Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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

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

share|cite|improve this answer
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.

share|cite|improve this answer

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 $\pi(0,1)=\pi(1,0)= \frac{3}{14}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.