We have a fair four-sided dice with faces $1, 2, 3$ and $4$. We roll this dice five times with the results being
the random variables $A, B, C, D, E$.
We define the following four new random variables:
• $W = 2A + B − 1$
• $X = A(A + C)$
• $Y = (A + B + C)$ mod $4$
• $Z = (A + C)(D + E)$
Determine:
- Which pairs and triples of random variables $W, X, Y, Z$ are independent?
- What is the expected value of $W + X + Y + Z$?
In first part I calculated pairs and got that $W,Y$ and $X,Y$ are independent and I couldnt calculate for Z because there are so $4^4$ options so I guess there has to be some trick how to calculate independence in this problem.
And triples I couldnt calculate because there are so many options. If there is some faster way I would like to hear.
I tried to compute 2. part but Im not sure if it is correct and I got $48.75$
Edit
From lulu's comment I checked extreme cases and Im not sure if it is correct but I got
$W,X$ are dependent because of $P(W = 2 | X = 3)$
$W,Y$ are independent
$W,Z$ are dependent $P(W = 2 | Z = 64)$
$X,Y$ are independent
$X,Z$ are dependent $P(X = 2 | Z = 64)$
$Y,Z$ cannot determine (need help with that)
For triples, it is computationally hard so if there is some trick I would like to know it.
Edit-2
$E[W] = E[2A + B - 1] = 2E[A] + E[B] - 1 = 2 \cfrac{1+2+3+4}{4} + \cfrac{1+2+3+4}{4} - 1 = 6.5$
$E[X] = E[A(A+C)] = \cfrac{1^2+2^2+3^2+4^2}{4} + \cfrac{(1+2+3+4)*(1+2+3+4)}{16} = \cfrac{220}{16} = 13.75$
$E[Y] = \cfrac{1+2+3+4}{4} + \cfrac{1+2+3+4}{4} + \cfrac{1+2+3+4}{4} = 3.5$ mod $4$
$E[Z] = E[(A+C)(D+E)] = E[AD] + E[AE] + E[CD] + E[CE] = 25$
$E[W+X+Y+Z] = 48.75$