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7

This is an interesting and not so easy question. Here's my 2 cents: First, you should distinguish between mathematical models for the dynamics of an underlying asset (Black-Scholes, Merton, Heston etc.) and numerical methods designed to calculate financial instruments' prices under given modelling assumptions (lattices, Fourier inversion techniques etc.). ...


6

$$\begin{array}{rcl} (1) & \partial_KC_t(T,K) & \leq 0 \\ (2) & \partial^2_KKC_t(T,K) & > 0 \\ (3) & \partial_T C_t(T,K) & \geq 0 \\ \end{array}$$ If $(1)$ doesnot hold, it exists $K_1<K_2$ such that $C_t(T,K_1)<C_t(T,K_2)$. Then as barrycarter said in his comment, you sell $C_t(T,K_2)$ and you buy $C_t(T,K_1)$, so your ...


4

Peter Jaeckel has written various papers on this. "by implication" and "Let's be rational" are the most recent ones. He also provides code on his website www.jaeckel.org. (Note: the question asked for literature.)


2

As I mentioned above, I am not sure what the variable $r$ is. If we ignore that, or assume the questioner wanted to say its the risk free interest rate, then it has no effect on the number of paths. Then it is clear that after 50 steps going from \$1024 to \$2500 requires a net of 4 up movements with the given $x=y^{-1}=1.25$. Thus the number of steps ...


1

There is a logical fallacy in your argument. The price of a European call expiring 1 day before a dividend payment may well be greater than that of a call expiring after it. In other words, claiming that $$ C_E (S_0,K,t_D-1\text {day}; D, t_D) < C_E (S_0,K,T; D, t_D) $$ is not necessarily true. Try the above inequality with a huge dividend (e.g. $D ...


1

Under the risk neutral measure, the expected present value of the butterfly payoff is: $$V_0 = e^{-rT} * \int_{S_T=K_1}^{K_3}P(T,S_T)f_{S_T}dS_T$$ And if we assume that $f_{S_T}$ is constant from $K_1$ to $K_3$, then: $$V_0 = e^{-rT} * \dfrac{1}{\Delta K} \int_{S_T=K_1}^{K_3}P(T,S_T)dS_T = e^{-rT} *\dfrac{\delta^2}{\Delta K} $$


1

Look on Google for Asymptotic behavior of Implied Volatility Near Infinity you will find results like : $$I(K) \stackrel{K\to\infty}{=} \sqrt{\frac{2}{T}}\left(\sqrt{\ln \frac{K}{C(K)}}-\sqrt{\ln\frac{1}{C(K)}}\right) +\text{O}_{K\to \infty}\left(\frac{\ln\ln\frac{1}{C(K)}}{\sqrt{\ln\frac{1}{C(K)}}}\right)$$


1

Two hints : The number of paths never going up to $3125$ when starting from $1024$ and stepping up by a multiplicative factor of $5/4$ and down by a multiplicative factor $4/5$ is the same as the number of paths starting from $0$ and and stepping up by an additive factor $+1$ and stepping down by an additive factor of $-1$ and never going up to $5$ Let ...


1

[Short Answer] You write $E [S_T]=S_0(1+r)^T $ but you actually compute the RHS as $X (1+r)^T$ in your numerical application. [Long Answer] The stock price is a martingale in an equivalent measure using the risk-free asset as numeraire i.e. $$ E [S(T)] = (S_0 u) q + (S_0 d) (1-q) = S_0 (1 + r ) \Delta t $$ In that case, dividing each member by $S_0$ and ...



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