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The risk-free rate used in the valuation of options must be the rate at which banks fund the cash needed to create a dynamic hedging portfolio that will replicate the final payoff at expiry. Dealers borrow and lend at a rate close to LIBOR, which is the funding rate for large commercial banks. The LIBOR swap curve is therefore the rate to be used when ...


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Page 3 of this document ad-co.com/analytics_docs/ALevin_QP_2012.pdf shows the result, originally given in Risk Magazine by Blyth and Uglum. The intuition for the formula is given in my comment above. The original motivation for such a formula was for interest rate options in the 1990s. Everyone had a lognormal pricing model, but traders understood that ...


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The proof is fine. For example, $D(t)S(t)$ is a martingale and then \begin{align*} E\big(D(t)S(t)\big) = S(0). \end{align*} Regarding the function $C(1, T-T_0, K)$, it is the value, at time $T_0$, of the option payoff \begin{align*} \left(\frac{S(T)}{S(T_0)} - K \right)^+. \end{align*} Here, you can treat $\frac{S(T)}{S(T_0)}$ as the normalized value or ...


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Your statement should be correct, the weights into the risky asset are not bounded between $0$ and $1$. Essentially, by setting $r=0$ you omit the term which shows that your weights always sum up to one, simply by choosing the weight for the risk-free asset to be $1-\pi^*$. In other words, obtaining $\pi^*>1$ simply implies you go short in the risk-free ...


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Besides the code's problem, I highly recommend the Brownian Bridge correction method which can compensate the pricing error resulting from discretization of the continuous path.


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There are many things wrong with your code. I'll leave aside the manner in which it is implemented, but note that it is: (1) not Matlab friendly with all the for loops (you should vectorise), (2) the fact that you have splitted the case j==0 in the main loop is a poor coding practice. for i=1:n I=1; for j = 0:(m-1); Z(j+1)= randn (1 ,1); dW=sqrt (T/...


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I think the chain of logic should be as follows: We have put value >= intrinsic. Therefore either put value > intrinsic or put value= intrinsic. If put value > intrinsic, then it is not optimal to exercise. If put value = intrinsic , it may be optimal to exercise.. Hence there is no contradiction.


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So, from this simple no-arbitrage argument, we see that the price of the option must always be at least its intrisic value. Yes indeed However, at this point I realized something strange: if this is true, why in the world should I exercise my put option before expiry?? The inequality seems to indicate that it would be an unwise decision to ever ...


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Under GBM $$ \frac {dS_t}{S_t} = \mu dt + \sigma dW_t $$ we get $$ S_T = S_0 e^{(\mu - \frac{1}{2}\sigma^2)T + \sigma W_T} $$ suggesting that $$ S_T \sim \text{ln}\mathcal {N} ( \tilde {\mu}, \tilde {\sigma}) $$ where \begin{align} \tilde {\mu} &= \ln S_0 + (\mu - \frac{1}{2}\sigma^2)T \\ \tilde {\sigma} &= \sigma \sqrt {T} \end{align} Now if $X \...



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