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7

I think this question has no easy answer but I'll give it a shot anyway (beware: oversimplification ahead!). The main idea of the Malliavin calculus is to be able to differentiate stochastic processes like Brownian motion (or more general martingales with bounded quadratic variation), which are not differentiable in the traditional sense (because of their ...


7

The part where you say that $$\frac{dS_t}{S_t} = d\ln(S_t)$$ is wrong, because $S$ is a stochastic variable. This is exactly what Itô tells you with his formula that you apply right do compute your $dZ$. The difference comes from the quadratic variation of the process $S$ which you express as $(dS)^2$. If you don't add this term when the variable are ...


6

If you consider $X_1$ a random variable which is normally distributed with mean $\mu$ and variance $\sigma^2$ them $S_1 = \exp(X_1)$ is log-normally distributed with mean $\exp(\mu + \sigma^2/2)$ and variance $(\exp(\sigma^2)-1)\exp(2\mu+\sigma^2)$. This follows from the definitions of the normal distribution and the log-normal distribution and deriving the ...


5

Multi-fractal models can be applied to the modeling and forecasting of volatility. I read the following book with much interest and actually setup couple models in order to compare performance vs Garch family models and the application of multi-fractals much better captures discontinuous regime-changes than traditional volatility models. ...


4

If the loss distribution is normal with mean $\mu$ and variance $\sigma^2$, then the Value-at-Risk and Expexted Shortfall (or CVaR) at level $\alpha \in (0, 1)$ are \begin{align*} \mbox{VaR}_\alpha & = \mu + \sigma \Phi^{-1}(\alpha) , \\ \mbox{ES}_\alpha & = \mu + \sigma \frac{\phi\{\Phi^{-1}(\alpha)\}}{1 - \alpha} , \end{align*} where $\phi$ ...


4

The actual problem one solves for American options is an optimal stopping time problem, so the value of the option is $$ V_0 = \max_\tau E_{\tau}\left[e^{-r \tau} (S_\tau-K)^+ \right] $$ where the maximum is taken over all stopping times (exercise strategies $\tau>0$ permissible in the contract). With a PDE operator such as you have, the instantaneous ...


4

The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[e^{\sigma W}]> e^{\sigma E[W]}=1$. $\sigma$ can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ...


4

Suppose that there are multiple martingale measures $Q_1$ and $Q_2$ that attain the minimal variance. Then the convex combination $Q_* := \frac{1}{2}Q_1 + \frac{1}{2}Q_2$ is also a martingale measure. Due to the strict convexity of $f(x) = x^2$, it can be shown that $$ E_P \left[\frac{dQ_*}{dP}^2 \right] < \frac{1}{2} E_P \left[ \frac{dQ_1}{dP}^2 ...


3

This is a good shorter reference: http://www.impan.pl/CZM/tankov.pdf. Cont and Tankov have also written a longer book about modelling with Levy processes that I think is really good. There's going to be a strong connection between the sequence of jump times and the Levy measure $\nu$. In a single unit of time, $ \nu(dx)$ is a measure (not necessarily a ...


3

So we have the identity $$g(S,\sigma, t, C,C_t,C_S,...)=g(S, t,\sigma, V,V_t,V_S,...)$$ where $S$, $\sigma$, and $t$ are independent variables and $V=V(S,\sigma,t)$, $C=C(S,\sigma,t)$ are some unknown functions. But we can also treat the above identity formally and assume that the functions $C,C_t,C_S,...,V,V_t,V_S,... $ are themselves independent ...


3

I will try to answer this a bit differently. The rigorous answer: because Ito calculus tells us that we need the second order term. Look at $$ S_t = S_0\exp(\mu t + \sigma B_t). $$ Assume that $S_0$ is known and fixed and look at by Ito's formula $$ d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt. $$ Then with some abuse of notation: $$ ...


3

If you allow $X_t$ to be two dimensional then a model with a stock price $X_t^1$ and its variance process $X_t^2$ (stochastic volatility) would fit your definition. In such cases to my knowledge we often don't have a closed form of the density of $X_T^1$ but in some cases we have a closed form of the Laplace transform. An example is the Heston model.


3

"Like" Ito: $$d (B^2) = B dB + B dB + dB dB$$ That is $$B dB = \frac{1}{2} d (B^2) - \frac{1}{2} dB dB$$ Integrate. Last term is 1/2 the quadratic variation. I understand the questions as follows: In iii) one has to define what $dB dB$ stands for and one has to "proof" the first line in my answer. In ii) one may use Ito to "know" that $dB dB = dt$.


