12}$.