The pearls of AP Statistics 27

Sep 16

The pearls of AP Statistics 27

category: Agresti & Franklin, Agresti and Franklin, AP Stats and Business Stats, ST104A, ST104B, Statistics 1, Statistics 2, The College Board 2 Comments

Independence of events: intuitive definitions matter

First and foremost: independence of an AP Statistics course from Math is nonsense. Most of Stats is based on mathematical intuition.

Independent events

The usual definition says: events $A,B$ are called independent if

(1) $P(A\cap B)=P(A)P(B).$

Figure 1. Independence illustrated - click to view the video

You can use it formally or you can try to find a tangible interpretation of this definition, which I did. In Figure 1, the sample space is the unit square. Let $A$ be the rectangle delimited by red lines, of width $a$ and height $1$ . Since in this illustration probability of an event is its area, we have $P(A)=a\times 1=a$ . Similarly, let $B$ be the rectangle delimited by blue lines, of width $1$ and height $b$ , so that $P(B)=b\times 1=b$ . Obviously, the intersection $A\cap B$ has area $ab$ which equals $P(A)P(B).$ Equation (1) is satisfied and $A,B$ are independent. When the rectangle $A$ moves left and right and/or the rectangle $B$ moves up and down, the independence condition is preserved. We have a visual illustration of the common explanation that "what happens to one event, does not affect the probability of the other".

In Mathematica, enter the command

Animate[ParametricPlot[{{0.2 + a, t}, {0.4 + a, t}, {t, 0.3 + b}, {t,
0.6 + b}}, {t, 0, 1}, PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True, Frame -> True,
PlotStyle -> {Red, Red, Blue, Blue}, Mesh -> False], {a, -0.15,
0.55}, {b, -0.25, 0.35}, AnimationRunning -> False]

Choose "Forward and Backward" and then press both Play buttons. Those who don't have Mathematica, can view my video.

The statement "If A and B are dependent events, then so are A and the complement of B" (Agresti and Franklin, p.237) is not so simple. Here is the formal proof of the complementary statement ("dependent" is replaced with "independent"; $S$ denotes the sample space): if (1) is true, then

$P(A\cap B^c)=P(A\cap (S\setminus B))=P(A)-P(A\cap B)=P(A)-P(A)P(B)=P(A)P(B^c).$

Reading equation (1) from left to right: in practice, if we know that events are independent, we can find the probability of the joint event $A\cap B$ by multiplying individual probabilities $P(A),P(B).$

Reading equation (1) from right to left: in theory, if we want our events to be independent, we can define the probability of the joint event $P(A\cap B)$ by multiplying individual probabilities $P(A),P(B).$

Why there is division in the definition of conditional probability?

Figure 2. Conditional probability

Golovkin crushed Brook, and I am happy. Let $A$ be the event that the fight did not end in the first round. Suppose we know that the fight did not end in the first round, we just don't know the score for the round. Let $B,C,D$ be the events that Golovkin scored more, Brook scored more and there was a tie, respectively. Our sample space, based on the information we have, is limited to $A$ but the probabilities we are interested in do not sum to one:

$P(A\cap B)+P(A\cap C)+P(A\cap D)=P(A)$

To satisfy the completeness axiom, we divide both sides by $P(A)$ :

$P(A\cap B)/P(A)+P(A\cap C)/P(A)+P(A\cap D)/P(A)=1.$

This explains why conditional probabilities are defined by

(2) $P(B|A)=P(A\cap B)/P(A),$

$P(C|A)=P(A\cap C)/P(A),$

$P(D|A)=P(A\cap D)/P(A).$

If $A,B$ are independent, from (1) we see that $P(B|A)=P(B)$ . The multiplication rule $P(A\cap B)=P(B|A)P(A)$ is a consequence of (2) and not an independent property.

Raising the bar

Ins and outs of quantitative courses of University of London