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I would bet on the slowness on Monte carlo method. Your volatlity is quite high. The error would decrease as var/sqrt(pathcount)= var *0.01, where var is the variance of the final price. 50*0.7*0.01=0.35 ~ 0.4


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You can also include variance reduction techniques in you monte carlo simulations, such as control variates or antithetic variates. Both aim at reducing the variability of your simulated option price and are very popular for monte carlo simulations. http://en.wikipedia.org/wiki/Antithetic_variates http://en.wikipedia.org/wiki/Control_variates Both are ...


0

Consider your question in the idealized case of zero transaction cost and where the underlying stock price follows geometric Brownian motion with constant volatility -- identical to the implied volatility used to price the option. If the delta hedge is rebalanced over time short time intervals of length $\Delta t,$ then the cost of hedging is a random ...


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The Black-Scholes price of this option is approximately $14.8$. When I run a Monte Carlo simulation with $10000$ paths and "exact" time stepping, I get results very close to this value. You are simulating the terminal asset price with the first-order Euler approximation over multiple time steps: $$S(t+\Delta t)= S(t) + rS(t)\Delta t + \sigma ...


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Emcor is correct, especially the part regarding the unmatched number of degrees of freedom between that of the known and unknown functions. We can make the problem even clearer. Your problem is essentially finding density $f(x,y)$ given $\int xf(x,y)dx, \forall y$. This obviously has infinitely many solutions.


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It is not possible to derive the joint distribution from the expectation under the given information here. The fact that you have the expectation for all $K$ says nothing about the joint distribution $f(x,y)$ because $K$ just shifts the mean of $Y$ but gives no information on the joint probability for $(x,y)$. You may particularly note if $f(x,y)$ have >1 ...


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The required probability is equivalent to asking: what is the probability that the geometric brownian motion of the underlying touches the strike for the first time before the given time $T$? A strategy for solving this related to Brownian motion - first passage time. After transforming from geometric brownian motion to brownian motion via the log, the ...


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Assume the price follows a lognormal process. We can convert it into a problem of finding the probability of a standard Brownian motion particle starting from $0$ and hitting $x$ before time $t$, or its first passage time $\tau_x$ being less than $t$. This can be derived through the reflection principle. The paths crossing $x$ are exactly paired up by the ...


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Q: What does the risk-neutral price represent if the option is not replicable? In an incomplete market, there is no unique martingale measure but instead a set $Q$ of equivalent martingale measures. Consequently, there is an interval of arbitrage-free prices: $ \Big( inf_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX], sup_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX] ...


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I think you misinterpreted what you read. The whole point of the frictionless market assumption is that you can forget about any cost or any bound on volumes or latency associated with transactions used to rebalance a self-financed portfolio. So you are right when you say that sustaining a replicating portfolio doesn't cost anything. This implies that ...



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