Hot answers tagged stochastic-calculus
17
A stationary process is one where the mean and variance don't change over time. This is technically "second order stationarity" or "weak stationarity", but it is also commonly the meaning when seen in literature.
In first order stationarity, the distribution of $(X_{t+1}, ..., X_{t+k})$ is the same as $(X_{1}, ..., X_{k})$ for all values of $(t, k)$.
...
14
This is pure speculation:
MFE's are really tailored toward valuation models (how can we develop a model to price x swap, etc.). You don't entirely have to worry about those details in order to trade them: you're just quoted a price based on these models. But if you go in-house at a bank and are working as a product quant (structured products, etc.), then ...
9
There are many numerical approaches to solving stochastic integrals such as the above. Assuming that there is no closed form slight-of-hand, the easiest approach is the Monte Carlo approach. I would recommend referring to Glasserman's excellent "Monte Carlo Methods in Financial Engineering"
If you are not familiar with MC, think of it as evaluating ...
9
Baxter and Rennie say it better than me, so I will summarize them.
Suppose that $N_t$ is not stochastic and $f(.)$ is a smooth function then the Taylor expansion is
$$
df(N_t) = f'(N_t)dN_t + \frac{1}{2}f''(N_t)(dN_t)^2 + \frac{1}{3!} f'''(N_t)(dN_t)^3 + \ldots
$$
and the term $(dN_T)^2$ and higher terms are zero. Ito showed that this is not the case in the ...
8
(1) You analytically solve a stochastic differential equation (SDE) using Ito's lemma. Your second equation (the discretized one) is how you could model one path over one step. To find the solution, you would model many of these paths over many steps and then take the expectation (i.e., Monte Carlo methods). The solution to the SDE models all of these paths ...
7
This is the separable differential equation for simple continuous compounding!
See this very accessible article for a step-by-step derivation (esp. under continuous compounding):
http://plus.maths.org/content/have-we-caught-your-interest
6
If $\alpha(t)$ is of finite variation, then the product rule is the same as in ordinary calculus:
$$
d(\alpha(t)X_t) = \alpha(t) dX_t + X_t d\alpha(t).
$$
If you had $X_t$ and $Y_t$ as processes, you would get
$$
d(X_t Y_t) = X_t dY_t + Y_t dX_t + d [X,Y]_t.
$$
If $Y$ has finite variation, the last quadratic covariation term is zero. The second equation ...
6
I like Richard's answer, but I think we can compute the mean and the variance of $\int_0^T W_t dt$ by ourselves using Ito's lemma. Let $f(W_t, t) = t W_t$.
$$
d( t W_t ) = W_t dt + t dW_t .
$$
Integrating both sides, and re-arranging the terms, we get
$$
\int_0^T W_t dt = T W_T - \int_0^T t dW_t \, .
$$
We'll be using Ito's isometry formula $\mathbb{E} ...
6
The model for the stock is the Bachelier model with the solution
$$
S(t) = S(0) + \sigma W(t)
$$
Thus the law of the stock $S(t)$ is Gaussian with mean $S(0)$ and variance $\sigma^2 t$.
For average process $Z(T)$ is thus the average of linear Brownian motion, we can rewrite this as
$$
Z(T) = \frac{1}{T} \int_0^T S(0) + \sigma W(t) dt = S(0) + ...
6
The classic argument using risk-neutral pricing is to assume that discounted stock prices are $\tilde{P}$-martingales where $\tilde{P}$ is the risk-neutral probability measure.
Then, you know that
$$\frac{S_t}{(1+r)^t}=\tilde{E}[\frac{S_T}{(1+r)^T} | \mathcal{F}_t]$$
by definition of a martingale process.
As the discounts are non-stochastic, you can ...
6
A process is defined here and is simply a collection of random variables indexed (in general) by time.
Otherwise I know the concept stated by Shane under the name of "weak stationarity", strong stationary processes are those that have probability laws that do not evolve through time.
More formally let $X_t$ be a given process, then let's call $P_X$ the ...
5
A very excellent discussion of stationarity as it relates to trading can be found in Sherry's (Sherrys'?) Mathematics of Technical Analysis (poorly organized, but very useful book). As he puts it, if the price changes of a stock, etc., are stationary over a time period, the underlying rules generating the price changes are effectively unchanged. The ...
5
In general, if you have a process that you can write under the form $F(B_t,t)$ where $F$ is $\mathcal{C}^{2,1}$ then Itô's lemma gives you the drift term and diffusion term of $dF$. Then if the resulting SDE has a null drift (that's where Black Scholes PDE comes from), and you get a only local martingale. For it to be a proper martingale you can look at ...
5
I think you should see the hint as follows:
$$d(W_t^{n+1})=d(f(W_t))$$ with $$f(x)=x^{n+1}$$
Apply Ito:
$$d(W_t^{n+1}) = f'(W_t)dW_t + \frac{1}{2} f''(W_t) d<W>_t$$
$$d(W_t^{n+1}) = (n+1) W_t^n dW_t + \frac{1}{2} n (n+1) W_t^{n-1} dt$$
If you integrate, you get:
$$W_{t_2}^{n+1}-W_{t_1}^{n+1}=(n+1) \int_{t_1}^{t_2} W_t^n dW_t+ ...
4
Okay so I'll take Jase answer and format it properly so that it answers your question and it will be useful for users in the future.
