# Can one infer independence by simple reasoning/intuition?

From my recent experience in probability, it feels as though independence is something we "discover" from the system via the equation:

$$P(A)*P(B)=P(A\cap B)$$

Could one ever conclude independence from the "system" by intuition? Is it wise to conclude independence for events that are "seemingly" independent? What would be some interesting examples where this would fail.

• No, you cant. There is two kind of independent random variables: if under a test the hypothesis " are independent" holds, or if we just create a theoretic problem where we, axiomatically, define two random variables as independent. Aug 21, 2016 at 2:36
• Sometimes it is obvious. But intuition can be hazardous in mathematics, even though it is a useful tool. So, even if it is "obvious" best to show your work. A very elementary example might be: A couple has two children $A$ and $B$. If one of $A$, $B$ is known to be a girl, what is the probability they are both girls? Without training, the average person will say $0.5$. It stems from the fact that the events "two girls" and "at least one girl" are dependent even though the sexes of the individual children are assumed independent. Aug 21, 2016 at 3:20
• @DavidP Sorry, the events "two girls" and "at least one girl" are dependent because the fact that we already have "at least one girl" gives us information about the "two girls" event? (I did a calculation and yes those are dependent events haha). Aug 21, 2016 at 8:24
• They are dependent because the occurrence of one changes the probability of the other. Having information is not equivalent to dependence. For example if I roll a die and I tell you it is a multiple of 3, what is the probability it is even? It's the same as if I hadn't told you it was a multiple of 3. "Rolling a multiple of 3" and "Rolling an even" are independent events even though the occurrence of one reduces the sample space a lot. Aug 21, 2016 at 8:59

The OP asks for an intuitive understanding of $$\mathbb{P}(A)\mathbb{P}(B)=\mathbb{P}(A\cap B)$$ for independent events $$A$$ and $$B$$. Here it is:

Let $$(\Omega, \mathcal{A}, \mathbb{P})$$ be a probability space. Assume $$\mathbb{P}(B) > 0$$. Since $$\mathbb{P}(\Omega)=1$$, we can write the above equation as

$$\displaystyle\frac{\mathbb{P}(A)}{\mathbb{P}(\Omega)}=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}.$$

The left hand side is the "proportion of $$A$$" in $$\Omega$$, while the right hand side is the "proportion of $$A$$" that is in $$B$$. From the equation above, we can see that the probability of the happening of $$A$$ remains unchanged inside $$B$$ (if you zoom in $$B$$ and treat it as a new probability space with $$\mathbb{P}_B(B)=1$$). Event $$A$$ will happen with a probability $$\mathbb{P}(A)$$ independent of whether your space is $$\Omega$$ or $$B$$. Thus is the independence of $$A$$ and $$B$$.

Imagine the $$\Omega$$ as a unit disk, and $$A$$ occupies the left semi-circle, so that $$\mathbb{P}(A)=\frac{1}{2}$$. Now let $$B$$ be a concentric circle inside $$\Omega$$. Then $$A$$ has exactly the same pattern in $$B$$ as in $$\Omega$$, and $$\mathbb{P}_B(A)=\frac{1}{2}$$ inside $$B$$. So in this respect $$B$$ is somewhat "irrelevant" for $$A$$: $$B$$ "looks like" $$\Omega$$, so its happening or not get unnoticed for $$A$$.

• Beautiful and highly underrated answer!
– BCLC
May 26, 2018 at 17:10

As @Masacroso has mentioned, independence in math is by definition the formula you've mentioned. If you're referring to real life systems, yes: Independence of events is a hypothesis that can be tested empirically - the relative frequency of a particular event may have been independent from that of another event.

Is it wise to conclude independence empirically? Depends on the payoffs. If you're a decision-maker, you don't only care about being right or wrong; you care a lot too about the size of the payoff if you're right, and that of the penalty if you're wrong. It has been speculated, for instance, that a large contributor to the housing crisis was banks' assumption that mortgage failures of one household are independent of those of others. While this had been true in the past, it changed very quickly. The payoff from assuming independence was a relatively small gain from being able to make more loans; these gains and more were wiped out from the penalty for being wrong. (See Nassim Taleb's books for related discussion.)

An interesting example is as follows

Suppose there are two fair six faced dice, say I and II. Both dice are rolled together, let $X$ denotes the face shown by die I and $Y$ denotes the face shown by die II.

Define $Z:= X+Y$ and $U :=$ remainder when $Z$ is divided by $6$.

Surprisingly in the above construction random variables $U$ & $X$ are independent.

For example, let $A= \{X= 1\}$ & $B=\{ U= 1\} = \{ (X,Y)= (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$

Then $\mathbb{P}(A)= \dfrac{1}{6}$, $\mathbb{P}(B)= \dfrac{1}{6}$ and $\mathbb{P}(A\cap B)= \mathbb{P}(\{ (X,Y)= (1,6)\})= \dfrac{1}{36}= \mathbb{P}(A) * \mathbb{P}(B)$.