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Measure, Integration & Real Analysis, 2021a

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384 Chapter 12 Probability <strong>Measure</strong>s<br />

12.8 Example independent events: coin tossing<br />

Suppose Ω = {H, T} 4 , where H and T are symbols that you can think of as<br />

denoting “heads” and “tails”. Thus elements of Ω are 4-tuples of the form<br />

ω =(ω 1 , ω 2 , ω 3 , ω 4 ),<br />

where each ω j is H or T. Let F be the collection of all subsets of Ω, and let<br />

P =(counting measure on Ω)/16, as we expect from a fair coin toss.<br />

Let<br />

A = {ω ∈ Ω : ω 1 = ω 2 = ω 3 = H} and B = {ω ∈ Ω : ω 4 = H}.<br />

Then A contains two elements and thus P(A) = 1 8 , corresponding to probability 1 8<br />

that the first three coin tosses are all heads. Also, B contains eight elements and thus<br />

P(B) = 1 2 , corresponding to probability 1 2<br />

that the fourth coin toss is heads.<br />

Now<br />

P(A ∩ B) =<br />

16 1 = P(A) · P(B),<br />

where the first equality holds because A ∩ B consists of only the one element<br />

(H, H, H, H) and the second equality holds because P(A) = 1 8 and P(B) = 1 2 .<br />

The equation above shows that A and B are independent events.<br />

If we toss a fair coin many times, we expect that about half the time it will be<br />

heads. Thus some people mistakenly believe that if the first three tosses of a fair<br />

coin are heads, then the fourth toss should have a higher probability of being tails,<br />

to balance out the previous heads. However, the coin cannot remember that it had<br />

three heads in a row, and thus the fourth coin toss has probability 1 2<br />

of being heads<br />

regardless of the results of the three previous coin tosses. The independence of the<br />

events A and B above captures the notion that the results of a fair coin toss do not<br />

depend upon previous results.<br />

12.9 Example independent events: product probability space<br />

Suppose (Ω 1 , F 1 , P 1 ) and (Ω 2 , F 2 , P 2 ) are probability spaces. Then<br />

(Ω 1 × Ω 2 , F 1 ⊗F 2 , P 1 × P 2 ),<br />

as defined in Chapter 5, is also a probability space.<br />

If A ∈F 1 and B ∈F 2 , then (A × Ω 2 ) ∩ (Ω 1 × B) =A × B. Thus<br />

(P 1 × P 2 ) ( (A × Ω 2 ) ∩ (Ω 1 × B) ) =(P 1 × P 2 )(A × B)<br />

= P 1 (A) · P 2 (B)<br />

=(P 1 × P 2 )(A × Ω 2 ) · (P 1 × P 2 )(Ω 1 × B),<br />

where the second equality follows from the definition of the product measure, and<br />

the third equality holds because of the definition of the product measure and because<br />

P 1 and P 2 are probability measures.<br />

The equation above shows that the events A × Ω 2 and Ω 1 × B are independent<br />

events in F 1 ⊗F 2 .<br />

<strong>Measure</strong>, <strong>Integration</strong> & <strong>Real</strong> <strong>Analysis</strong>, by Sheldon Axler

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