Three random variables, X1, X2 and X3 are normally distributed such that X1-N(0,4), X2-N(1,3), and X3-N(3,2).
The Covariance between X1 and X2 is -0.35, the covariance between both X2 and X3 and also X1 and X3 is = 0.
A new variable, Y is created such that Y= 2X1 + 3X2 - 4X3. Define the distribution Y will follow as completely as possible.
Not sure what to do with normal distribution and covariances??!!
Since $Y$ is a linear combination of Normally distributed Random variables, $Y$ too will be normally distributed with the following parameters:
$\mu_Y = E(Y) = E(2X_1 + 3X_2 - 4X_3) = 2E(X_1) + 3E(X_2) - 4E(X_3) = 2\mu_{X_1} + 3\mu_{X_2} - 4\mu_{X_3} = 2*0 + 3*1 - 4*3 = -9$
$\sigma^2_Y = var (Y) = var(2X_1 + 3X_2 - 4X_3) = 4var(X_1) + 9var(X_2) + 16var(X_3) + 12Cov(X_1,X_2) - 24Cov(X_2, X_3) - 16Cov(X_3,X_1) = 4\sigma_{X_1}^2 + 9\sigma_{X_2}^2 + 16\sigma_{X_3}^2 + 12\sigma_{X_1 X_2}^2 - 24\sigma_{X_2 X_3}^2 - 16\sigma_{X_3 X_1}^2 = 4*4 + 9*3 + 16*2 + 12 * -0.35 - 24*0 - 16*0 = 70.8$
So $Y \sim \mathscr N(-9,70.8)$