I am working on a problem from probability theory and seem to be getting stuck. The problem is the following:
P is a distribution {1, 0, 1} all with probability 1/3
Let (X, Y, Z) ~ p as random variables.
My question: How do you set up the distribution for this? Is this just
P(X = -1) = 1/3 P(X = 0) = 1/3 P(X = 1) = 1/3
or is it
P(X = -1) = 1/3 P(Y = 0) = 1/3 P(Z = 1) = 1/3
The next question is to find E(X + Y + Z) so I imagine it is the latter?
It's more like your first one. For the random variable $X$ alone, it holds that $$X = \begin{cases} -1, & \text{w/ probability }1/3 \\ 0, & \text{w/ probability } 1/3 \\ 1, & \text{w/ probability } 1/3\end{cases}$$ and the same is true for $Y$ and also for $Z$.
Here's a useful way to think of it: imagine that $X$ is the result of rolling a 3-sided die, but its three sides were $-1$, $0$, and $1$. Suppose now that you have three of such dice (let's say a white one, a black one, and a green one); in this way, $X$ is the result of rolling the white one, $Y$ the result of rolling the black one, and $Z$ the result of rolling the green one.
Your question is a bit unclear, so I'm not sure if you're asked for the joint distribution of $X, Y,$ and $Z$, or if you're asked about the distribution of $X + Y + Z$. In either case, your task is essentially the same; you must explore all $3^3 = 27$ combinations of what the three dice can be. If you're asked about the distribution of $X + Y + Z$, note that it can be any value between $-3$ and $3$ inclusively, but not with equal probability; it's much more likely to be $0$ than $3$, for instance.
If you have the distribution of $X + Y + Z$, you can use it directly to compute $\mathbb E[X + Y + Z]$, or you can use the linearity of expectation and instead compute the equivalent expression $\mathbb E[X] + \mathbb E[Y] + \mathbb E[Z]$.