Q.7.9 - A First Course in Probability

Question

A coin having probability $p$ of landing on heads is flipped $n$ times. Compute the expected number of runs of heads of size 1 , of size 2 , and of size $k,1\le k\le n$.

Step-by-Step Solution

Answer

The expected number of runs of heads of size 1 , of size 2 , and of size $k,1\le k\le n$ is $(n-k)p^k(1-p{)}^2+p^k\left(1-p^2\right).$

Step 1: Given Information

Coin's probability $=p$

Number of flipping times $=n,$

Step 2: Explanation

Define random variable $I_j$ that marks whether the run of Heads of size $k$ started from the trial number $j$ or not, $j=1,2,\dots n-k+1$.

Now define an indicator variable as follows:

$I_j=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ (if a }\mathrm{k}\text{ sized run begins at }{\mathrm{j}}^{\text{th }}\text{ position }\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ (otherwise) }\end{array}\right\}$

So, if we define $X$ as the random variable that marks the total number of $k-$runs of Heads, So we have that:

$X=\sum _{j=1}^{n-k+1}{I}_j$

Using the linearity of the expectation, so:

$E\left(X\right)=\sum _{j=1}^{n-k+1}E\left(I_j\right)$

$=E\left(\sum _{j=1}^{n-k+1}{I}_j\right)$

$=P\left(I_1=1\right)+\sum _{j=2}^{n-k}P\left(I_j=1\right)+P\left(I_{n-k+1}\right)$

Step 3: Explanation

Now solving above expression:

$=P\left(I_1=1\right)+\sum _{j=2}^{n-k}P\left(I_j=1\right)+P\left(I_{n-k+1}\right)$

$=p^k(1-p)+(n-k-1)p^k(1-p{)}^2+p^k(1-p)$

$=2p^k(1-p)+(n-k-1)p^k(1-p{)}^2$

$=(n-k)p^k(1-p{)}^2+p^k\left(1-p^2\right)$

Step 4: Final Answer

Hence, the expected number of runs of heads of size as per question is:

$(n-k)p^k(1-p{)}^2+p^k\left(1-p^2\right)$