# Tag Info

## New answers tagged arbitrage

0

well generally only the discrete bonds associated to the ends of the forward rates are modelled. to make these be martingales the drifts of the rates are chosen to make them driftless. for an extension to all bonds, see http://ssrn.com/abstract=1461285

0

If there are no arbitrage opportunities there is no dominant hedge or long position. Why would there be an arbitrage opportunity if everything was priced correctly? There may be arbitrage opportunities even if there are no dominant hedges or long positions. Put-call parity shows arbitrage opportunities of badly priced options regardless of long position ...

0

Please clarify rigorously what you mean by each term. It is not true that no dominance is a consequence of no arbitrage. Think of the put-call parity: $C-P=S-K$, assuming $r=0$ since it's inconsequential. If there is no short selling then we can have: $C-P \geq S-K$ without arbitrage but No Dominance would not hold. If you think very deeply about this, ...

0

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.

1

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 ...

-3

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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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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Mark Joshi has pretty much solved it. To add to it, you can see that from Feynamn Kac (see remarks in http://en.wikipedia.org/wiki/Feynman–Kac_formula ) it follows that $$F(t,S,v) = B_t \mathbf{E}\left[ \frac{ \sqrt{ S_T } }{ B_T } \big \vert S_t = S, v_t = v \right],$$ where the expectation is taken with respect to a measure where $W$ and $Z$ are ...

1

Show that the discounted expectation price of the new security is the same as the solution of the PDE. Once this is done all three assets have discounted price processes which are martingales so there can be no arbitrage.

0

Absence of statistical arbitrage is a stronger condition that usual NA condition $$\nexists \varphi\in \Phi: V(\varphi)>0 \text{ and } \mathbb EV_T(\varphi)>0 \text{ for some } T\geq0 \tag{1}$$ since as soon as there exists an admissible trading strategy $\varphi\in \Phi$ satisfying $(1)$, it readily provides statistic arbitrage. Now, in BS the ...

5

Neither. Black--Scholes says nothing about the parameter values: $\mu$ and $\sigma.$ A very large $\mu$ and very small $\sigma$ is very unlikely to actually occur in the market and if it did you could make money with high probability without using option contracts. BS simply says that if the market follows a certain process then a certain option price is ...

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