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Question: Let H(x)=1/x be the payoff function for a European style derivative security. Find a closed form expression for the price: $$ u(t,x)=e^{-r(t-t)}E[H(S_T)|S_t=x] $$ for this claim using Black Scholes dynamics, with r and sigma constants.

My attempt: Under Black Scholes dynamics: $$ dS_t=S_t({\sigma}dW_t+rdt) $$ Using Ito's on this, with f(x)=1/x, I get that: $$ \frac 1{S_T} =\frac 1{S_t}exp((\sigma^2-r)(T-t)-\sigma\sqrt{T-t}Z) $$ Where Z is standard normal. Finally, plugging into the pricing formula: $$ V_t=B_tE[B^{-1}_TX|S_t] $$ Which gives me a final answer: $$ \frac 1{S_t}exp((T-t)(\frac32\sigma^2-2r)) $$ However, not sure if this is correct, anyone have any ideas? I don't have the solutions to this question unfortunately..

Cheers

WeakLearner
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    It is unclear to me what you are doing. Your B&S dynamics are incorrect, it should be $\mathrm dS_t=S_t(r\mathrm dt+\sigma\mathrm dW_t)$. Additionally, your solution to the SDE is also incorrect, it should be $S_T=S_t\exp\left((r-\sigma^2/2)(T-t)+\sigma(W_T-W_t)\right)$. Lastly, why does your final answer depend on $S_T$ which is a random variable? – Ian May 21 '14 at 07:43
  • Hi Ian, sorry, these were typos, I used the BS dynamics you just pointed out, and I got the solution in my initial question. The final answer is dependent on $$ S_t$$ not $$S_T$$, thanks for pointing out that mistake, was just a typo – WeakLearner May 21 '14 at 13:03
  • Is this now correct? – WeakLearner May 21 '14 at 13:04
  • Also note the solution to the SDE that I mentioned. You are missing a factor $1/2$. – Ian May 21 '14 at 14:00
  • $$ dS_t=S_t({\sigma}dW_t+rdt) $$ Using Ito's on this, with f(x)=1/x, I get that: $$ d(\frac 1{S_t})=dt(0+(\frac {-1}{S_t^2})rS_t+\frac {2}{S_t^3}\frac {S_t^2\sigma^2}{2})+{\sigma}S_tdW_t(\frac {-1}{S_t^2}) $$ which gives: $$ dt(-r(\frac {1}{S_t})+{\sigma^2}(\frac {1}{S_t}))-(\frac {1}{S_t}){\sigma}dW_t $$ shouldn't this give me: $$ \frac 1{S_T} =\frac 1{S_t}exp((\sigma^2-r)(T-t)-\sigma\sqrt{T-t}Z) $$ Where Z is standard normal I'm not seeing where that extra 0.5 comes from?

    Thanks again

     – WeakLearner May 21 '14 at 14:16
  • okay, took me a while but I see where I went wrong now, didn't apply Ito's that second time. Thanks again Ian – WeakLearner May 24 '14 at 03:39
  • Ok, good to hear. I think the end result should be $\frac 1{S_t}exp((T-t)(\sigma^2-2r))$ if I'm not mistaken. – Ian May 24 '14 at 03:42
  • Yes, that's what I got too, cheers! – WeakLearner May 24 '14 at 03:47

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