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It is known that in an arbitrage-free continuous time market, the price of every asset is evaluated as the corresponding price in the replicating strategy using risk-neutral valuation.

I want to know is the converse true? There is actually a question that asks to show that for a Black-Scholes model with a bank account ($dB_t = B_t r dt$) and a stock satisfying $dS_t = r S_t dt + \sigma S_t dW_t$, then there is no arbitrage if the time-t price of a call with maturity $T$ and strike $K$ is \begin{equation} X_t= S_t \mathbb{E} \big\{ \big( e^{-\frac{(T-t)\sigma^2}{2} + \sqrt{T-t} \sigma Z} - \frac{K e^{-r(T-t)}}{S_t} \big)^{+} \big\}. \end{equation} Any ideas of how to show this? I don't know how to evaluate this expectation, as it involves two random variables and I don't know the joint density of $S_t$ and $Z$.

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  • $\begingroup$ That's kinda a strange question: given only the formula for the call price, there can be multitude of other assets that allow for arbitrage. $\endgroup$ – Ulysses Dec 1 '14 at 9:22
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I do this question to death in Concepts and ...

If (discounted price of) everything is a martingale then every trading strategy is a martingale. Therefore any self-financing portfolio of initial value zero and has expectation zero. Therefore there are no arbitrages (since these have positive expectation and initial value zero).

So there is no arbitrage in the martingale measure.

However, the pricing measure and the real-world measure are equivalent so they have the same set of arbitrages. So there is no arbitrage in the real-world measure either.

So if we set the price to be its expectation price, we get a martingale in the pricing measure and the price is arbitrage free.

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  • $\begingroup$ Please see my edit above.... $\endgroup$ – Richard Nov 29 '14 at 21:46
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The fact that one claim has an arbitrage-free price, does not imply that the entire market (for all claims) is arbitrage-free. E.g. $C_T=0$ is always arbitrage-free.

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  • $\begingroup$ Please see my edit in question above. $\endgroup$ – Richard Nov 29 '14 at 21:35
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You need to replace Z by Brownian motion at time t. Also, the expectation should be conditional expectation with respect to the sigma-algebra at time t. See http://kalx.net/fms/fms.html for a more complete explanation.

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  • $\begingroup$ Z represents a $N(0,1)$ random variable. Also, this is a past exam question, so I suppose that it shouldn't be wrong. $\endgroup$ – Richard Nov 29 '14 at 21:51
  • $\begingroup$ It's wrong. As is your SDE for S. Should be dS/S = r dt + sigma dW, where W is Brownian motion. See the link I gave you for all the details. $\endgroup$ – Keith A. Lewis Nov 29 '14 at 22:06
  • $\begingroup$ Yes, this was a typo. $\endgroup$ – Richard Nov 29 '14 at 22:08
  • $\begingroup$ I fixed it. But I still can't evaluate the expectation. $\endgroup$ – Richard Nov 29 '14 at 22:08
  • $\begingroup$ Use the fact exp(-sigma^2 t/2 + sigma W_t) is a martingle. $\endgroup$ – Keith A. Lewis Nov 29 '14 at 22:12
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put call parity implies no arbitrage , along with a expected value of stock price of $pe^{rt}$ . working out the integrals yields this outcome

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