# Random variables defined by other random variables?

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!

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

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

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

• My I ask how you computed $P(B=1|A=0)$? Commented Aug 1, 2021 at 18:44
• Yea (btw, fixed that to $B=0$), $P(B=0|A=0)=\frac{P(B=0\cap A=0)}{P(A=0)} = \frac{P(X=0\land Y=0}{P(X=0\land Y=0}=1$
Commented Aug 1, 2021 at 21:13
• the only possible case if $A=0$ is given is that $X=Y=0$, but that means $B=0$ in a probability of $1$
Commented Aug 1, 2021 at 21:15
• Yeah, my friend has confirmed this as well when I asked him about the reasoning. Thank you! Commented Aug 1, 2021 at 21:16
• No problem, good luck
Commented Aug 1, 2021 at 21:17

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.

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 $$B = A\mod 2.$$