Question

For another approach to Theoretical Exercise 7.33, let Tr denote the number of flips required to obtain a run of r consecutive heads.

(a) Determine $E\left[T_r\mid T_{r-1}\right]$.

(b) Determine $E\left[T_r\right]$in terms of $E\left[T_{r-1}\right]$.

(c) What is $E\left[T_1\right]$?

(d) What is $E\left[T_r\right]$?

Short answer

1. It has been determined that $E\left[T_r\mid T_{r-1}\right]=T_{r-1}+1+(1-p)E\left[T_r\right]$
   
2. It has been determined that$E\left[T_r\right]=\frac{1}{p}+\frac{1}{p}E\left[T_{r-1}\right]$.
3. It has been found that $E\left[T_1\right]=\frac{1}{p}$.
4. It has been found that$E\left[T_r\right]=\sum _{i=1}^r {\left(\frac{1}{p}\right)}^i$.

Step-by-step solution

Given information (Part a)

Tr denote the number of flips required to obtain a run of r consecutive heads

Solution (Part a)

$p=$the probability that a coin lands on heads

$E\left[T_r\right]=$the number of flips required to obtain a run of r consecutive heads.

Find $E\left[T_r\mid T_{r-1}\right]$

$E\left[T_r\mid T_{r-1}\right]=T_{r-1}+1+(1-p)E\left[T_r\right]$

Final answer (Part a)

It has been determined that$E\left[T_r\mid T_{r-1}\right]=T_{r-1}+1+(1-p)E\left[T_r\right]$

Given information (Part b)

Tr denote the number of flips required to obtain a run of r consecutive heads

Solution (Part b)

Find $E\left[T_r\right]$

If expectations on both sides of (a) yields,

$E\left[T_r\right]=E\left[T_{r-1}\right]+1+(1-p)E\left[T_r\right]$

$=\frac{1}{p}+\frac{1}{p}E\left[T_{r-1}\right]$

Final answer (Part b)

It has been determined that$E\left[T_r\right]=\frac{1}{p}+\frac{1}{p}E\left[T_{r-1}\right]$̣

Given information (Part c)

Tr denote the number of flips required to obtain a run of r consecutive heads

Solution (Part c)

Find $E\left[T_1\right]$

Substitute 1 for r in part (b).

$E\left[T_1\right]=E\left[T_{1-1}\right]+1+(1-p)E\left[T_1\right]$

$=E\left[T_0\right]+1+(1-p)E\left[T_1\right]$

$=\frac{1}{p}+\frac{1}{p}E\left[T_0\right]$

$=\frac{1}{p}$

$\left[E\left[T_0\right]=0\right]$

Final answer (Part c)

It has been found that$E\left[T_1\right]=\frac{1}{p}$

Given information (Part d)

Tr denote the number of flips required to obtain a run of r consecutive heads

Solution (Part d)

Find $E\left[T_r\right]$

$E\left[T_r\right]=\frac{1}{p}+\frac{1}{p}E\left[T_{r-1}\right]$

$=\frac{1}{p}+\frac{1}{p}\left[\frac{1}{p}+\frac{1}{p}E\left[T_{r-1}\right]\right]$

$=\left(\frac{1}{p}\right)+{\left(\frac{1}{p}\right)}^2+{\left(\frac{1}{p}\right)}^{2}E\left[T_{r-2}\right]$

$=\left(\frac{1}{p}\right)+{\left(\frac{1}{p}\right)}^2+{\left(\frac{1}{p}\right)}^3+{\left(\frac{1}{p}\right)}^{3}E\left[T_{r-3}\right]$

$=\sum _{i=1}^r {\left(\frac{1}{p}\right)}^i+{\left(\frac{1}{p}\right)}^{r}E\left[T_0\right]$

$=\sum _{i=1}^r {\left(\frac{1}{p}\right)}^i$

Final answer (Part d)

It has been found that$E\left[T_r\right]=\sum _{i=1}^r {\left(\frac{1}{p}\right)}^i$