Variance for uniform distribution

Question

I have to calculate the variance:

$\mathsf{Var}(aX + bY + cZ + d)$

I know that a,b,c,d are positive and that $X, Y$ and $Z$ have a common uniform distribution on $\big\{(x,y,z) \in \{0,3\}^3: x + y + z\text{ is a twofold}\big\}$.

The question is to calculate $\mathsf{Var}(aX + bY + cZ + d)$

I used the common rules to calculate the variance but I don't see why the solution must be

${\mathsf{Var}(aX + bY + cZ + d) } = {a^2\mathsf{Var}(X) + b^2\mathsf{Var}(Y) + c^2\mathsf{Var}(X) \\+ ab\mathsf{Cov}(X,Y) + ac\mathsf{Cov}(X,Z) + bc\mathsf{Cov}(Y,Z)}$

Can anyone help me with this? I assume that X, Y, and Z are numbers and dependent.

Answer 1

Your expression isn't quite right. There should be $2$'s in front of every covariance term. It might help to first derive the following expression for two random variables:

Var($aX + bY$) = $a^2$Var($X$) + $b^2$Var($Y$) + $2ab$Cov($X,Y$)

As a start,

Var($aX + bY$) = $\mathbb{E}[(aX + bY - a\mu_X - b\mu_Y)^2] \\ = \mathbb{E}[(aX - a\mu_X)^2] + \mathbb{E}[(bY - b\mu_Y)^2] + 2\mathbb{E}[(aX - a\mu_X)(bY - b\mu_Y)]$

Answer 2

Hinging on the interpretation of "$x+y+z$ is a twofold" means "the sum of the terms is even".

$$\{(x,y,z)\in\{0,3\}^3: (x+y+z)\in2\Bbb Z\} = \{(0,0,0),(0,3,3),(3,0,3),(3,3,0)\}$$

Of course, if "twofold" means something else, you will have a different set.   Once you have determined what set you have, you can evaluate the variance of each variable, and covariances of each pair.

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From there it is a matter of using the bilinearity of covariance, which infers: $$\def\Var{\mathsf{Var}} \def\Cov{\mathsf{Cov}} \Var(aX+bY+cZ+d)~{=~{a^2\Var(X)+b^2\Var(Y)+c^2\Var(Z) \\+ 2ab\Cov(X,Y)+2ac\Cov(X,Z)+2bc\Cov(Y,Z)} }$$