0
$\begingroup$

I am currently reading the PP BOYLE's article ' Option Replication in Discrete Time with Transaction Costs' written in 1992. Here is one place i couldn't figure out:

enter image description here

enter image description here

Where does that $\widehat{p}$ come from? And the $\bar{P} = \begin{pmatrix} \bar{p_{u}} & \bar{p_{d}}\\ 1-\bar{p_{u}} & 1-\bar{p_{d}} \end{pmatrix}$

What are $X_{i}$ and $\bar{P}^{i-1}$ ?

Thank you, this is my first time reading academic paper, i tried to google the transition matrix and Markov chains with expectations but the result is not useful.

$\endgroup$
3
  • $\begingroup$ Commonly, if the transition matrix has no absorbing states, the limiting distribution of the states is a non-degenerate vector, I think this is $\hat{p}$ in your case - but I am guessing here, as the author did not define it in the paper (I did not find it, at least). Furthermore, the transition matrix is defined in equation 9 of that paper. $\endgroup$ Jul 7 at 7:02
  • $\begingroup$ Thank you. Do you know what does the $\bar{P}^{(i-1)}$ mean? Based on the author's definition of $\bar{P}$, can i say that the first column of $\bar{P}^{(i-1)}$ means the probability distribution of $X_{i}$ if $X_{i-1} = ln(u)$ and the second column means the probability distribution of $X_{i}$ if $X_{i-1} = ln(d)$ $\endgroup$ Jul 7 at 8:31
  • $\begingroup$ I'd argue that this is the matrix $\bar{P}$, as defined in equation 9, raised to the $(i-1)$st power. The interpretation is, roughly, for each column, the probability to being in state $u$ or $d$ after $i-1$ steps given today's state (i.e. first or second column). $\endgroup$ Jul 7 at 9:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.