# With $X$ and $Y$ being two independent $\text{Bernoulli(1/3)}$ rvs, show whether $U = |Y-X|,~V = X+Y$ are independent or not

Let $$X$$ and $$Y$$ be two independent $$\text{Bernoulli(1/3)}$$ random variables. Define random variables $$U$$ and $$V$$ as $$U = |Y-X|, \hspace{5mm} V = X+Y$$ Are $$U$$ and $$V$$ independent?

I am new to the field of Statistics and very excited to work on this site.

One way is, I calculate the joint PMF table of $$U$$ and $$V$$ and Marginal PMF using joint PMF and check for each value in table. But since the question just asked for "Are $$U$$ and $$V$$ independent or not?", Is there any short way of answering the question?

Table looks something like this which shows that $$U$$ and $$V$$ aren't independent $$\begin{array}{|c|c|c|c|c|} \hline \frac{V}{U} & 0 & 1 & 2 & f_U(u)\\ \hline 0 & 4/9 & 0 & 1/9 & 5/9\\ \hline 1 & 0 & 4/9 & 0 & 4/9\\ \hline f_V(v) & 4/9 & 4/9 & 1/9 & 1\\ \hline \end{array}$$

• You don't need the entire table, once you check that $P(U=0,V=0)\ne P(U=0)P(V=0)$. Commented Jun 17 at 16:01

## 1 Answer

Your method is a good way of showing lack of independence and will work in many other examples.

Another way in this particular example, which would work in some other cases when $$X$$ and $$Y$$ are integers, is to see that $$U=|Y-X|$$ and $$V=X+Y$$ are integers which can each be even or odd. But $$U$$ and $$V$$ must have the same parity as each other, and so are not independent.

• I didn’t get what you are trying to say Commented Jun 17 at 17:54
• He says that if $U$ is odd, then $V$ is odd (same if you replace odd with even). So knowing somethingh about $U$ tells you something about $V$. That is dependence! Commented Jun 17 at 19:38
• @kjetilbhalvorsen Ah... OK, Thanks! Commented Jun 19 at 5:54