Example 4.2.2. A miner is trapped in a mine containing 3 doors. The first door leads to a tunnel that will take him to safety after 3 three hours walking. The second door leads to a tunnel that will return him to the mine after five 5 hours of walking. The third door leads to a tunnel that will return him to the mine after 7 seven hours. If we assume that the miner is at all times equally to close any of the doors, what is the expected length of time until he reaches to safety?

Example 4.2.3. An urn contains \(n\) white balls and \(m\) black balls. One ball is selected
randomly at a time until the first white ball is selected. Find the expectation of the black balls selected.

Let \(X\) be the number of black balls selected.

We want \(E(X)\) \[\text {Let}\hspace {0.5cm} Y= \begin {cases} 1, &\text {if the white ball is selected first time.}\\\\ 0, &\text {if the black ball is selected first time}\\ \end {cases} \] \[P(Y=1)=\frac {n}{n+m}\hspace {0.5cm} \text {and}\hspace {0.5cm} P(Y=0)=\frac {m}{n+m}\] Denote \(E(X)\) by \(N(n,m)=\) expected number of black balls selected. \[E(X \mid Y=1)=0,\hspace {0.5cm} E(X \mid Y=0)=1+N(n,m-1).\] \[E(X)=N(n,m)=E(X \mid Y=1)P(Y=1)+E(X \mid Y=0)P(Y=0)\] \[N(m,n)=0.\frac {n}{n+m}+\Big (1+N(n,m-1)\Big )\frac {m}{n+m}\] \[\implies \hspace {0.5cm} N(m,n)=\Big (1+N(n,m-1)\Big )\frac {m}{n+m}................... (*)\] \[\text {if}\hspace {0.4cm} m=1,\hspace {0.4cm} N(n,1)=\Big (1+N(n,0)\Big )\frac {1}{n+1}=\frac {1}{n+1}\] \[N(n,1)=\frac {1}{n+1}\] \[\text {if}\hspace {0.4cm} m=2,\hspace {0.5cm} N(n,2)=\Big (1+N(n,1)\Big )\frac {2}{n+2}=\frac {2}{n+1}\] \[\text {if}\hspace {0.4cm} m=3,\hspace {0.5cm} N(n,3)=\Big (1+N(n,2)\Big )\frac {3}{n+3}=\frac {3}{n+3}\]
\[\therefore \hspace {0.5cm} N(n,m)=\frac {m}{n+1}\]
\[\therefore \hspace {0.5cm} E(X)=N(n,m)=\frac {m}{n+1}\]

Example 4.2.4. Let \(X\) be the number of independent trials required to get first success where the probability of success is constant in each trial is \(P\). Then \(X\) has geometric distribution. Find the variance of \(X\). \[f_X(x)=P(X=x)=P(1-P)^{x-1},\hspace {0.5cm} x=1,2,........\] \[E(X)=\frac {1}{P}\] \[\text {if}\hspace {0.3cm} Y= \begin {cases} 1, &\text {if the first trial is a success}\\\\ 0, &\text {if the first trial is a failure}\\ \end {cases} \]

\[P(Y=1)=P,\hspace {0.5cm} P(Y=0)=1-P\] \[\Var (X)=E(X^2)-\Big (E(X)\Big )^2\] \begin {align*} E(X^2) &=E(X^2 \mid Y=0)P(Y=0)+E(X^2 \mid Y=1)P(Y=1)\\ &=E(1+X)^2(1-P)+(1)^2P\\ &=E\Big (1+2X+X^2\Big )(1-P)+P\\ &=\Big (1+2/P+E(X^2)\Big )(1-P)+P \end {align*}

\[E(X^2)=\frac {(P+2)}{P}\hspace {0.1cm}(1-P)+(1-P)E(X^2)+P\] \begin {align*} \implies \hspace {0.5cm} P\,E(X^2) &=\frac {(P+2)(1-P)+P^2}{P}\\ &=\frac {P-P^2+2-2P+P^2}{P} =\frac {2-P}{P}\\\\ \implies \hspace {0.5cm} E(X^2) &=\frac {2-P}{P^2} \end {align*}

\begin {align*} \therefore \hspace {0.5cm} \Var (X) &=E(X^2)-\Big (E(X)\Big )^2\\ &=\frac {2-P}{P^2}-\frac {1}{P^2}\\ &={\frac {1-P}{P^2}} \end {align*}

OR \begin {align*} E(X^2) &=\sum ^{\infty }_{x=1}x^2(1-P)^{x-2}\\ &=P(1-P)\sum ^{\infty }_{x=1}\Big (x(x-1)(1-P)^{x-2}\Big )+\sum ^{\infty }_{x=1}xP(1-P)^{x-1}\\ &=P(1-P)\sum ^{\infty }_{x=1}\frac {d^2}{dP^2}(1-P)^x-P\sum ^{\infty }_{x=1}\frac {d}{dP}(1-P)^x\\ &=P(1-P)\frac {d^2}{dP^2}\sum ^{\infty }_{x=1}(1-P)^x-P\frac {d}{dP}\sum ^{\infty }_{x=1}(1-P)^x\\ &=P(1-P)\frac {d^2}{dP^2}\Bigg (\frac {1}{1-(1-P)}-1\Bigg )-P\frac {d}{dP}\Bigg (\frac {1}{1-(1-P)}-1\Bigg )\\ &=P(1-P)\frac {d^2}{dP^2}\Bigg (\frac {1}{P}-1\Bigg )-P\frac {d}{dP}\Bigg (\frac {1}{P}-1\Bigg )\\ &=P(1-P).\frac {2}{P^3}+\frac {1}{P}=\frac {2(1-P)}{P^2}+\frac {1}{P}=\frac {2(1-P)+P}{P^2}\\ \implies \hspace {0.5cm} E(X^2) &=\frac {2-P}{P^2}\\ \end {align*}