For clarity, let me restate the dynamics of the Modified Ornstein-Uhlenbeck model using the more common notation:
$$dS_t = \theta (\mu-S_t)dt + \sigma S_t dW_t$$
This blog post provides a closed form solution:
$$ S_t = S_0 ...
4
This is not a finance concept. Augmented data is related to Bayesian inference. It's essentially a way to improve maximum likelihood estimation from incomplete data. For details see the article "The calculation of posterior distributions by data augmentation" by Martin A. Tanner and Wing Hung Wong (it's referred in the paper you are reading).
3
Just following Musiela Rutkowski (the link redirects to Amazon). The risk neutral measure is derived form imposing that the present value of a self financed portfolio (i.e.; no infusion or withdraw of money) is a martingale. A portfolio can be seen as a stochastic process where its value at time $t$ is given by
$$
V_t = \phi^0_tP_t + \phi^1_tS_t\ ,
$$
...
3
Well the problems where Malliavin Calculus is applicable are mostly regarding greeks of exotic derivatives where some non smoothness in the payoff function creates trouble when trying to get this by finite difference methods. The thing is in my opinion that Malliavin Calculus is only an opening as it gives you basically an infinite number of ways to get ...
3
Do his first step first; integrate both sides:
$$\displaystyle \ \ \int_0^T \frac{dS(t)}{S(t)} = \mu T - 0 \,\,\,\,\,\,\,\,\,\,\,(1)$$
With zero diffusion, we know that $\langle S_.\rangle_t = 0$. Therefore, by applying Ito's lemma (or actually normal calculus):
$$d\ln{S(t)} = \frac{1}{S(t)}dS(t)\,\,\,\,\,\,\,\,\,\,\,(2)$$
Sub this into $(1)$:
...
3
1 ) A first-order Taylor expansion gives $\ln \left(\frac{S_{t+\Delta_t}}{S_t}\right)\approx \frac{S_{t+\Delta_t}-S_{t}}{S_t}+o(\Delta_t)$ , thus unless $\Delta_t$ is not small you can drop the residual term and consider $Z_t\overset{law}{=}\frac{S_{t+\Delta_t}-S_{t}}{S_t}$.
2 ) Calculation of the moments: we can proceed by using the classical Dynkin way
...
3
Hi the forward rate equation is not dependent on the model it is calculated upon the prices of zero coupon bonds by the following equation :
$$
P(t,T)=exp{-\int_t^T f_t(u).du}
$$
If you have a continuum of zero coupon bond prices which are sufficiently smooth then you can deduce from it that :
$$f_0(T)=-\frac{\partial Ln(P(0,T))}{\partial T}$$
Anyway, ...
3
(1)
You can easily solve it in the case of constant coefficients. The answer will be $\infty$.
In fact, this equation has no solution on any interval. The intuition is the following. For the SDE like
$$
dS_t = \mu(t,S_t)dt+ \sigma(t,S_t)dw_t
$$
you can mention that $dw_t = \xi_t\sqrt{t}$, where $\xi_t\sim\mathcal{N}(0,1)$ are standard gaussian i.i.d. random ...
2
1) This last DE is implicit equation. This can't be solved analytically. I guess you can solve it by finite difference method.
2) The last term is indeed a differential term of order 1/2. However, it is the term for time difference and it can remain in the equation as it is. In the final formula as well it will come out to be as difference term, implying ...
2
For Itô Processes $dX(t) = \mu(t) \mathrm{d}t + \sigma(t) \mathrm{d}W(t)$ you have the result that (under appropriate assumptions which ensure that the local martingale is a martingale, e.g. $E( (\int \sigma(t)^2 \mathrm{d}t )^{1/2} ) < \infty$, etc.): $X$ is a martingale $\Leftrightarrow$ $\mu(t) = 0$.
So in order to check if a process $X$ is a ...
2
ad) "Is it normal to assume no other drift?"
Under measure P you might have drift. You could use it as a working assumption, but in general indices drift every now and then. So, no, usually you do not assume away the drift.
"The index is described as "following a geometric Brownian motion", which to me says that the there is no other drift going on" ...
2
As far as I can tell, you've essentially written the model that you are concerned with. The only difference is that you would instead have $\theta_{i}$ when $s_{t}=i$ where $s_{t}$ is a latent variable that reflects the probability of being in state $i$. You would also need to include the dynamics that drive the probability transitions as another part of ...
2
The standard method to manage your kind of problem (i.e. dealing with stochastic processes that are note presented or built thanks to a Brownian motion) is to use a measure change.
The power of Brownian motion is that you have a lot of representation theorems (Doob-Meyer theorem, Wold theorem, etc) that allows to (thanks to a change of measure or a ...
2
What you have to start with is:
$$dS_t=\mu S_t dt + \sigma S_t dW_t$$
where $W_t$ is a standard brownian motion (SBM).
You want to solve for $S_t$, so how would you proceed?
If you integrate both sides of the equation between 0 and $T$, you get:
$$S_T - S_0= \mu \int_0^T S_t dt + \sigma \int_0^T S_t dW_t$$
Okay and then what? The fact that you have ...
2
it depends on how applied the class is. A deep understanding of stochastic calculus is not required for "P-Quants", the type of person that lives in the physical word of forecasting and risk. That being said understanding the type of models that get used by the Q-Side (requiring lots of stochasic theory) is a useful skill to have.
Like John said, if you ...
2
"Like" Ito:
$$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 $dB dB = dt$.
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