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In a roll of die, if $X$ is the number on the first die and $Y$ is the number on second die, then determine whether the random variable $X+Y$ and $X-Y$ are independent.

The covariance between the two turned out to be $\mathrm{Var}(X) + \mathrm{Var} (Y)$. So zero covariance would mean that $\mathrm{Var}(X)$ and $\mathrm{Var}(Y)$ is zero, that is, no spread. But we also know that zero covariance does not imply independence. I really cannot think of a way to prove independence between the two.

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  • $\begingroup$ The contrivance is Var(X) - Var(Y). Also, do the dies have the same number of sides? $\endgroup$
    – Kozolovska
    Commented Oct 21, 2018 at 7:47
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    $\begingroup$ Independence is P(XY)=P(X)P(Y). You can calculate all probabilities for 36 outcomes and show that the equation holds. $\endgroup$
    – keiv.fly
    Commented Oct 21, 2018 at 7:59
  • $\begingroup$ @keiv.fly Do we calculate X-Y and X+Y probabilities for different cases and then use P((X-Y)(X+Y))=P(X-Y)P(X+Y)? Isn't there a shorter, more formal method to do the same? $\endgroup$
    – S.Rana
    Commented Oct 21, 2018 at 8:12
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    $\begingroup$ @t.f I'm not aware of the term 'contrivance' yet. Yes, they have same no of sides. $\endgroup$
    – S.Rana
    Commented Oct 21, 2018 at 8:13
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    $\begingroup$ If your dice are known, e.g. with standard numbering from 1 to n, then (X+Y) and (X-Y) are not independent. A simple way of thinking about disproving independence is that you only need to show that there exists at least one outcome of X+Y such that X-Y is known with absolute certainty. if X+Y = 2, then (X-Y) is known and it has to be 0. $\endgroup$
    – NofP
    Commented Oct 21, 2018 at 11:07

2 Answers 2

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They're not: If $X+Y=12$ then both rolls were sixes, so $X-Y=0$. So you have:

$$1 = \mathbb{P}(X-Y =0|X+Y=12) \neq \mathbb{P}(X-Y =0) = \frac{1}{6}.$$

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When checking if the random variables are independent or not, the first thing needing to check is the range of the random variables. If the range of one random variable varies according to values of other random variables, then they are not independent, and stop. Otherwise, need to check further.

In this problem, when A+B = 12, A-B can have only one value 0. if A+B = 11, A-B can be -1 and 1. So they are not independent.

This skill is very useful for test/exam in probability and statistics, because it can save you a lot of time.

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    $\begingroup$ +1. One might extend your rule to replace "range" by "any easily computed property." $\endgroup$
    – whuber ♦
    Commented Oct 21, 2018 at 20:36

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