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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.

Thanks!

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  • $\begingroup$ 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 '14 at 9:00
  • $\begingroup$ Hint: $\log(1/S_t) = -\log(S_t)$ $\endgroup$ – Brian B Apr 1 '14 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 '14 at 1:01
  • $\begingroup$ Is this under Black-Scholes world? $\endgroup$ – emcor Jul 30 '14 at 20:24
  • $\begingroup$ If you do not know, this is the dynamic hedge for a variance swap. $\endgroup$ – Dom Oct 29 '16 at 17:07
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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.

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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

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  • $\begingroup$ Why is there a factor of 2 after the second equality? $\endgroup$ – James Jun 1 '18 at 2:59
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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: $100\,(B(0,t) - d\cdot Q_1)$

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  • $\begingroup$ What is $Q_1$ ? $\endgroup$ – emcor Aug 11 '14 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 '14 at 16:22

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