Q. 3.22 - A First Course in Probability

Question

As a simplified model for weather forecasting, suppose that the weather (either wet or dry) tomorrow will be the same as the weather today with probability $p$. Show that the weather is dry on January $1$, then $P_n$, the probability that it will be dry $n$ days later, satisfies

$P_n=(2p-1)P_{n-1}+(1-p)$  $n\ge 1$

$P_0=1$

Prove that:

  $P_n=\frac{1}{2}+\frac{1}{2}(2p-1{)}^n$  $n\ge 0$

Step-by-Step Solution

Answer

By getting recursion, use the formula of total probability on $P_n$ using $P_{n-1}$, $P_{n-1}^c$

The explicit formula is proved by the principle of  mathematical induction.

Step1: Mathematical induction.

Mathematical induction is a type of mathematical proof. It is mostly used to demonstrate that a proposition $P\left(n\right)$ holds for every natural number $n=0,1,2,3,4,5,...,$ i.e., that the overall assertion is a series of infinitely many examples.

Step2: Evaluate P n

$P_n$, After January $1$, the weather will be dry for a few days.

Define the probabilities:

$P\left(A_n\right)=P_n$

$P\left(A_n\mid A_{n-1}\right)=p$

$P\left(A_n^c\mid A_{n-1}^c\right)=p$

$P_0=1$

Total probability formula:

$P\left(A_n\right)=P\left(A_n\mid A_{n-1}\right)P\left(A_{n-1}\right)+P\left(A_n\mid A_{n-1}^c\right)P\left(A_{n-1}^c\right)$     $n\ge 1$

first term on right hand is namely p

formula for the probability of a complement gives:

$$\begin{gathered} P\left(A_{n-1}^c\right)=1-P\left(A_{n-1}\right) \\ =1-P_{n-1} \end{gathered}$$

$$\begin{gathered} P\left(A_n\mid A_{n-1}^c\right)=1-P\left(A_n^c\mid A_{n-1}^c\right) \\ =1-p \end{gathered}$$

Transferring total probability formula into:

$P_n=p{P}_{n-1}+(1-p)\left(1-P_{n-1}\right) n\ge 1$

which is equal to

$P_n=(2p-1)P_{n-1}+(1-p) n\ge 1$

$P_n=\frac{1}{2}+\frac{1}{2}\times (2p-1{)}^n, n\ge 0$

Step3: Prove by mathematical induction

For, $n=0$

$P_0=\frac{1}{2}+\frac{1}{2}\times (2p-1{)}^0=1$

By the first half of the exercise, it is true for $n+1$ that:

$P_{n+1}=(2p-1)P_n+(1-p) n\ge 0$

and because the formula holds for $n$

$P_{n+1}=(2p-1)\left[\frac{1}{2}+\frac{1}{2}\times (2p-1{)}^n\right]+(1-p)$

The result of algebraic multiplication is,

$P_{n+1}=\frac{1}{2}\times (2p-1{)}^{n+1}+\frac{1}{2}$

This formula is valid for every $n\in N$ since this statement is true.