# Use inclusion-exclusion to find explicit formula for $P(A_k)$. Where is independence used?

Probability with Martingales

This was answered here using bounds and here using PGF. I would like to try inclusion-exclusion.

Let $H_1, H_2, ...$ be independent events with $P(H_k) = p$ where $H_k$ means the kth toss is heads.

It seems that ($\cap$ omitted)

$$A_0 = H_1$$

$$A_1 = H_2 \cup H_3$$

$$A_2 = (H_4H_5) \cup (H_5H_6) \cup (H_6H_7) := \bigcup_{i=1}^{3} E_{2,i}$$

$$A_3 = (H_8H_9H_{10}) \cup ... \cup (H_{13}H_{14}H_{15}) := \bigcup_{i=1}^{6} E_{3,i}$$

$$\vdots$$

$$A_k = (H_{2^k}...H_{2^k + k-1}) \cup ... \cup (H_{2^{k+1} - k}...H_{2^{k+1} - 1}) := \bigcup_{i=1}^{2^k - k + 1} E_{k,i}$$

So by inclusion-exclusion, we have

$$P(A_0) = p$$

$$P(A_1) = p + p - p^2$$

$$P(A_2) = 3p^2 - 2p^3$$

$$P(A_3) = \ldots$$

It gets tricky starting $A_3$

$$P(A_3) = P\left(\bigcup_{i=1}^{6} E_{3,i}\right)$$

Attacking that directly is going to lead to problems.

$$P\left(\bigcup_{i=1}^{6} E_{3,i}\right)$$

$$= P\left(\bigcup_{i=1,4} E_{3,i} \cup \bigcup_{i=2,5} E_{3,i} \cup \bigcup_{i=3,6} E_{3,i} \right)$$

$$:= P(F_{3,1} \cup F_{3,2} \cup F_{3,3})$$

Each $F$ is a union of disjoint sets.

1. Also, I notice independence isn't being used. I guess the $H_k$'s and the $A_k$'s are independent. What about some subset of the the $E$'s? The $F$'s?
• You can get proper spacing by using \ldots instead of periods. – joriki Jun 10 '16 at 16:35
• Not checking all the calculations. For $P(A_2)$, is it $p^2 + p^2 + p^2 - p^3 - p^3 - p^4 + p^4$ instead? – BGM Jun 11 '16 at 5:11
• @BGM I'm a dumbass. Thanks! ^-^ I edited my post – BCLC Jun 14 '16 at 22:01

The events you are uniting are not independent, and they intersect in intricate ways, so using inclusion-exclusion on them will quickly lead to a mess. Take for instance the union $$H_8H_9H_{10} \cup H_9H_{10}H_{11}\cup\cdots\cup H_{13}H_{14}H_{15},$$ also known as $A_3$. Taking the sets $H_iH_{i+1}H_{i+2}$ in the union one at a time, they each have probability $p^3$, but intersecting them two at a time will yield events like $H_8H_9H_{10}\cap H_9H_{10}H_{11}=H_8H_9H_{10}H_{11}$ which have probability $p^4$ but also events like $H_8H_9H_{10}\cap H_{10}H_{11}H_{12}=H_8H_9H_{10}H_{11}H_{12}$ with probability $p^5$ and also events like $H_8H_9H_{10}\cap H_{13}H_{14}H_{15}$ with probability $p^6$. You will need to count how many there are of each type. Intersecting them three at a time will yield events with probabilities ranging from $p^5$ to $p^8$, and you're not even done with intersections four at a time and five at a time. It is not obvious whether the general form for $P(A_k)$ is compactly expressible in terms of $p$.
Better to use the bounding approach in the first answer you cited. The $p\ge\frac12$ case of the bounding argument uses the events $E_i$ defined in the hint; the $E_i$ have the advantage of being independent events, so an inclusion-exclusion argument (or finding the probability of a complement event) leads to a tidy expression. It is true that the bounding approach is not calculating the exact prob of $A_k$; rather, it is calculating upper and lower bounds on $P(A_k)$, which are sharp enough to apply the Borel-Cantelli lemma.
• @BCLC Your edit gets you closer to the bounding approach. In your expression $F_{3,1}\cup F_{3,2}\cup F_{3,3}$, the advantage is that each $F$ is the union of independent events. However, the three $F$'s are not disjoint from each other and they still intersect messily. If you throw away all but the first of these $F$'s, keeping just $F_{3,1}:= H_{8,9,10}\cup H_{11,12,13}$ then the two events $E_{3,1}:=H_{8,9,10}$ and $E_{3,2}:=H_{11,12,13}$ you've retained are exactly as in the Hint and you still have the advantage that the two $E$'s are independent. (Note that the $E$'s are not disjoint.) – grand_chat Jun 14 '16 at 23:46
• For $p<\frac12$ the bounding approach to $P(A_k)$ uses the crude bound $P(A_k)= P(\bigcup F_{k,i})\le \sum P(F_{k,i})$. You can think of the inequality $P(\cup F_i)\le \sum P(F_i)$ as a one-term truncation of inc-exc. For $p\ge\frac12$ the bounding approach computes $P(A_k)\ge P(\bigcup E_{k,i}) \stackrel{(*)}= 1-P(\bigcap E_{k,i}^c)$. Note that it is possible to prove (*) using inc-exc. – grand_chat Jun 16 '16 at 22:56