Part 1

Consider $ n $ independent flips of a coin having probability $ p $ of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if $ n=5 $ and the sequence $ HHTHT $ is observed, then there are 3 changeovers. Find the expected number of changeovers for $ n $ flips. Hint: Express the number of changeovers as a sum of Bernoulli random variables.

Solution 1

The number of changeovers $ Y $ can be expressed as the sum of n-1 Bernoulli random variables:

$ Y=\sum_{i=1}^{n-1}X_i $.

Therefore,

$ E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i) $.

For Bernoulli random variables,

$ E(X_i)=p(E_i=1)=p(1-p)+(1-p)p=2p(1-p) $.

Thus

$ E(Y)=2(n-1)p(1-p) $.

Comments Solution 1:

Good solution with a appropriate explanation.

Solution 2

For n flips, there are n-1 changeovers at most. Assume random variable $ k_i $ for changeover,

$ P(k_i=1)=p(1-p)+(1-p)p=2p(1-p) $

$ E(k)=\sum_{i=1}^{n-1}P(k_i=1)=2(n-1)p(1-p) $

Critique on Solution 2:

The solution is correct. However, it's better to explicitly express $ k_i $ as a Bernoulli random variable. This makes it easier for readers to understand.

Solution 3

First, we define a Bernoulli random variable

$ X = \left\{ \begin{array}{ll} 0, & the change over does not occur\\ 1, & the change over occurs \end{array} \right. $

Then we can compute

$P(X = 1) = P(1-P)+(1-P)P = P-{P}^{2}+P-{P}^2=2P-2{P}^{2} $

$P(X = 0) = P•P+(1-P)(1-P) = {P}^{2}+1-2P+{P}^2 $

Define Y as the number of changes occurred in n flips, there exists at most n-1 changes

$E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i) $

$E(X_i)=p(X_i=1)=p(1-p)+(1-p)p=2p(1-p) $

Therefore, we have a final solution as

$ E(Y)=2(n-1)p(1-p) $.

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