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Assume the price follows a lognormal process. We can convert it into a problem of finding the probability of a standard Brownian motion particle starting from $0$ and hitting $x$ before time $t$, or its first passage time $\tau_x$ being less than $t$. This can be derived through the reflection principle. The paths crossing $x$ are exactly paired up by the ...

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I'll outline how you can estimate the (implied) real-world density function from (observed) option prices. Having found this real-world density, you can then compute all sorts of probabilities and quantify the market's expectation of future prices. Recall firstly that (European-style) options are priced as risk-neutral expectation of the discounted payoff. ...

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$\mathbb{P}$ is the true probability measure. Measure $\mathbb{Q}$ is a measure of convenience that allows risk neutral pricing. Stochastic discount factor $M$ takes you between the two. If you care about prices you can either: (1) work under $\mathbb{Q}$ or (2) work under $\mathbb{P}$ with a stochastic discount factor $M$. There's an isomorphic ...

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If you want to address interesting problems that are interesting for financial mathematics, I do not believe you have the good list. Pricing. For instance, most of explicit formulas for pricing that are not available yet will never be. In this direction, you should have a look at simulation techniques. See for instance Nonlinear Option Pricing. Interesting ...

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In the derivatives context, "arbitrage free" means almost surely for the probability measure under consideration. This is in opposition with statistical arbitrage used at high frequencies for example. More precisely the assumption is that there is no $T\geq 0$ and self-financed portfolio $V$ such that $V_0 = 0$, $P(V_T < 0) = 0$ and $P(V_T > 0) > ... 12 In order to estimate a probability of an event with small probability, you might want to try to estimate the probability for a changed random variable that allocates a larger probability mass for the event happening. So, in your case, you might want to change the original$N(0, 1)$to$N(100, 1)$because for the second r.v. the probability of it being higher ... 10 The way that I understand your question is that you are looking to fit the market prices of European plain vanilla options of a single maturity and then back out the corresponding implied probability density function. There are multiple ways that you could approach your problem. 1) Modelling the Market Prices The market prices of European plain vanilla ... 9 the information you provided is not sufficient to deduce risk neutral probabilities. You have to provide something like a price process from which risk neutral probabilities can be computed. Here are some examples: Example1: Consider a game where you pay 1 and you win 6 in case a six is thrown and 0 otherwise. So in financial mathematics terms we have a ... 8 I can clarify 100% that$(dw)^2$=$dt$and recommend you to accept it as a fact. Like any other differential, this differential is defined in terms of its integral: $$\int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2}$$ Where$t_{k}=t_{0}+k(t_{1}-t_{0})/n. Since W(t_{k+1})-W(t_{k})=\sqrt{t_{k+1}-t_{k}}\... 7 Not sure about all of the complicated math and programming above, but I can tell you that, if you want to calculate for 1 Standard Deviation from the current stock price X days away, the following calculation will give you a +/- value from the current stock price. 1 StdDev Move = (Stock Price X Implied Volatility X the Square Root of 'how many days') all ... 7 Is this the proof you are looking for? -- from Shreve, S. E.'s book "Stochastic calculus for finance II, continuous-time Models", chapter 5. 7 you should have a look at implied probability densities. They do exactly what you are asking - extracting the pricing density from option prices. This is done by differentiating the option price with respect to the call. Here are two links. The first one explains the procedure the second one deals with where such densities can be applied Implied state ... 7 You cannot use negative probabilities in this context. When there is no unique probability measure, there can be no unique price. You only know that it is in [0, 0.6] range, if you want to tighten this interval you need to make further assumptions/tweak inputs I agree with your conclusion that there no suitable probability measure. But I am not sure about ... 7 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}\, ... 7 IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure \mathbb{Q} equivalent to \mathbb{P} and under which X_t is a martingale. Just take\frac {d\mathbb{Q}}{d\mathbb{P}} =\mathcal{E}\left(-\int_0^t \cos(s) dW_s + a\right)for any a \in \... 