6 Independence



RECOMMENDED READING:

B & H Chapters 2.5 - 2.9.





NEW NOTATION

  • Simplify a long product of \(n\) numbers \(x_1,x_2,...,x_n\): \[\prod_{i = 1}^n x_i = x_1 \cdot x_2 \cdot \cdots \cdot x_n\]

  • Simplify the intersection of \(n\) events, \(\{A_1,A_2,...,A_n\}\): \[\cap_{i=1}^n A_i = A_1 \cap A_2 \cap \cdots \cap A_n\]







6.1 Discussion

We’ve been exploring the concept and calculation of conditional probabilities, i.e. the re-calculation of the plausibility of an event \(A\) in light of information/evidence/data \(B\):

\[P(A|B) = \frac{P(A \cap B)}{P(B)}\]

Of course, the recalculation depends upon the relationship between \(A\) and \(B\). Consider an example.









EXAMPLE 1: warm-up on your own

I have 6 cool pairs of shoes - 4 pairs are blue, 2 pairs are red. Thus there are 12 shoes total: (red shoe image, blue shoe image)

Pick a shoe at random and define events: \(B\) = blue, \(R\) = red, \(L\) = goes on the left foot.

  1. Calculate the following probabilities: \(P(B)\), \(P(B|R^c)\), \(P(B|L)\).
  2. Does knowing that the shoe goes on the left foot give you any additional information about the shoe being blue?













Independence of 2 events

Events \(A\) and \(B\) are independent if \[P(A|B) = P(A)\] Equivalently, \(A\) and \(B\) are independent if \[P(A \cap B) = P(A)P(B)\]









EXAMPLE 2

  1. Prove that the two definitions of independence are equivalent:
    • show that the second definition (\(P(A \cap B) = P(A)P(B)\)) follows from the first (\(P(A|B) = P(A)\))
    • show that the first definition (\(P(A|B) = P(A)\)) follows from the second (\(P(A \cap B) = P(A)P(B)\))
  2. Identifying independence
    According to a December 2019 poll: Bernie Sanders (who is running for the Democratic presidential nomination) has an overall approval rating of 54.6% but a 70% approval rating among people between the ages of 18 and 29. Is Sanders’ approval independent of voter age?













Independence of multiple events

Multiple events \(\{A_1,A_2,...,A_n\}\) are independent if they satisfy the following 2 properties.

  1. \(\{A_1,A_2,...,A_n\}\) are pairwise independent. That is, each pair of events \(A_i\) & \(A_j\) are independent: \[P(A_i \cap A_j) = P(A_i)P(A_j)\]

  2. \(\{A_1,A_2,...,A_n\}\) are mutually independent: \[P(\cap_{i=1}^n A_i) = \prod_{i=1}^nP(A_i)\]

NOTE:

  • It’s possible for \(\{A_1,A_2,...,A_n\}\) to be pairwise independent but not mutually independent. (Not all rectangles are squares.)
  • If \(\{A_1,A_2,...,A_n\}\) are mutually independent, then they are pairwise independent. (All squares are rectangles.)









6.2 Exercises

  1. Independence of 3 events (Example 2.5.5 from the book)
    Flip 2 independent, fair coins. Thus \(S = \{HH, HT, TH, TT\}\) where each outcome is equally likely. Define events \(A\) = 1st coin is Heads, \(B\) = 2nd coin is Heads, and \(C\) = both tosses have the same result.

    1. Prove that \(\{A,B,C\}\) are pairwise independent.
    2. Prove that even though \(A,B,C\) are pairwise independent, they’re not independent. That is, prove that \(P(A\cap B \cap C) \ne P(A)P(B)P(C)\).
    3. Explain in words why \(A,B,C\) are pairwise independent, but not independent.


    Solution

    1. \(P(A \cap B) = P(HH) = 1/4 = P(A)P(B)\)
      \(P(A \cap C) = P(HH) = 1/4 = P(A)P(C)\)
      \(P(B \cap C) = P(HH) = 1/4 = P(B)P(C)\)
    2. \(P(A \cap B \cap C) = P(HH) = 1/4 \ne 1/8 = P(A)P(B)P(C)\)
    3. Knowing any 1 event doesn’t give us information about another. However, knowing the combination of 2 events tells us about the third. For example, \(P(C | (A \cap B)) = 1 \ne P(C)\).



