1) Suppose S is the stock price, how to hedge a derivative that pays $1/S_t$ at time $t$?

2) Suppose there will be a dividend of amount $d$ between $t$ and $T$, how to hedge a derivative that pays $100 $*$ S_T/S_t$ at time $T$?

The person who asked me the question said we don't need to assume the distribution of S here.


  • $\begingroup$ You should clarify your questions: First, are you looking for a static or a dynamic hedge? Second, is the dividend paid continuously between t and T, or is it a lump-sum at a random time between t and T? $\endgroup$
    – pbr142
    Apr 1, 2014 at 9:00
  • $\begingroup$ Hint: $\log(1/S_t) = -\log(S_t)$ $\endgroup$
    – Brian B
    Apr 1, 2014 at 12:42
  • $\begingroup$ The dividend is a lump sum of amount of $d$ at a fixed time, say $t < t_1 < T$ $\endgroup$
    – benh
    Apr 2, 2014 at 1:01
  • $\begingroup$ Is this under Black-Scholes world? $\endgroup$
    – emcor
    Jul 30, 2014 at 20:24
  • $\begingroup$ If you do not know, this is the dynamic hedge for a variance swap. $\endgroup$
    – Dom
    Oct 29, 2016 at 17:07

3 Answers 3


Note that, for a smooth function and constant a $$f(S_t) = f(a) + f'(a) (S_t-a) + \int_a^{\infty}(S_t-x)^+f^{''}(x)dx + \int_{0}^a(x - S_t)^+f^{''}(x)dx.$$ Then, the payoff $1/S_t$ can be approximately hedged by call and put options: $$\frac{1}{S_t} = \frac{1}{a} -\frac{1}{a^2}(S_t-a)+ 2\bigg[\int_a^{\infty}\frac{(S_t-x)^+}{x^3}dx + \int_{0}^a\frac{(x - S_t)^+}{x^3}dx \bigg], $$ where $a = E(S_t)$.

As for $S_T/S_t$, let $d$ be the dividend paid at $t_1$, where $t<t_1<T$. Note that $$E(S_T \mid \mathcal{F}_t) =S_t \exp\Big(\int_t^T r_s ds \Big) - d\exp\Big(\int_{t_1}^T r_s ds \Big). $$ We replicate the payoff $1/S_t$ at time $t$. Then we replicate by forwards and bonds.


We can explicitly value the Inverted Option under Black-Scholes Model as follows:

enter image description here

Then the delta-hedging ratio is given as:

enter image description here

  • $\begingroup$ Why is there a factor of 2 after the second equality? $\endgroup$
    – James LT
    Jun 1, 2018 at 2:59

For question 2): At time $T$, we need to pay $100\cdot\frac{S_T}{S_t}$ (in domestic currency, say \$). To do this, we need to buy 100\$ worth of shares at time $t$: that gives us $N=100\cdot \frac{1}{S_t}$ shares, with the desired final value of $$N\cdot S_T = 100\cdot\frac{S_T}{S_t}$$ at expiry. Needless to say, today's PV of 100\$ at time $t$ is $100\,B(0,t)$.

However, then at time $t_1$, we hold $N$ shares, so we get a dividend of $d$ per share, so we receive $d\,N = 100\cdot d/S_t$. Being the nice investment bankers that we are, we charge the client correspondingly less, namely the PV of that, which is $100\,d$ times the answer to question 1.

Thus, final answer: $100\,(B(0,t) - d\cdot Q_1)$

  • $\begingroup$ What is $Q_1$ ? $\endgroup$
    – emcor
    Aug 11, 2014 at 23:28
  • $\begingroup$ Hi @emcor, $Q_1$ is the answer to the previous question 1), namely the price of a derivative that pays $1/S_t$ at time $t$... It's European, some sort of hyperbola, and replicable with calls and puts and what have you, and you and Gordon answered that question :-) $\endgroup$
    – Fab
    Aug 12, 2014 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.