I know the derivation of the Black-Scholes differential equation and I understand (most of) the solution of the diffusion equation. What I am missing is the transformation from the Black-Scholes differential equation to the diffusion equation (with all the conditions) and back to the original problem.All the transformations I have seen so far are not very clear or technically demanding (at least by my standards).My question:Could you provide me references for a very easily understood, step-by-step solution?...Read more

Let's consider standard Black-Scholes model with price process $S_t$ satisfying SDE $$dS_t = S_t(bdt + \sigma dB_t)$$, where $B_t$ is standard Brownian Motion for probability $\mathbb{P}$. I understand the proof of existence of martingal measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ based on Girsanov theorem, but I can't see how to derive uniqueness of $\mathbb{Q}$. Can you help?Edit: In Jeanblanc, Yor, Chesney $\textit{Mathematical Methods for Financial Markets}$ I found the following proof: If $\mathbb{Q}$ is equivalent to $\mathbb{P}$ the...Read more

Given:Consider a two-asset, continuous time model (B,S) where $$dB_t = B_t r dt, \quad dS_t = S_t ( \mu dt + \sigma dW_t)$$Clearly, the martingale deflator is:$$Y_t = e^{(-r - \frac{\lambda^2}{2})t - \lambda W_t}$$There is a theorem that states the following: For a claim with payout $\xi_T$, $T>0$, if $\xi_T>0$, $\mathcal{F}_T$-measurable and such that $\xi_TY_T$ is integrable, then there exists an admissible strategy $(\pi_t)_{t \in [0,T]}$ that replicates the European claim with payout $\xi_T$.Problem:Suppose that $\xi_T = g(X_T)$, whe...Read more

I'm trying to understand probability distributions implied from market prices and was reading through this reference explaining the interpretation of $N(d_1)$ and $N(d_2)$ in the log-normal vol Black-Scholes model. I have two sets of questions:If I buy a call at strike $K$ and write a call at strike $K+\Delta K$, can I back into the risk-neutral probability of the underlying rising to $[K, K + \Delta K]$ (which I would work out as $N(d_2)$) from observed vol and market prices? Is this algebraic rearrangement of the Black-Scholes equation mean...Read more

I am new to Black Scholes, and trying to use it to model a clawback in private equity. Essentially, a clawback gives the "limited" partners in the deal the option to pull some funds away from the "general" partner. I am essentially trying to value a put option, to assess the likelihood of the put option being in the money (i.e. $\mathcal{N} \left( -d_2 \right)$).I believe the option in this case would rely on:Underlying value (i.e. "S"): net asset value of the investments + the distributions earned by the "limited" partnersExercise ("K"): the c...Read more

I'm stuck trying to analytically prove that a partial derivative of a specific, lower defined function $C$ is negative. The context of this problem is actually a Black-Scholes market situation, where a price of a call option decreases as its strike increases. For given positive constants $S, K, r, \sigma$ and $T$, we have:$$C(S,K,r, \sigma,T)=S \Phi(d_1)-Ke^{-rT}\Phi(d_2),$$where $$d_1=\frac{\ln \frac{S}{K}+(r+\frac{1}{2}\sigma^2)T}{\sigma \sqrt{T}},$$$$d_2=d_1-\sigma \sqrt{T}.$$I have to prove that the function $C$ is decreasing if $K$ is incr...Read more

I am trying to derive the differential of the product of two processes, but I got stuck. This is what I have until now: We have the following two stochastic processes:$dX_t= \mu_t dt +\sigma_t dW_t$ and $dY_t = \eta_t dt + \vartheta_t d \bar{W}_t$, with the correlation between the two Brownian motions equal to $\rho$, that is E[($W_t - W_s$)($\bar{W}_t - \bar{W}_s$)|$F_s$] = $\rho(t-s)$ for s $\leq$ t.Then I can get an expression for $d(X_t Y_t)$ with the following trick:I start with decomposition: $(X_t+Y_t)^2=X_t^2+Y_t^2+ 2X_t Y_t$,Which lead...Read more

