1. Impossible Question

Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out.

if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day?

It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas?

2. Originally Posted by megno
Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out.

if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day?

It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas?
Suppose that the probability that it snows on any day is p.

Now consider today. There is a probability p it snowed yesterday, and (1-p)
that it did not. Therefore the probability that it snows today is:

p*0.3+(1-p)*0.2=p

Solve this for p to find p=2/9~=0.222.

RonL

3. Originally Posted by megno
Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out.

if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day?

It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas?
This is an example of a two states Markov chain. Let X(n) be the state of the process, i.e the weather, at day n. Then we have a state space S={0,1}

0: Not snowing
1: Snowing

Transition probabilities between the states from day to day is

$P(X(n+1)=0|X(n)=0)=p_{00}=0.8$
$P(X(n+1)=1|X(n)=0)=p_{01}=0.2$
$P(X(n+1)=1|X(n)=1)=p_{11}=0.3$
$P(X(n+1)=0|X(n)=1)=p_{01}=0.7$

That is we have a transition probability matrix P

$P=\left( \begin{array}{cc}
0.8 & 0.2 \\
0.7 & 0.3 \end{array} \right)$

What we're interested in is to find the long run behaviour of the process. (There are certain conditions to be satisfied in order for the process to have a long run behaviour, i.e to have a limiting stationary distribution. We don't need to worry too much about that, cause in this case they're all fullfilled.)
We are looking for a limiting distribution <> stationary distribution $\pi=(\pi_0, \pi_1)$ satisfying

$\pi=\pi P$

That is, if we act upon the distribution according to the transition probabilities we will still have the same distribution.
Now solving the above equation for $\pi$ with the condition that $\pi_0+\pi_1=1$ yields

$\pi=\left(\frac{7}{9},\frac{2}{9}\right)$

That is, the probability that it snows on a given day is $\pi_1=2/9$

For this example there is maybe an easier approach... but for the general case with more than 2 states this one is quite useful...