GO Classes DPP | STATISTICS |EXPECTATION

Answer 1

We want to find $E[X]$, where $X$ is the total number of runs in $n=10$ tosses.

A "run" is a sequence of identical outcomes. A new run begins at a position $i$ if the outcome at $i$ is different from the outcome at $i-1$.

• The first toss (at position $i=1$ ) always starts the first run. So, we have 1 run guaranteed.
   
• For every subsequent toss $i$ (from $2$ to $10$), we can define an indicator variable $I_i$ :
   
• $I_i=1$ if $\operatorname{Toss}_i \neq \operatorname{Toss}_{i-1}$ (a new run starts at position $i$ )
   
• $I_i=0$ if Toss $_i=$ Toss $_{i-1}$ (the previous run continues)

The total number of runs $X$ is the first run plus all the new runs that start:

$$
X=1+\sum_{i=2}^{10} I_i
$$

The expected value of the sum is the sum of the expected values:

$$
\begin{gathered}
E[X]=E[1]+E\left[\sum_{i=2}^{10} I_i\right] \\
E[X]=1+\sum_{i=2}^{10} E\left[I_i\right]
\end{gathered}
$$

The expected value of an indicator variable is just the probability of the event it indicates:

$$
E\left[I_i\right]=P\left(I_i=1\right)=P\left(\text { Toss }_i \neq \text { Toss }_{i-1}\right)
$$

Since the coin is fair, there are four equally likely possibilities for the pair of tosses $\left(\right.$ Toss $_{i-1}$, Toss $\left._i\right)$ :

1. $(\mathrm{H}, \mathrm{H})$
    
2. $(\mathrm{H}, \mathrm{T}) \leftarrow$ New run
    
3. $(\mathrm{T}, \mathrm{H}) \leftarrow$ New run
    
4. ($\mathrm{T}, \mathrm{T}$ )

The probability of a new run (H, T or T, H) is $\frac{2}{4}=\frac{1}{2}$. Therefore, $E\left[I_i\right]=\frac{1}{2}$ for all $i$ from $2$ to $10$.

 

\begin{aligned}
&\text { We have } 9 \text { indicator variables (from } I_2 \text { to } I_{10} \text { ), and each has an expected value of } \frac{1}{2} \text {. }\\\\
&\begin{gathered}
E[X]=1+\sum_{i=2}^{10} \frac{1}{2} \\\\
E[X]=1+\left(9 \times \frac{1}{2}\right) \\\\
E[X]=1+4.5 \\\\
E[X]=5.5
\end{gathered}
\end{aligned}

Comments

Alternate way

Here coin is fair. so it is uniform distribution

for uniforn distribution E[x] = (min + max) / 2 

Min possible runs 1 for all tosses are either H or T  [ H ,H, H, H ,H, H, H, H, H, H ] or  [ T,T,T,T,T,T,T,T,T,T]

Max possible runs 10 for alternate H and T (HTHTHTHTHT)

so , E[x] is 5.5

@GO ClassesOP  if this worng method or misconception please correct me?

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Bro lets take example of n = 4 runs thas means either 4 T or 4 H but as wek know its max of no of consecutive T or H so 4 T or H can appear twice which has more probab than having 1 run with either have alternating TH or HT pattern ie 2 cases in comparsion 4 cases for the above (n=4) so its not uniform

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ie TTTTHTTTTH = still 4 and TTTTHHHTTT which is 4 or TTTTHHHHTH which is 4