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6

Let $\tau = T-t$. Then \begin{align*} S_T = S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z}, \end{align*} where $Z$ is a standard normal random variable, independent of $\mathcal{F}_t$. Moreover, \begin{align*} E\left(S_T 1_{\{S_T >K\}}\mid \mathcal{F}_t \right) &= E\left(S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, ...


3

Use Dynkin's formula to write the expectation: $\mathbb{E}[e^{-r\tau} \phi(S_\tau)]= g(S_0)+\mathbb{E}[\int_ 0 ^ \tau (A g -rg) dt]$ where $\phi$ is the payoff. Use the infinitismal generator $A$ to derive an ODE which describes the solution Use the fact that American options must be equal to or greater than their intrinsic value to derive boundary ...


2

Another take on the question which uses stochastic calculus [Digression] Assume deterministic and constant rates without loss of generality. Also assume the absence of arbitrage opportunities and market completeness Let $B_t$ denote the time-$t$ value of a risk-free money market account in which 1 unit of currency $C$ has been invested at $t=0$: ...


2

For this type of question, you basically need only to write the payoff with certain indicator functions. In particular, for the above payoff, we have that \begin{align*} \textrm{Payoff} &= K\, 1_{S_T \le K} + (2K-S_T)\,1_{K < S_T \le 2K}\\ &=K\, 1_{S_T \le K} + (2K-S_T)\big(1_{S_T \le 2K} - 1_{S_T \le K} \big)\\ &=(2K-S_T)\,1_{S_T \le 2K} - ...


1

I assume $r>0$. Let look at a) Let $v$ be the solution. $v$ is increasing (easy to prove, take $x<y$ and show that $v(x)<v(y)$ due $(S^x_t-K)^++a<(S^y_t-K)^++a$ on the continuity region $C$, i.e $x:v(x)>(x-K)^++a$, you have : $$\text{Black Scholes PDE perpetual case : }\frac{1}{2}\sigma^2x^2v''(x)+rxv'(x)-rv(x)=0$$ solutions are of the ...


1

I think you mix up marginal law, and law of the process. Your $Z_k$ must have three values, you just have to write the value of $\ln(H_k)$ for each possible value Let $X$ taking values $(x_1,x_2,...,x_n)$ and $p_i=P(X=X_i)$, then $P(f(X)=f(x_i))=\sum_{j=1}^n p_j\mathbf{1}_{f(x_j)=f(x_i)}$ if $f$ is a one-to-one mapping, you get $f(x_j)=f(x_i)\Rightarrow ...


1

your statement is quite imprecise. See https://en.wikipedia.org/wiki/Central_limit_theorem With : $(Z_k)_{k=1\dots n}$ i.i.d with $\mathbb{E}\left[Z_1\right] = \mu$ and $\text{Var}(Z_1)=\mathbb{E}\left[Z_1^2\right] -\mu^2=\sigma^2$ and by denoting $\mathcal{N}(m,v)$ a normal variance with mean $m$ and variance $v$ we have : $$ ...


1

In your answer, you don't include dividend. I am sorry to say it is wrong. Payoff function is $$ g(S_T) = (S_T - K_1)_+ - 2(S_T - \frac{K_1+K_2}{2})_+ + (S_T - K_2)_+ $$ BS pricing formula with dividend gives $$ V(t=0,S) = e^{-r}E(g(\tilde{d}S_T)) = \tilde{d} \left(BS_{call}\left(\frac{K_1}{\tilde{d}}\right) - 2BS_{call}\left(\frac{K_1+K_2}{2 ...


1

Let \begin{align*} V(t, S_t) = E\Big(e^{-r(T-t)} g(S_T)\mid \mathcal{F}_t \Big) \end{align*} be the risk-neutral value at time $t$ of the option payoff $g(S_T)$. Then $\{e^{-rt}V(t, S_t), 0 \le t \le T\}$ is a martingale. Consequently, \begin{align} -rV + \frac{\partial V}{\partial t} + (r-q)S\frac{\partial V}{\partial S_t}+\frac{1}{2}\sigma^2 S_t^2 ...



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