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9 votes

Why does Black Scholes formula give inconsistent dimensional analysis result?

$C= S_0 N(d_1) - K e^{-rT} N(d_2)$ $C$, $S_0$ and $K$ have units of currency (e.g. USD). $N(d1)$ and $N(d_2)$ are unit-less (dimensionless), the formula is dimensionally correct. Considering, $d1 ...
bhutes's user avatar
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6 votes
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Brennan-Schwartz algorithm for pricing American options

Ikonen and Toivanen don't say that the LCP is solved exactly, they simply say that the modified back-substitution is a valid algorithm to solve the LCP. A numerical error may arise around the ...
jherek's user avatar
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6 votes
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Any book which is intro to PDEs but prioritises techniques useful for solving Black-Scholes?

Ok I'll answer this from a practioner's perspective rather than a purist's (which I am solidly not). Below are the books that influence my understanding of this space, listed in chronological order of ...
James Spencer-Lavan's user avatar
6 votes
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Since $S = e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t}$, why treat it as a constant when calculating the greek Theta (dC/dt) for a European call option?

Let me heed @Bob's suggestion and turn my comment into a full answer: Like other disciplines, finance uses lots of shortcuts to achieve brevity and convenience. That can be awfully confusing for ...
Kevin's user avatar
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5 votes

Pricing log-contract with Black-Scholes PDE

The B/S PDE for a contingent claim $V(S, t)$ is \begin{equation} \frac{\partial V}{\partial t} + r S \frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial V^2}{\partial S^2} - r V = ...
LocalVolatility's user avatar
5 votes
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Boundary Conditions for Call Spread

Time $T$ boundary condition is correct $u(T,x)=(x-K_1)^+-(x-K_2)^+$. Time $x\to 0$ boundary condition is known and is equal to $0$. Time $x\to\infty$ boundary condition is also known and is correct $...
M. Jeunesse's user avatar
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5 votes
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Errors on Finite Differences + Implicit Scheme + Black & Scholes

The PDE is defined for $x \in ]-\infty, +\infty[$ but the finite difference scheme requires a truncated domain $[x_{\min}, x_{\max}]$, and the choice of $x_{\min}$ and $x_{\max}$ will affect the ...
Antoine Conze's user avatar
4 votes

Why does it make sense that $S$ and $e^{rt}$ are solutions to the Black-Scholes PDE?

These are well known trivial solutions to the Black-Scholes PDE. The first one is just the price of the underlying stock and the second is interest bearing money in a bank. These are trivially true ...
vonjd's user avatar
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4 votes
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Why does it make sense that $S$ and $e^{rt}$ are solutions to the Black-Scholes PDE?

Under the standard assumptions, generally speaking, any contract that depends on the current values of t and S, and which are paid for at the start satisfy this PDE. In the financial context, the ...
Magic is in the chain's user avatar
4 votes

Alternative derivation of the Black Scholes formula

The option pricing formula must satisfy the PDE you have derived for all values of $K$. The only way this can be the case is if the two parts that you separate are both equal to zero. Suppose the ...
RLK's user avatar
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4 votes
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Proof Black Scholes Theta

There is a well known identity for the Black Scholes model: $S_0 n(d_1)-X e^{-rT} n(d_2) = 0$ (proof). Using this allows you to combine these two terms: $$S_0 n(d_1)\frac{\partial d_1}{\partial t} - ...
Alex C's user avatar
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4 votes

Boundary condition issues for Black-Scholes PDE using finite-differences

Alright, so I solved my issues here. Since it might be useful to others stuck on the same thing, I mention it here. The problem I described about assuming the second order term was equal to zero is ...
Jesper Tidblom's user avatar
4 votes
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Hedging gamma, theta or other risks

In the Black-Scholes model Gamma and theta do not need to be hedged because the BS PDE says that they balance each other (I'll take $r = 0$): $$ \frac{\partial f}{\partial t} + \frac12 \sigma^2 S^2\...
Frido's user avatar
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3 votes
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Cash deposit in replicating portfolio for BS equation unnecessary?

