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Let $X$ and $Y$ be two random variables with a Bernoulli Distribution of $\frac{1}{2}$. Now, define random variables $A = X + Y$ and $B = |X - Y|$

Are $A$ and $B$ dependent vs. independent?

Now, obviously if they are independent, then $$P(A=0, B=0) = P(A=0)*P(B=0)$$ and $$P(A=1, B=1) = P(A=1)*P(B=1)$$ but I'm not quite sure how to compute $P(A=0, B=0)$ nor am I sure how to compute $P(A=0)$, $P(A=0)$ or $P(B=1)$ for example. The notion of a random variable defined by another random variable is confusing to me, would appreciate some help, thanks in advance!

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  • $\begingroup$ You are asking about independence of $A$ and $B$, but the second half of your post tries to assess independence of $X$ and $Y$ $\endgroup$
    – angryavian
    Commented Jul 31, 2021 at 21:04
  • $\begingroup$ @angryavian thank you! Corrected. $\endgroup$
    – JakeDrone
    Commented Jul 31, 2021 at 21:19
  • $\begingroup$ Don't forget $A$ can equal $2$. $\endgroup$
    – Brian Tung
    Commented Jul 31, 2021 at 21:22

3 Answers 3

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$A=0\iff X=0\land Y=0,\\ A=1\iff (X=0\land Y=1)\lor(X=1\land Y=0),\\ A=2\iff(X=1\land Y=1)$

Therefore: $P(A=0)=P(A=2)=0.25, P(A=1)=0.5$

$B=0\iff (X=0\land Y=0)\lor (X=1\land Y=1),\\ B=1\iff (X=1\land Y=0)\lor(X=0\land Y=1)$

Therefore $P(B=0)=P(B=1)=0.5$

But $P(A=0,B=0)=P(B=0|A=0)P(A=0)=0.25\neq P(A=0)\cdot P(B=0)=0.25\cdot 0.5$

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  • $\begingroup$ My I ask how you computed $P(B=1|A=0)$? $\endgroup$
    – JakeDrone
    Commented Aug 1, 2021 at 18:44
  • $\begingroup$ Yea (btw, fixed that to $B=0$), $P(B=0|A=0)=\frac{P(B=0\cap A=0)}{P(A=0)} = \frac{P(X=0\land Y=0}{P(X=0\land Y=0}=1$ $\endgroup$
    – e.ad
    Commented Aug 1, 2021 at 21:13
  • $\begingroup$ the only possible case if $A=0$ is given is that $X=Y=0$, but that means $B=0$ in a probability of $1$ $\endgroup$
    – e.ad
    Commented Aug 1, 2021 at 21:15
  • $\begingroup$ Yeah, my friend has confirmed this as well when I asked him about the reasoning. Thank you! $\endgroup$
    – JakeDrone
    Commented Aug 1, 2021 at 21:16
  • $\begingroup$ No problem, good luck $\endgroup$
    – e.ad
    Commented Aug 1, 2021 at 21:17
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I'll assume $X$ and $Y$ are independent, which you didn't state in your question.

In this case, an easy way to see that $A$ and $B$ are dependent is the following: suppose $B = 0$, then that means $X = Y$ (since $\lvert X - Y \rvert = 0$), so either $X=Y=0$ or $X=Y=1$, which implies $A = 0$ or $A = 2$, so in particular, $A$ cannot be $1$. This means that $P(A = 1, B = 0) = 0$: there is no possible pair $(X, Y)$ such that $A=1$ and $B=0$.

However, the product $P(A = 1) P(B = 0)$ is not zero. If you are unsure about this, we can compute explicitly $P(A = 1) = P(X + Y = 1) = P(X=0, Y=1) + P(X=1, Y=0) = 1/4 + 1/4 = 1/2$ and $P(B = 0) = P(X = Y) = P(X = 0, Y = 0) + P(X = 1, Y = 1) = 1/4 + 1/4 = 1/2$ so therefore $P(A=1)P(B=0) = \frac12 \cdot \frac12 = \frac14$.

So we have $P(A = 1, B = 0) = 0$ while $P(A=1)P(B=0) = \frac14$, which shows that they $A$ and $B$ are dependent.

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We can establish the dependence between variables with no explicit computation of the probabilities. Consider the following small table containing all possible outcomes.

$$XYAB$$ $$0 0 0 0$$ $$0 1 1 1$$ $$1 0 1 1$$ $$1 1 2 0$$

This immediately shows that $B$ is a function of $A$, therefore dependent. The reverse is not true, because $B=0$ does not determine A, which could be $0$ or $2$.

We may express the relationship with particular functions containing the three points $(A,B)$ $$(0,0),(1,1),(2,0).$$ Some options are a parabola $$B=2A-A^2,$$ a sinus $$B=\sin\left(A\frac{\pi}{2}\right),$$ or a remainder $$B = A\mod 2.$$

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