# Prove that the sum and the absolute difference of 2 Bernoulli(0.5) random variables are not independent

Let $$X$$ and $$Y$$ be independent $$Bernoulli(0.5)$$ random variables. Let $$W = X + Y$$ and $$T = |X - Y|$$. Show that $$W$$ and $$T$$ are not independent.

I know that I have to show that $$P(W, T)$$ is not equal to $$P(W)P(T)$$, but finding the joint distribution is hard. Please help.

• Re: "finding the joint distribution is hard:" have you made a table? Label the rows with values of $X$, the columns with values of $Y$, and in the cells put the values of $T,$ $W,$ and the associated probabilities. Collect your results into a new table with rows labeled with $T$ and columns labeled with $W:$ put the total probabilities into the entries. That depicts the entire joint distribution of $(T,W).$ You can then draw your conclusion with a visual inspection. No operation is any more difficult than computing $1/2\times 1/2.$
– whuber
Nov 27 '18 at 23:19
• I get it now. I was looking for an elegant, mathematical expression for the joint distribution, but I now realize that I can just enumerate the sample space and the probabilities easily. Thanks, @whuber.
– MSE
Nov 28 '18 at 0:35

The product of the marginal distributions is defined on $$\{0,1,2\} \times \{0,1\}$$. You can plug in any of the $$6$$ possible pairs, and get a nonzero number out.

However, the joint density is defined on a smaller space: $$\{0,0\} \cup \{1,1\} \cup \{2, 0\}.$$

To disprove independence, take any $$(w,t)$$ pair not in the above, and plug it in to $$P(W,T)$$ and $$P(W)P(T)$$. You will see that, for that particular pair: $$P(W,T) = 0 \neq P(W)P(T).$$

Alternatively, because you're dealing with a small space, you can just go ahead and compute every probability and just check every possible pair.

When T = 0, W = 0 or 2; when T = 1 then W = 1. So T and W are not independent. See Independence of $$X+Y$$ and $$X-Y$$

• I want to mark your solution as correct, too! Thanks, @user158565.
– MSE
Nov 28 '18 at 0:39