In one of my lecture notes, I stumble across this exercise question:

Consider Low Exercise Price Options, LEPOs, (with dividends) in Australia. Using the value at the outset, explain why such options are sometimes priced above $S_0$.

I am confused how to continue to answer the above question. Is the following is the statement we are trying to show?$$(S_0 - D)e^{rt}>>S_0$$ If yes, how do I show such statement holds? In case of $D=0$ it is obvious that the above holds, but how to prove such statement for some $D>0$?

On a side note, I am also confused with this question, isn't it violating the upper bound of call options, i.e. $c_t \leq S_t$?

Thank you, any help is highly appreciated



\begin{equation} \left( S_0 - D \right) e^{r T} > S_0, \end{equation}

you see that this is true when

\begin{equation} r > \frac{1}{T} \ln \left( \frac{S_0}{S_0 - D} \right), \end{equation}

i.e. when the interest rate is sufficiently large relative to the present value of future dividends.

When you replicate the LEPO, and ignoring that $X$ is not actually equal to 0, you buy one stock initially. The cost of financing this position until maturity is $S_0 \left( e^{r T} - 1 \right)$ and the returns for holding it is $D e^{r T}$. When the former is higher than the latter, then the overall cost of replication are positive and the initial price of the LEPO is above $S_0$.

Having the LEPO priced above $S_0$ is not a contradiction to the no-arbitrage bounds of a call option $C_0 < S_0$. The latter applies to call options where the premium is paid upfront i.e. at $t = 0$. In case of the LEPO (ignoring margining) the premium is settled forward-style at the maturity date $t = T$.

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