According to the european put lower bound condition:

$ p \geq max(D + K \cdot e^{-r(t_2-t_0)} - S_0, 0)$

where $t_0$ is now and $t_2$ is maturity. Say $t_1$ is the dividend release time where $t_0<t_1<t_2$.

Shouldn't it be:

$ p \geq max(D \cdot e^{-r(t_1-t_0)} + K \cdot e^{-r(t_2-t_0)} - S_0, 0)$ ?

The common proof when it comes to portfolio B consists of investing $D + K \cdot e^{-r(t_2-t_0)}$ at the risk-free rate. (Portfolio A is going long on a put and the asset. The 2 portfolios' cashflows are compared and it is shown that since $P_A \geq P_B$ at $t_2$ that should stand for $t_0$ as well.)

Shouldn't we invest instead:

$D \cdot e^{-r(t_1-t_0)} + K \cdot e^{-r(t_2-t_0)}$ in order to get equal cashflows $(=D)$ at time $t_1$? I.e. discounting the divident in the same fashion with the strike but at its respective time-frame?

I struggle to get my head around this, thank you!

  • 1
    $\begingroup$ Be very careful how $D$ is defined in the book/article you are referencing. Sometimes $D$ is defined as the present value of the dividend, i.e. it has already been discounted from the ex-dividend date to the present. $\endgroup$
    – noob2
    May 18 '19 at 16:44
  • $\begingroup$ You are absolutely right, I found it mentioned in some sources and then it is clear. Thanks! Not in every source that someone can come across though and that can confuse a student, e.g. some lecturing notes! $\endgroup$
    – ChrisB
    May 18 '19 at 23:21

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.