7 Firstly, m_T=\min\limits_{t\in[0,T]} B_t = -\max\limits_{t\in[0,T]} -B_t \overset{Law}{=} -\max\limits_{t\in[0,T]} B_t = -M_T. So, you can either consider the running maximum or minimum. Let \tau be a stopping time and (B_t) a Brownian motion. Then, \begin{align*} W_t =\begin{cases} B_t & t\leq \tau, \\ 2B_\tau - B_t & t\geq \tau, \end{cases} ... 6 One approach is to take the entire option chain, and calculate the prices for adjacent butterflies along the chain. The risk / reward of each of the butterflies represent the empirical probability that the market is pricing for the underlying to move between the strikes of the butterfly. To make sure it is a proper probability distribution, you will want ... 6 Since W_{2t}-W_{t} is independent of W_t and has the same law as W_{2t-t}=W_t we only have to computeP(X(X+Y)<0)where (X,Y) follows a bivariate normal distribution (with zero correlation). From there you can split the probability in two cases : either X<0 and X+Y>0 or the opposite. The two events have the same probability since (-... 6 I believe there is not a unique price if you can't short. Say, instead of buying the option you spent 0.5 on a half a unit of the asset S^2_1 This asset pays out [0.4, 0.6, 0.8] which first order stochastically dominates the option. So, no matter your probability beliefs about the states, in that setting you'd never pay 0.5 for the option which pays ... 6 Consider two jointly normal random variables X_1 \sim N(u_1, \sigma_1^2) and X_2 \sim N(u_2, \sigma_2^2). Note that, \begin{align*} \max(X_1, X_2) = X_2 + \max(X_1-X_2, \ 0). \end{align*} Moreover, X_1-X_2 is a normal random variable with mean \mu=\mu_1-\mu_2 and variance \begin{align*} \sigma^2 = \sigma_1^2+\sigma_2^2 - 2 \rho\sigma_1\sigma_2, \end{... 6 Quantiles are preserved under monotonic transformations, hence the quantile for Y is simply the exponential of the quantile of X, no need for corrections whatsoever (see here for instance). Put otherwise, let q denote the quantile \alpha of X i.e.\Bbb{P}(X \leq q) = \alphathen \begin{align} \Bbb{P}( X \leq q ) &= \Bbb{P}( \underbrace{\... 6 The question requires you to provide a method which uses uniform random variables and transforms them to generate realizations of the described asset values. To give a bit more general answer: this is solved by the inverse transform sampling method. The main idea is to obtain realizations of a random variablex$with any given distribution function$F(x)$, ... 5 When I run this simulation I see the same results, and it makes sense. For the straight 50%/50%, I found that my win ration was about 38% and my loss ratio 61%. The reason it wasn't 50/50 was that if I had consecutive up flips my value could keep going up, but if I had consecutive down flips I would 0 out and the sequence would have to end as I had lost ... 5 I asked this question 6 years ago, and in the meantime I came across this little volume: Lévy Processes in Finance: Pricing Financial Derivatives by Wim Schoutens (2003). 5 Orthogonality and independence are different concepts. The concepts are the same for Wiener processes because in the context of normal random variables, independence is equivalent to orthogonality (i.e. uncorrelatedness) Independence is the standard definition for probability. Let$\mathcal{F}, \mathcal{G}$be the sigma algebras generated by two processes,... 5 Let's consider a random process$X$. If$X$is an adapted process, then we know, without any uncertainty, what its value is at the present time. This idea is formalized with measure theory. For$X$to be a martingale, it needs to have the following property: at any given time, our best estimate of the value at some point in the future (i.e. forecast), is ... 5 I have had a read through the paper that you quoted and have the following comments which you might find helpful: (I am formally trained in QM, so hopefully there shouldn't be any errors in the physics portions of the answer, but if there are any questions then please comment). A few comments about Quantum Mechanics (QM): Quantum mechanics is a physical ... 5 In this case it is just the notion that your payoff function should not explode at some point - made mathematically rigorous. Have a look at the following picture from wikipedia: Intuitively the Lipschitz condition (or Lipschitz continuity) ensures that your payoff function always remains entirely outside the white cone, so it cannot e.g. become infinitely ... 5 I understand that Moody's uses an empirical distribution while KMV uses a normal distribution in order to calculate these probabilities KMV doesn't use a normal distribution to map distance to default to a probability of default (EDF in the KMV model). It uses a proprietary database. By a strict structural interpretation,$EDF\$, the expected default ...

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