# Finding the transition probability matrix, two switches either on or off..

Each of two switches is either on or off during a day. On day n, each switch will independently be on with probability [1+number of on switches during day n-1]/4 For instance, if both switches are on during day n-1, then each will independently be on with probability ¾. What fraction of days are both switches on? What fraction are both off?

I am having trouble finding the transition probabilities. I know what they all are (I have looked at the solution), but I don't understand how you get the values for each P_ij... I can easily find the stationary probabilities after finding the transition probability matrix.. Can anyone help me guide through the transition steps?

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If the probability for a switch to be on is $p$, the probabilities for $0$, $1$ or $2$ switches to be on are $(1-p)^2$, $2p(1-p)$ and $p^2$, respectively. If $0$, $1$ or $2$ switches were on on the previous day, the corresponding values of $p$ for this day are $1/4$, $2/4$ and $3/4$, respectively. Thus the transition matrix is

$$\pmatrix{ \left(\frac34\right)^2&2\cdot\frac14\cdot\frac34&\left(\frac14\right)^2\\ \left(\frac24\right)^2&2\cdot\frac24\cdot\frac24&\left(\frac24\right)^2\\ \left(\frac14\right)^2&2\cdot\frac34\cdot\frac14&\left(\frac34\right)^2\\ }=\frac1{16}\pmatrix{9&6&1\\4&8&4\\1&6&9}\;.$$

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Thank you, you cleared this up extremely well for me. I can't thank you enough! I can finally go to bed :) – Wooooop Feb 25 '13 at 12:24

IMO the given information is sufficient.

2 bulbs, 2 states - ON & OFF each. That means total 4 states are possible per day:

OFF-OFF, ON-OFF,OFF-ON, ON-ON. Therefore, the transition matrix will be a 4*4 one.

Given that

$P(any one switch=open next day)= \frac {(1+ number of on switches during previous day)}{4}$

Therefore, the Transition probability matrix will be as follows.

$00$ $01$ $10$ $11$

$00$ $\frac{9}{16}$ $\frac{3}{16}$ $\frac{3}{16}$ $\frac{1}{16}$

$01$ $\frac{4}{16}$ $\frac{4}{16}$ $\frac{4}{16}$ $\frac{4}{16}$

$10$ $\frac{4}{16}$ $\frac{4}{16}$ $\frac{4}{16}$ $\frac{4}{16}$

$11$ $\frac{1}{16}$ $\frac{3}{16}$ $\frac{3}{16}$ $\frac{9}{16}$

Hope this helps.

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The state space is $(x_n^1,x_n^2)$ which are the states of bulbs 1 and 2 being on/off at time $n$. The transition probability matrix is found as follows, where the ordering of the states is $(0,0), (0,1), (1,0)$ and $(1,1)$.

$$P=\left[\begin{array}\ \frac{9}{16} & \frac{3}{16} & \frac{3}{16} &\frac{1}{16}\\ \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4}\\ \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4}\\ \frac{1}{16} & \frac{3}{16} & \frac{3}{16} &\frac{9}{16}\\ \end{array}\right]$$

To find the fraction of time both bulbs are off/on, you need to solve $\pi P=\pi$ and find $\pi(0,0)$ and $\pi(1,1)$. The stationary probabilities turn out to be $\pi(0,0)=\pi(1,1)= \frac{2}{7}$ and $\pi(0,1)=\pi(1,0)= \frac{3}{14}$.

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Differentiating between all $4$ states is unnecessarily complicated because the two states $(0,1)$ and $(1,0)$ are indistinguishable for the purposes of the problem. Also your stationary probabilities are wrong, which you can see from the fact that they don't add up to $1$. – joriki Feb 25 '13 at 12:30
Thanks @joriki. The answer now seems a bit intuitive too, given the transition probabilities. – Bravo Feb 25 '13 at 12:44
They're still wrong. – joriki Feb 25 '13 at 13:02
@joriki: Huh! Corrected again... – Bravo Feb 25 '13 at 13:24
OK, that looks good :-) – joriki Feb 25 '13 at 13:39