Problem
I am trying to solve the following question (3.34 from Paul Meyer's "Introductory probability and Statistics", 2nd ed.):
The following (somewhat simple-minded) weather forecasting is used by an amateur forecaster. Each day is classified as "wet" or "dry" and the probability that any given day is the same as the preceding one is assumed to be a constant $\boldsymbol{p}$ ($0<p<1$). Based on past records, it is supposed that January 1 has a probability of $\beta$ of being "dry". Letting $\beta_n$ equal the probability that the $n$-th day of the year is "dry", obtain an expression for $\beta_n$ in terms of $\beta$ and $p$. Also evaluate $\lim\limits_{n \to \infty}\beta_n$ and interpret your result.
(Hint: Express $\beta_n$ in terms of $\beta_{n-1}$.)
My attempt of a solution
So far I've managed to write an equation
$$P(\text{dry}\mid\text{Jan 2})=P(\text{dry}\mid\text{Jan 1}) \cdot p + P(\text{wet}\mid\text{Jan 1}) \cdot (1-p)$$
or
\begin{gather*}
\beta_2=\beta \cdot p +(1-\beta)\cdot (1-p)=2\beta p +1-\beta-p\\
=1-p+\beta(2p-1)=p+(2p-1)(\beta-1)
\end{gather*}
By the same logic, I can write $\beta_n=p+(2p-1)(\beta_{n-1}-1)$. But after that I stumbled. The answer at the back of the book is $\beta_n=\frac{1}{2}+(2p-1)^n(\beta – \frac{1}{2})$. I do not have a clue as to how to arrive at that. What am I missing?
Best Answer
We start from $P_0=\beta$. The recursive relationship is $$\begin{aligned}P(D_n)&=P(D_n|D_{n-1})P(D_{n-1})+P(D_n|W_{n-1})P(W_{n-1})=\\ &=pP(D_{n-1})+(1-p)(1-P(D_{n-1}))\end{aligned}$$ If a limit exists, then for $n$ sufficiently large $P_n\approx P_{n-1}$ so $$P_{\infty}=\frac{1}{2}$$ Now observe: $$\begin{aligned} P_1&=\beta(2p-1)^\color{blue}{1}+(2p-1)^\color{red}{0}(1-p)\\ P_2&=\beta(2p-1)^\color{blue}{2}+(2p-1)^\color{red}{1}(1-p)+(2p-1)^\color{red}{0}(1-p)\\ P_3&=\beta(2p-1)^\color{blue}{3}+(2p-1)^\color{red}{2}(1-p)+(2p-1)^\color{red}{1}(1-p)+(2p-1)^\color{red}{0}(1-p)\\ &(...) \end{aligned}$$ The pattern is clear: $$\begin{aligned}P_n&=\beta(2p-1)^{n}+(1-p)\sum_{1\leq k\leq n}(2p-1)^{n-k}=\\ &=\beta(2p-1)^{n}+(1-p)(2p-1)^n\frac{(2p-1)^{-1}-(2p-1)^{-n-1}}{1-(2p-1)^{-1}}=\\ &=\beta(2p-1)^{n}+(1-p)(2p-1)^n\frac{1-(2p-1)^{-n}}{2(p-1)}=\\ &=\beta(2p-1)^{n}-\frac{(2p-1)^{n}-1}{2}\end{aligned}$$