I always thought equity options where quoted with implied volatility, the price being given by the Black-Scholes price of the option with volatility equal to the implied volatlity.But apparently there is another way to quote option : with reference implied vol (or reference price) and delta for an underlying at a price near the spot price. I also heard about "target price" in this context.I cannot find any reference about this, so more more information or an explanation would be great...Read more

We know Black-Scholes is an imperfect model for options pricing. Why is so much of the analysis of its defects focused on implied volatility? The fact that IV varies for the same stock at the same time proves that BS is faulty (and cannot be fixed by making volatility time-dependent). It also proves that $\sigma$ is not actually the standard deviation of the stock price, but just some fudge factor that is used to tweak results.Why all the attention on this fudge-factor? Once you know the model is defective I would think you should go back t...Read more

The following is a summary of the derivation of the Black-Scholes equation as given on wikipedia (http://en.wikipedia.org/wiki/Black-Scholes_equation#Derivation) - I have a question regarding the assumption that the specified portfolio is self-financing.We have a two asset market:$dB_t = B_t r dt $$dS_t = S_t (\mu dt + \sigma dW_t)$We introduce a European option with price $v(t,S_t)$ at time $t$. We now consider a portfolio consisting of one option and -$\frac{\partial v}{ \partial S}$ stocks. Therefore if $X_t$ is our wealth at time $t$, we mu...Read more

A sequence of transformations can be used to turn the Black-Scholes PDE into the heat equation.Let $C(S, t)$ be the price of a vanilla European option at time $t$, maturing at time $T$, where the underlying stock's price is $S$.$C(S, t)$ satisfies the Black-Scholes equation:$$ \frac{\partial C}{\partial t}+ \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{\partial S^2}+ (r-q)S\frac{\partial C}{\partial S} - rC = 0$$By introducing new variables $\tau = \frac{\sigma^2}{2}(T-t)$ and $x = \ln(S/K)$ and a suitable choice of constants $\alpha$ and $\bet...Read more

I am trying to solve the equation $\frac{d}{dt}V(t)=r(t)V(t)+\pi-\mu(x+t)(b_d-V(t))$numerically using the R function 'ode'. This is a Thiele differential equation for a life insurance reserve with premium rate $\pi$, mortality intensity $\mu$ (for an $x+t$ year old), death benefit $b_d$ and interest rate process $r$.I want to do this in a so called unit-linked setting, where the returns on the policy are generated by investments in stocks, hence I have assumed a Black Scholes model for simplicity. I generate a Geometric Brownian Motion. As seen...Read more

By the Black-Scholes operator I mean the following.$$L_{BS}u(x) = \frac{1}{2}\sigma^2x^2\frac{\partial^2}{\partial x^2}u(x) + rx\frac{\partial}{\partial x}u(x) - ru(x)$$Obviously, the domain of $L_{BS}$ must be a subset of twice continuously differentiable functions on some interval but making a definitive statement about the domain requires something specific about the application at hand (option pricing in this instance) and knowledge about the origin of the operator. So what is what I am looking for.Feynman-Kac says something about which sol...Read more

I'm having troubles with the transformation from the Black-Scholes PDE and transforming it to the diffusion equation. I read this other stackexchange post (Here) and I understand most of the process, except where they changed the initial condition.I got \begin{equation}\begin{split}u(x,0) &= e^{r\tau}C(S,T)\\&=e^{r\tau}\text{max}(S-K,0)\\&=e^{r\tau}\text{max}(e^y-K,0)\\&=e^{r\tau}\text{max}(e^{x-(r-\sigma^2/2)\tau)}-K,0)\\&=\text{max}(e^{x+\sigma^2\tau/2}-e^{r\tau}K,0)\\\end{split}\end{equation}Which is different from their...Read more

By Transformation from the Black-Scholes differential equation to the diffusion equation - and back, we are able to transform vanilla European option into a heat equation.And we know that the arithmetic-average strike continuous sampling Asian Black-Scholes equation is $$\frac{\partial V}{\partial t} +\frac{1}{2}\sigma^2S^2\frac{\partial ^2 V}{\partial S^2} +rS\frac{\partial V}{\partial S} + S\frac{\partial V}{\partial J}- rV=0$$i.e., only one more term $S\frac{\partial V}{\partial J}$ compared with original BS equation.Since this equation is...Read more