Question

On any given day Gary is either cheerful (C), so-so (S), or glum (G). If he is cheerful today, then he will be C, S, or G tomorrow with respective probabilities 0.5,0.4,0.1 . If he is feeling so-so today, then he will be \(C\), S, or G tomorrow with probabilities \(0.3,0.4,0.3 .\) If he is glum today, then he will be \(\mathrm{C}, \mathrm{S},\) or \(\mathrm{G}\) tomorrow with

\(X_{n}\) denote Gary's mood on the nth day, then \(\left\{X_{n}, n \geq 0\right\}\) is a threeprobabilities \(0.2,0.3,0.5 .\) Letting \(X_{n}\) denote Gary's mood on the nth day, then state Markov chain (state \(0=C\), state \(1=S\), state \(2=G\) ) with transition probability matrix

\(P=\begin{array}{rrr}0.5 & 0.4 & 0.1 \\ 0.3 & 0.4 & 0.3 \\ 0.2 & 0.3 & 0.5\end{array}\)

a) Gary is currently in a cheerful mood. What is the probability that he is not in a glum mood on any of the following three days?

b) Gary was in a glum mood four days ago. Given that he hasn't felt cheerful in a week, what is the probability he is feeling glum today?

Answer 1

Given transition matrix

\(P=\left[\begin{array}{lll}0.5 & 0.4 & 0.1 \\ 0.3 & 0.4 & 0.3 \\ 0.2 & 0.3 & 0.5\end{array}\right]\)

a) Gary is currently in a cheerful mood.

What is the probability that he is not in a glum mood on any of the following three days. We have to find the matrix of 3 stages we get \(p^{3}=\left[\begin{array}{ccc}\frac{89}{250} & \frac{189}{500} & \frac{133}{500} \\ \frac{42}{125} & \frac{37}{100} & \frac{147}{500} \\ \frac{161}{500} & \frac{91}{250} & \frac{157}{500}\end{array}\right]\)

Here current state is cheerfull mood,probability that glum

mood on any of the 3 days.

That is \(\frac{133}{500}\)

Probabilty that not in glum mood is \(=1-\frac{133}{500}=\frac{367}{500}\)

b)

Gary was in a glum mood four days ago. Given that he hasnt felt cheerful in a week, what is the

probability he is feeling glum today. The four stages transition probability is \(B=\left[\begin{array}{lll}\frac{1723}{5000} & \frac{1867}{5000} & \frac{141}{500} \\ \frac{1689}{5000} & \frac{1853}{5000} & \frac{729}{2500} \\ \frac{333}{1000} & \frac{1843}{5000} & \frac{373}{1250}\end{array}\right]\)

4 days ago he is glum mood and we have to find the

probability that he is glum mood today

That is \(\frac{373}{1250}\)

Answer 2

>>> Here current state is cheerfull mood,probability that glum mood on any of the 3 days.

>>> That is $\frac{1$$$
$$
$$
$sdfsdf33}{500}$

The above statement from Yuiai  for part a) is not correct.

In fact, the probability that he is in glum mood on any of the following 3 days is:

Sum of the 3 probabilities P_CG + P²_CG + P³_CG

So the answer has to be: 1 - (P_CG + P²_CG + P³_CG)

Answer 3

I stand corrected, 

 1 - (PCG + P2CG + P3CG)  => not correct

The correct answer should be:

 ( 1 - PCG )( 1-  P2CG )( 1 - P3CG)