3

A key property of Brownian motion is independent increments. So if $x-1 > y$, then $$ \mathbb{E}[\Delta W_x \Delta W_y] = 0 $$ because the time intervals [x-1,x] and [y-1,y] do not overlap. If they do overlap, i.e. $x-1 \leq y < x$, then \begin{align} \mathbb{E}[\Delta W_x \Delta W_y] =&\ \mathbb{E}[(W_x - W_{x-1}) (W_y-W_{y-1})] \\ =&\ ...


2

This will be the inverse process $$\frac{1}{S_t}$$ Applying Itô's formula the dynamics are then given by $$d\frac{1}{S_t}=\frac{-1}{S_t^2}dS_t+\frac{1}{S_t^3}dS_tdS_t$$ some simple algebra then leads to $$d\frac{1}{S_t}=\frac{1}{S_t}(\sigma^2 -r)dt+\frac{1}{S_t}\sigma dW_t$$


2

I don't know what you did when you tried pulling out $1-\alpha$, the correct expression would be $\lim_{\alpha \to 1} \frac{\mu(1-\alpha) + \sigma {\phi^{-1}(\alpha)}}{(1-\alpha)(\mu + \sigma \phi^{-1}(\alpha))}$. Anyhow, you can try using the substitution $\Phi^{-1}(\alpha) = x$, $x \to \infty$ and $\alpha = \Phi(x)$. Then the expression becomes ...


2

For Itô Processes $dX(t) = \mu(t) \mathrm{d}t + \sigma(t) \mathrm{d}W(t)$ you have the result that (under appropriate assumptions which ensure that the local martingale is a martingale, e.g. $E( (\int \sigma(t)^2 \mathrm{d}t )^{1/2} ) < \infty$, etc.): $X$ is a martingale $\Leftrightarrow$ $\mu(t) = 0$. So in order to check if a process $X$ is a ...


2

$$S_t = S_0\exp((r-\frac{\sigma^2}{2})t+\sigma W_t)$$ is not yet a martingale for it is not dirftless. From a probabilistic point of vew the "drift adjustment" comes into play so that the expected value of $S_t$ will be $e^{rt}$ rathern than $e^{(r+0.5\sigma^2)t}$. For the expected value of a log-normaly distributed variable with mean $\mu$ and vol ...


1

I would calculate it this way, $⟨[W_s+W_t−2W_0]^2⟩=\mathbb{E}[(W_s+W_t−2W_0)^2]\\ \hspace{4cm}= \mathbb{E}[((W_s-W_0)+(W_t-W_0))^2]\\ \hspace{4cm}=\mathbb{E}[(W_s-W_0)^2]+\mathbb{E}[(W_t-W_0)^2]+2\mathbb{E}[(W_s-W_0)(W_t-W_0)] \\ \hspace{4cm}=s+t+2\mathbb{E}[W_sW_t]\\ \hspace{4cm}=s+t+2min(s,t)$


1

I haven't read all the paper, just the section you mentioned. The previsible/predictable strategy $\pi_t$ represents the number of shares of the asset $S$ held at time $t$. The paper looks to use power utility in some way, and as is common in those types of problems, generally you want to think of $\tilde{\pi}_t$, which is the percentage of wealth at time ...


1

Any of a wide variety of local vol models, where (from your equation) $b(\cdot,\cdot)$ is some fitted surface, are unlikely to have closed-form solutions for the terminal distribution. Indeed it's well-known that these models tend to have very unusual forward term structures of volatility. As a specific example, take $b(\cdot,\cdot)$ to be an approximation ...


1

To complete the perfect answer of Richard, I would add that pretending that the expected value of the GBM at $t$ is $X_0\exp(\mu t)$ amounts to claim that $E(exp X) = exp(EX)$ which is wrong “because the exponential is not linear.” This is why there is this $\sigma^2/2$ term popping up, it is sometimes known as the “convexity correction”—the exponential ...


1

i picked this off from Shreve. Start with the definition of sampled quadratic variation: (1) $\frac{1}{2}Q_\pi = \frac{1}{2}\sum\nolimits_{j=0}^{n-1} (W_{j+1}) - W_j)) ^2$ where $\pi$ = {0,1,2...,n} is a partition of $[0,T]$ (Note we took $\frac{1}{2}$ of both sides for reasons that will be clear in the next line.) Now we know (1) is equal to ...


1

Similar to the answer aleady given. We can use a measure $Q$ such that $E_Q[A_n] = 0$. Let's reformulate the sequence as $X_0 =x$ and $X_{n+1} = X_n + A_{n+1}$. First, beause expectation is linear: $$ E_Q[X_{n+1}|F_n] = E_Q[X_n|F_n] + E_Q[A_{n+1}|F_n]. $$ Now assume that $\{F_n\}_{n=0}^\infty$ is the filtration that represents the information of ...



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