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A coin with heads probability $p$ is flipped $n$ times. A "run" is a maximal sequence of consecutive flips that are all the same. For example, the sequence HTHHHTTH with $n=8$ has five runs, namely H, T, HHH, TT,H. Show that the expected number of runs is $$1+2(n-1)p(1-p).$$

I have tried to use some generating function on this but calculus got pretty messy and didn't work.

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2 Answers 2

up vote 6 down vote accepted

I'd use indicator random variables. For $1\leq j\leq n-1$, let $Z_j$ take the value $1$ if the $j$th and $j+1$st coin tosses are different, and take the value $0$ otherwise. Then the number of runs is $R=1+\sum_{j=1}^{n-1}Z_j$ and its expected value is $$\mathbb{E}(R)=1+ \sum_{j=1}^{n-1}\, \mathbb{E}(Z_j)=1+(n-1) 2p(1-p).$$ I will leave the calculation $\mathbb{E}(Z_j)=2p(1-p)$ as an exercise!

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Using indicator random variable is a good way to solve this question(see the solution above). What's more, we can calculate the variance of the number of runs:$$Var\left( R \right) =Var\left( \sum _{ j=1 }^{ n-1 }{ { Z }_{ j } } \right) =E\left\{ { \left( \sum _{ j=1 }^{ n-1 }{ { Z }_{ j } } \right) }^{ 2 } \right\} -{ E\left( \sum _{ j=1 }^{ n-1 }{ { Z }_{ j } } \right) }^{ 2 } $$ $Z_j$ is a Bernoulli distribution, $Z_j$ and $Z_k$ are independent if $\left| j-k \right| \ge 2$(notice that $Z_j$ and $Z_{j+1}$ are dependent).

We can get $Var\left( R \right)=2pq\left( 2n-3-2pq\left( 3n-5 \right) \right) $.

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