# How to reason about independence of combinations of events?

Suppose sets $$A , B,$$ and $$D$$ are independent. Is it guaranteed that $$A \cap B^c \cap D$$ is independent from $$B^c \cup D^c$$?

Isn't $$B^c$$ ($$B$$ complement, or $$B$$ not happening) giving me information about $$B^c \cup D^c$$? (A) I know that $$B$$ complement happened so it should increase the chances of $$B^c \cup D^c$$ happening, thus they are not independent.

Is my statement (A) correct?

• Hint: What is the set $(A\cap B^c\cap D)\,\cap\,(B^c\cup D^c)$? I believe you can simplify this expression. Drawing a Venn diagram may be helpful.
– whuber
Commented May 25, 2020 at 19:28
• I found that the intersection is empty. Is that right?If so, this would mean that they are dependent? Because they are disjoint events. Commented May 25, 2020 at 20:31
• Any element that is in $A$ and $D$ but not in $B$ lies in this set.
– whuber
Commented May 25, 2020 at 20:39
• But to intersect it should be outside D, so D and Dc never intersect , right? Commented May 25, 2020 at 21:11
• You have forgotten that $B^c\cup D^c$ includes elements of $B^c,$ which may be in $D.$ The Venn diagram will make this clear.
– whuber
Commented May 25, 2020 at 21:14

For inspiration, let's examine Venn diagrams of each set.

Elements of $$\mathcal{X}=A\cap B^c\cap D$$ are (a) in $$A;$$ (b) not in $$B;$$ and (c) in $$D$$. This region is highlighted in yellow.

Elements of $$\mathcal{Y}=B^c\cup D^c$$ are either not in $$B$$ or not in $$D$$ (or both). This includes a lot of the diagram, so instead I have highlighted its complement -- namely, everything that is not in this set. The complement of $$\mathcal Y$$ is $$B\cap D$$ (this evident equality is one of DeMorgan's Laws):

Now, two sets $$\mathcal{X}$$ and $$\mathcal Y$$ are independent if and only if $$\mathcal{X}$$ and $$\mathcal{Y}^c$$ are independent. But since $$\mathcal{X}$$ and $$\mathcal{Y}^c$$ are disjoint, the probability of their intersection is zero. When both $$\mathcal X$$ and $$\mathcal{Y}^c$$ have nonzero probability, the rule for independence won't hold.

From this insight we may construct an example where the two sets $$\mathcal X$$ and $$\mathcal Y$$ are not independent. (Once we have found this example, we needn't repeat any of the foregoing analysis: the example alone is enough to resolve the problem. But I thought you might want to understand how such examples can be cooked up and what kind of thinking goes into that.)

For this example I will put one element into $$\mathcal{X}$$ and one into $$\mathcal{Y}^c:$$ since neither $$\mathcal X$$ nor $$\mathcal{Y}^c$$ can be empty, yet are disjoint, that's as simple as an example possibly could be.

Looking at the two diagrams, then, let's put (say) the number $$0$$ into $$A$$ and $$D$$ but not $$B$$ and (say) the number $$1$$ into both $$B$$ and $$D.$$ One way to do this is to define

$$A = \{0\},\quad B = D = \{1\}.$$

Define a probability measure for $$A\cup B \cup D = \{0,1\}$$ by giving $$\{0\}$$ and $$\{1\}$$ equal chances of $$1/2.$$ Note that any set in this probability space can have probability $$0,$$ $$1/2,$$ or $$1,$$ but no other value is possible. Here's a Venn diagram showing the set elements:

Computing

$$\Pr(\mathcal{X}) = \Pr(A\cap B^c\cap D) = \Pr(\{0\}) = \frac{1}{2},$$

$$\Pr(\mathcal{Y}) = \Pr(B^c\cup D^c) = \Pr(\{0\}) = \frac{1}{2},$$

we see that the independence of $$\mathcal{X}$$ and $$\mathcal{Y}$$ would mean (by definition)

$$\Pr(\mathcal{X}\cap\mathcal{Y}) = \Pr(\mathcal{X})\Pr(\mathcal{Y}) = \frac{1}{2}\times \frac{1}{2} = \frac{1}{4}.$$

Since a probability of $$1/4$$ is impossible, $$\mathcal{X}$$ and $$\mathcal{Y}$$ cannot be independent.

• Also since Pr(X) = Pr(Y) is different from Pr(X and Y), events are dependent.BTW Thank you, Whuber. Commented May 26, 2020 at 13:46
• Correct. But I didn't have to compute $\mathcal{X}\cap\mathcal{Y}$ :-).
– whuber
Commented May 26, 2020 at 14:01