Problem 1

Solution:

We start with the solution we already⁸ obtained for $E(Y_{i})$ which is

\[ E\left ( Y_{i}\right ) =1+{\displaystyle \sum \limits _{k=1}^{i-1}} E\left ( Y_{k}\right ) P_{ik}\]

Let $E\left ( Y_{i}\right ) =x_{i}$ hence the above can be written as

\[ x_{i}=1+{\displaystyle \sum \limits _{k=1}^{i-1}} x_{k}P_{ik}\]

But $P_{ik}=\frac {1}{i}$ then the above becomes

\[ x_{i}=1+\frac {1}{i}{\displaystyle \sum \limits _{k=1}^{i-1}} x_{k}\]

Multiply by $i$ the above becomes

\[ i\ x_{i}=i+{\displaystyle \sum \limits _{k=1}^{i-1}} x_{k}\]

Therefore, we obtain the following equations for $i=1\cdots r$

For $i=1$

\begin{equation} x_{1}=1 \tag {1}\end{equation}

For $i=2$

\begin{equation} 2x_{2}=2+x_{1} \tag {2}\end{equation}

For $i=3$

\begin{equation} 3x_{3}=3+x_{1}+x_{2} \tag {3}\end{equation}

For $i=4$

\begin{equation} 4x_{4}=4+x_{1}+x_{2}+x_{3} \tag {4}\end{equation}

etc...

Now evaluate (2)-(1) and (3)-(2) and (4)-(3), etc... we obtain the following equations

(2)-(1) gives

\begin{align} 2x_{2}-x_{1} & =2+x_{1}-1\nonumber \\ x_{2} & =\frac {1+2x_{1}}{2} \tag {5}\end{align}

(3)-(2) gives

\begin{align*} 3x_{3}-2x_{2} & =3+x_{1}+x_{2}-2-x_{1}\\ 3x_{3}-2x_{2} & =1+x_{2}\\ x_{3} & =\frac {1+3x_{2}}{3}\end{align*}

etc... Hence we see that for the $r^{th}$ term we obtain

\begin{equation} x_{r}=\frac {1+r\ x_{r-1}}{r}\nonumber \end{equation}

Hence

\begin{equation} x_{r}=\frac {1}{r}+x_{r-1} \tag {6}\end{equation}

Now replace $r$ by $r-1$ in the above we obtain

\[ x_{r-1}=\frac {1}{r-1}+x_{r-2}\]

replace the above in (6) we obtain

\begin{equation} x_{r}=\frac {1}{r}+\left ( \frac {1}{r-1}+x_{r-2}\right ) \tag {7}\end{equation}

And again, in the above, $x_{r-2}=\frac {1}{r-2}+x_{r-3\text { }}$hence (7) becomes

\begin{equation} x_{r}=\frac {1}{r}+\left ( \frac {1}{r-1}+\left ( \frac {1}{r-2}+x_{r-3\text { }}\right ) \right ) \tag {8}\end{equation}

and so on, until we get to $x_{1}=1$, hence we obtain

\[ \fbox {$x_{r}=\frac {1}{r}+\frac {1}{r-1}+\frac {1}{r-2}+\cdots +1$}\]

Hence

\[ x_{r}={\displaystyle \sum \limits _{k=1}^{r}} \frac {1}{k}\]

Which is the harmonic series. Now, it is known that⁹

\[ \lim _{r\rightarrow \infty }x_{r}=\log \left ( r\right ) -\gamma \]

Where $\gamma $ is Euler Gamma constant given by

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