# Markov Chain Exercise Help

In working with a particular gene for fruit flies, geneticists classify an individual fruit fly as $\small \text{dominant, hybrid or recessive}$. In running an experiment, an individual fruit fly is crossed with a hybrid, then the offspring is crossed with a hybrid and so forth. The offspring in each generation are recorded as dominant,hybrid or recessive

(a) What is the probability the third generation offspring is dominant given the first generation offspring is recessive?

Transition Table is :- $$\begin{bmatrix} .5 & .5 & 0 \\ .25 & .5 & .25 \\ 0 & .5 &.5 \\ \end{bmatrix}$$

where states 1, 2, and 3 are “dominant”, “hybrid” and “recessive” respectively.

According to book , solution for (a) is to find $p(2)13$, means starting from state "dominant" and finishing at state "recessive", but I think it should be $p(2)31$ means starting from state "recessive" and finishing at state "dominant" !

In short the question is , what should be the initial state for this problem ? and why ?

thanks!

here are the snap shots from the book

• Yes, you are correct. If $Z(t)$ is the Markov chain then $P_{31}^{(2)}=P[Z(t+2)=1|Z(t)=3]$. If $1\leftrightarrow dominant$ it means ending up dominant at time $t+2$. Likely there was a minor typo in the book due to mixing the mapping between $\{1,2,3\}$ and $\{dom, hybrid, rec\}$. Commented Oct 5, 2016 at 17:02
• So here the solution for (a) will be P(2)13 right? Commented Oct 5, 2016 at 17:05
• Is that what you intended to write? You can edit your comment if needed, if there was a typo. Commented Oct 5, 2016 at 17:07
• Okay, overall, after you read my first comment, I do not understand the logic of how you can write your comment after mine. My comment says "yes, you are correct" and then you somehow conclude by changing your answer [which does not make sense if your answer was correct]. I expected you had a typo in your comment, but apparently you did not want to fix your comment...? Commented Oct 5, 2016 at 17:15
• At the OP, how does the author define the transition matrix? Is $T(i,j)$ the probability of transitioning from state $i$ to state $j$, or the probability of transitioning from state $j$ to state $i$? I have seen both conventions used. Which convention is used determines what the correct answer is. Commented Oct 5, 2016 at 17:28

Answer edited to fix the multiplication order error noted by Michael.

If I understand your notation correctly, the answers you are considering are $P^2_{(1,3)}$ and $P^2_{(3,1)}$. You are correct. As we begin with the first generation offspring, a recessive fly, the initial state vector is $$v_1=\begin{bmatrix} 0,0,1 \end{bmatrix}$$

Then we compute the state vector for two generations in the future as

$$v_3=v_1P^2=\begin{bmatrix} P^2_{(3,1)}, P^2_{(3,2)}, P^2_{(3,3)} \end{bmatrix}$$

Finally, as we are interested in the probability the third generation is dominant, we are interested in the first entry of $v_3$, which is $P^2_{(3,1)}$.

• No. Same reason as my comment in other answer. Commented Oct 5, 2016 at 17:14
• This would be correct if you use as $P$ the transpose of the matrix in the OP. Commented Oct 5, 2016 at 17:32
• @Michael You're completely correct, I forgot that it should be left multiplication (or the transpose of $P$). I'll edit it in a second. Commented Oct 5, 2016 at 17:40
• Looks good. In general, books have typos/mistakes because authors are human. Commented Oct 5, 2016 at 18:09

I hope you don't mind it if I dumb it down (mainly to make sure I get it)...

So we have

$$P=\overset{\begin{matrix}\color{blue}{\text{Dom}}&\color{blue}{\text{Hyb}}&\color{blue}{\text{Rec}}\end{matrix}}{\begin{bmatrix} &.5 & .5 & 0 & \\ &.25 & .5 & .25 & \\ &0 & .5 & .5 & \end{bmatrix}}$$

with every row spelling a PMF of the distribution among the three possible states after one step. For example, for the first row, if the distribution at zero time only included $\small\text{Dominant}$, after one step $.5$ of the population would remain $\small\text{Dominant}$, while $.5$ would transition to $\small\text{Hybrid}$, and none to $\small\text{Recessive}$.

After two steps,

$$\small P^2=\begin{bmatrix} .5 & .5 & 0 \\ .25 & .5 & .25 \\ 0 & .5 & .5 \end{bmatrix} \begin{bmatrix} .5 & .5 & 0 \\ .25 & .5 & .25 \\ 0 & .5 & .5 \end{bmatrix} =\begin{bmatrix} .375 & .5 & .125 \\ .25 & .5 & .25 \\ .125 & .5 & .375 \end{bmatrix}$$

so that with an initial state of only $\small\text{Recessive}$, i.e. $v1=\begin{bmatrix}0&0&1\end{bmatrix}$, the distribution of interest would be found in the third row of $P^2$, and the probability of $\small\text{Dominant}$ in $P^2_{(\color{red}{3,1})}$:

$$v3 = v1\,P^2=v1=\begin{bmatrix}0&0&1\end{bmatrix}\begin{bmatrix} .375 & .5 & .125 \\ .25 & .5 & .25 \\ \color{red}{.125} & .5 & .375 \end{bmatrix}=\begin{bmatrix}\color{red}{.125}&.5&.375\end{bmatrix}$$