Lets consider a hypothetical stock with current price of $S_t$ at time t and it can take any positive value with a strictly positive probability.

There exists a derivative that pays $ e^{S_T}$ at maturity $T$ such that the writer pays $ e^{S_T}$ to holder, but holder pays nothing. $C_0$ is price of this derivative at time $0$.

Additionally, there is a risk-free zero-coupon bond with a face value of 1 and maturity $T$, priced at $Z_0$ at time $0$.

How can I show that the derivative cannot be super replicated if only the stock and bond are available in the market?

I am thinking in terms of the payoff structure of combination of bonds and stock (which is linear) vs payoff structure of derivative which is non linear. But depending on the units of stock or the bonds, the linear line is above the $e^{S_T}$ curve for some interval of $S_T$ right? so I can super replicate after all?

Or prove by contradiction in someway?

  • 2
    $\begingroup$ Upvoting because I am interested in the answer to this question as well. $\endgroup$
    – KaiSqDist
    Apr 15 at 9:59

1 Answer 1


I'm not familiar with super-replicating portfolios, but what I've gathered is that the idea is to find a static hedge that is greater than or equal to the asset with probability 1, ie find $a, b$ such that $aS_T + b \geq e^{S_t}$ almost surely.

I think you have generally the right idea in comparing the linear-in-$S_T$ static payoff with the exponential-in-$S_T$ derivative, and note that all we need to conclude that a portfolio is not super-replicating is some positive probabilty that that portfolio will be worth less than the derivative. In other words, when you note that the line is above the exponential for the interval, that doesn't make it a super-replicating portfolio -- having no intervals (with positive probability) where the line is below the exponential would.

  • $\begingroup$ Do you know how can I show that a and b doesnt exist $\endgroup$
    – CountDOOKU
    Apr 16 at 13:45
  • $\begingroup$ Any argument showing that an exponential function will grow larger than any affine function will do, I'd personally use a L'Hopital's type argument if I needed to prove that. $\endgroup$
    – Rylan
    Apr 16 at 13:59

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