  1. Utilizing independence
    In honor of Joe DiMaggio’s 56 game hitting streak, mlb.com runs an annual “Beat the Streak” competition. Consider a simplified setting:

    • Pick a player with batting average \(p\). That is, the player hits the ball with probability \(p\).
    • Assume the player gets 4 at bats (chances to hit the ball) per game.
    • Assume that the outcomes of each at bat are independent.

    Use these assumptions when answering the questions below and use the following notation: \(H_i\) = at their \(i\)th at bat, the player gets a hit.

    1. How reasonable are our simplifying assumptions? How might we relax these? (Your answers might differ depending on how well you know the game of baseball!)
    2. In a single given game, what’s the probability that the player…
      • gets a hit on both their 1st and 2nd at bat
      • gets at least one hit (HINT: what’s the complement of this event?)
    3. What’s the probability the player gets at least one hit per game for 57 games in a row?
    4. Calculate part c exactly for a pretty solid player with a batting average of \(p=0.3\).
    5. How good does a player’s batting average need to be for them to have even a 2.5% chance of beating the streak?


    Solution

    1. The number of at bats per game can vary (it’s not reasonable to assume it’s always 4). It’s maybe unreasonable to assume that at bats are independent.
    2. .
      \[\begin{split} P(H_1 \cap H_2) & = P(H_1)P(H_2) = p^2 \\ P(\text{at least 1}) & = 1 - P(\text{none}) \\ & = 1 - P(H_1^c \cap H_2^c \cap H_3^c \cap H_4^c) \\ & = 1 - P(H_1^c)P(H_2^c)P(H_3^c)P(H_4^c) \\ & = 1 - (1-p)^4 \\ \end{split}\]
    3. Let \(A_i\) be the event that the player gets at least one hit in game \(i\)
      \(P(A_1 \cap A_2 \cap ... \cap A_{57}) = P(A_1)P(A_2)\cdots P(A_{57}) = (1 - (1-p)^4)^{57}\)
    4. \((1 - (1-0.3)^4)^{57} \approx 0.000000160\)
    5. \(0.025 = (1 - (1-p)^4)^{57}\) when \(p = 1 - (1-0.025^{1/57})^{1/4}\approx 0.4997\)



PAUSE: We’ll talk about 1 and 2 as a class before moving on to 3.



  1. Truel!
    A, B, C are in a 3-cornered paintball competition. A has the worst aim, hitting their target only 30% of the time. C is on-target 50% of the time, while B never misses. The players aim paintballs at targets of their choice in succession, in the order A, B, C until only 1 remains unstained.

    Let’s play! In doing so, make the following assumptions:
    • B and C never miss on purpose.
    • All attempts are independent.
    • If A doesn’t first aim a paintball at B, B won’t first aim a paintball at A (Same goes for C.)
    • If A aims at B and misses, B will try to aim at A on their turn. (Same goes for C.)

    Now that we’ve played, let’s think about the optimal strategy. Who should A aim at in their first shot: B, C, or the sky? The following property might come in handy: \[\sum_{i=0}^\infty r^i = 1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r} \;\;\; \text{ for $0 < r < 1$}\] For example, if \(r = 0.5\):

    \[\sum_{i=0}^\infty 0.5^i = 1 + 0.5 + 0.25 + 0.125 + \cdots = \frac{1}{1-0.5} = 2\]



  1. Extra practice: Independent \(\ne\) Disjoint
    NOTE: This exercise will appear on your next homework, so we won’t discuss as a class.
    Draw one card from a standard 52-card deck. Let \(A\) be the event that card is an ace. Let \(B\) be the event that the card is a heart.

    1. Are \(A\) and \(B\) disjoint events? Provide evidence.
    2. Are \(A\) and \(B\) independent events? Provide evidence.
    3. In general, prove the following: If \(A\) and \(B\) are disjoint and \(P(A)>0\), \(P(B)>0\), then \(A\) and \(B\) are dependent (not independent).
    4. In words, provide an intuitive explanation of why disjoint events are not independent.