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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.

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  • $\begingroup$ 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 $1/2\times 1/2.$ $\endgroup$ – whuber Nov 27 '18 at 23:19
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    $\begingroup$ 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. $\endgroup$ – MSE Nov 28 '18 at 0:35
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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.

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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 $X-Y$

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

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