The key point here is that the portfolio must be self-financing, namely the initial option premium $V_0$ should be enough to allow you to hedge it throughout its life. If not, the option price $V_0$ ...
Daneel Olivaw's user avatar
3 votes
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Finite Difference Method for Black-Scholes-Formula

It depends on the type of payoff you want to price. If it is a call option, you know that $V(0,t) = 0$ and $V(x,t) \approx x$ when $x \rightarrow +\infty$ so you can use a dirichlet condition $V(a,t) =...
Antoine Conze's user avatar
3 votes
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Solve Black scholes PDE without using any transformation

Yes it can be done. However, bear in mind that a naive explicit FD scheme is not unconditionally stable (see CFL stability condition). As far as your initial/boundary conditions issue is concerned: [...
Quantuple's user avatar
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3 votes
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Monte Carlo and PDE results are different for a Call Option!

I would definitely recommend Volopta as a reliable source of self-contained and commented financial engineering source codes (useful for prototyping/understanding but clearly not production code). I ...
Quantuple's user avatar
  • 14.7k
3 votes

What's the intuition behind the transformation of Black-Scholes into Heat equation?

The intuition is that the price process is diffusive in nature. Over time the distribution of possible prices for the underlying spreads out (i.e. the variance in the possible price 1 year from now is ...
roz's user avatar
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3 votes
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Linear Or nonlinear Black Scholes Equation

The linear/non-linear classification is concerned about the dependent variables, and its derivatives. To verify whether the equation is linear, you should be checking that the equation is linear in ...
Magic is in the chain's user avatar
3 votes
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Black Scholes PDE

Your derivation is right and as Alex said, the only difference between the equations is how you measure time. There are two possibilities Time going forward, $t\in[0,T]$, Time going backwards, $\tau\...
Kevin's user avatar
  • 16k
3 votes

Nonlinear Black-Scholes model Vs linear Black-Scholes

In general you have don't have an exact solution for the non linear equation - so you have to use numerical methods. Have you seen this Non linear option pricing book. It covers the topics that you ...
Magic is in the chain's user avatar
3 votes
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Implicit finite difference method always guarantees positive and stable price of derivative?

For the original PDE, the positivity can be deduced from the maximum principle for a parabolic operator. There is also a discrete version of the maximum principle for the finite difference parabolic ...
Hans's user avatar
  • 2,806
3 votes
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Boundary Conditions for Call Option in Black Scholes Model

Note that: $$ C(t,S) =S-K{\rm e}^{-r(T-t)} $$ as $S\rightarrow \infty$, for all $t$. Basically because one can easily accept $$ P(t,S) =0 $$ as $S\rightarrow \infty$, for all $t$, and one still ...
ir7's user avatar
  • 5,043
3 votes

How to derive a pricing PDE for an asset that follows a mean-reverting process?

In a Black-Scholes-Merton-style hedge portfolio, we'd get: $$dS_t=\kappa\left(\mu-\ln S_t\right)S_tdt+\sigma S_t dW_t $$ with a hedged portfolio $$\Pi_t\equiv V_t-\Delta_tS_t$$ and $$ d\Pi_t=\frac{\...
Kermittfrog's user avatar
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3 votes

How to derive a pricing PDE for an asset that follows a mean-reverting process?

Let $\{r_t, \, t\ge 0\}$ be the interest rate process. For maturity $T$ and $0\le t \le T$, note that \begin{align*} V(S_t, t) = e^{\int_0^t r_s ds}\,\mathbb{E}\left(e^{-\int_0^T r_s ds}V(S_T, T) \...
Gordon's user avatar
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3 votes
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Black-Scholes PDE derivation gap

Ito process representation uniqueness states that, if $$ \int_0^t Y_u du + \int_0^t Z_u dW_u = 0 $$ for all $T\geq t\geq 0$, then $$ Y = Z = 0 $$ almost surely (aka up to a set of measure $0$). In ...
ir7's user avatar
  • 5,043
3 votes

What are the parallels between the Black-Scholes equation and the heat equation?

For heat equation, it describes how heat diffuses (usually measured by temperature) through the length of the material and over time. For Black Scholes, it describes how the value of the option ...
mirmo's user avatar
  • 131
2 votes
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Can someone check this boundary condition for me?

Judging from the oscillations near $S=0$, it looks like the payoff function is causing these problems. Your payoff should go towards -1 as $S$ goes towards zero, but your computer might just ...
werki's user avatar
  • 36

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