Joint PMF of 3 Discrete RVs

Question

I have 3 joint PMF $(X,Y,Z)$ respectively as follow:

$(0,0,1)$

$(0,1,0)$

$(1,0,0)$

$(1,1,1)$

Each with $p = 1/4$

I am asked to compute the $COV(X,Y), V(X+Y)$ and find out whether only $X,Y$ are independent or $X,Y,Z$ are all independent

My problem is I do not know how to extract $X,Y$ from those $4$ points since $Z$ always stick around. I can do this problem with joint PMF of 2 RVs. I have looked around but can't see any similar problem such as Joint PMF of more than $2$ RVs. Also, i don't know how to deal with independence of $X,Y,Z$ once at a time

Any help would be appreciated. Thank you very much :)

Answer 1

Guide:

To find $P(X=x, Y=y)$, you can sum up the $z$ variables:

$$P(X=x, Y=y)= \sum_{z} P(X=x, Y=y, Z=z)$$

Similarly,

$$P(X=x)= \sum_{y,z} P(X=x, Y=y, Z=z)$$

For example, $P(X=0)=p(0,0,1)+p(0,1,0)=\frac12$

Intuitively for independence of the $3$ random variables, think of whether having $2$ of them give u information about the third.

Answer 2

Outcome space is $\Omega=\{(0,0,1),(0,1,0),(1,0,0),(1,1,1)\}$.

$\sigma$-algebra is the powerset of $\Omega$.

Probability measure $P$ on $\langle\Omega,\wp(\Omega)\rangle$ is prescribed by $A\mapsto\frac14|A|$ for $A\in\wp(\Omega)$ so that $P(\{\omega\})=\frac14$ for every $\omega\in\Omega$.

We have the following random variables:

$X:\Omega\to\mathbb R$ is prescribed by $(\omega_1,\omega_2,\omega_3)\mapsto\omega_1$

$Y:\Omega\to\mathbb R$ is prescribed by $(\omega_1,\omega_2,\omega_3)\mapsto\omega_2$

$Z:\Omega\to\mathbb R$ is prescribed by $(\omega_1,\omega_2,\omega_3)\mapsto\omega_3$

Now everything is well modeled and we can start.

E.g. we can find $\mathbb EXY$ by stating that $$\mathbb EXY=\sum_{\omega\in\Omega}X(\omega)Y(\omega)P(\{\omega\})=\sum_{\omega\in\Omega}\omega_1\omega_2\frac14=[0\cdot0+0\cdot1+1\cdot0+1\cdot1]\frac14=\frac14$$

I leave the rest to you.