11
Here's my favorite example of an intraday strategy on S&P500 futures that at least used to work:
Intraday Share Price Volatility and Leveraged ETF Rebalancing
I pull it out whenever people start talking about market efficiency. The strategy is very simple: if S&P500 futures are up or down more than 2% on the day with two hours left until close, ...
7
Such a complex question...
Geometric Brownian Motion (GBM) will not typically work to aid one finding strategies based on technicals, as the pursuit of the technical trader is to find market deviations from a random walk.
However, some strategies, for example a "take profit/stop loss" strategy can work, (or at a minimum one can change the risk/reward ...
7
What a great question! I've had a go at it below, I'd say I'm about 75% sure of the result I've got to but I'd love feedback from others.
I'm going to use the definition of the Ito integral,
\begin{align}
\int^t_0 \sigma_s dW_s = \lim_{n \to \infty} \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( W_{t_i} - W_{t_{i-1}} \bigr)
\end{align}
where $t_n = t$.
Then, using the ...
6
In order to apply Ito's lemma, your function needs to be a twice-differentiable function. There is no issue with the non-differentiability of the Wiener process. $\frac{dF}{dX}$ involves differentiating F, not the Wiener process X.
Using a simple analogy: instantaneous velocity ($\frac{dD}{dt}$) is the derivative of position (D) over time; what is ...
5
A volatility innovation is the difference between our best prediction of future volatility and what is actually observed. Say that we predict the volatility at the next time step as $E_t[ \sigma_{t+1}] = \hat{\sigma}_{t+1}$, but instead observe $\sigma_{t+1} = \hat{\sigma}_{t+1} + \varepsilon_{t+1}$. Here $\varepsilon_{t+1}$ is the innovation, the ...
5
Note that
\begin{align*}
\int_0^t e^{B_s}ds &= t\int_0^1 e^{B_{tu}}du\\
&=t\int_0^1 e^{\sqrt{t}\frac{1}{\sqrt{t}}B_{tu}}du\\
&=t\int_0^1 e^{\sqrt{t}W_u}du,
\end{align*}
where $\{W_u=\frac{1}{\sqrt{t}}B_{tu}, \, u\ge 0\}$ is a standard Brownian motion. Then
\begin{align*}
\frac{1}{\sqrt{t}} \ln \int_0^t e^{B_s}ds &= \frac{\ln t}{\sqrt{t}} + \...
4
I have been through your confusion myself for the last five years. Until recently, my account started to get some consistent performance.
First, I started with Technicals, Spent $$$ on a automated trading platform. From there I created common strategies. The results is not promising. The strategy doesn't consists parameters and if one strategy works on one ...
4
While Richard's answer is technically correct, just saying the result can be obtained using Ito's formula doesn't make the issue much clearer. So let me go into the microscopics of the issue.
The Ito integral is defined in the following way. Suppose we divide the time interval $[0,t]$ into $n$ pieces with $t_i = i~dt$ where $dt=\frac{t}{n}$ then we define ...
4
Apply Ito's lemma to $f(W_t) = W_t^2$ then
$$
f(W_T) = f(W_0) + \int_0^T f'(W_t) dW_t + \frac{1}{2} \int_0^T f''(W_t) dt.
$$
Thus
$$
W_T^2 = 2 \int_0^T W_tdW_t + \frac12 2 T = 2 \int_0^T W_tdW_t + T.
$$
If we rearrange terms then we get
$$
\int_0^T W_tdW_t = (W_T^2-T)/2.
$$
4
Note that, for $0 \leq s < t$,
\begin{align*}
W_t^3 &= (W_t-W_s+W_s)^3\\
&= (W_t-W_s)^3 + 3(W_t-W_s)^2 W_s + 3 (W_t-W_s) W_s^2 + W_s^3.
\end{align*}
Moreover,
\begin{align*}
E\big( (W_t-W_s)^3 \mid \mathcal{F}_s\big) &= E\big( (W_t-W_s)^3\big)\\
&= 0,\\
E\big((W_t-W_s)^2 W_s \mid \mathcal{F}_s\big) &= W_s E\big( (W_t-W_s)^2\big)\\
&...
4
Let us define the auxiliary process $\Lambda_t=e^{\kappa t}\lambda_t$. Note that:
$$ \Lambda_t = \kappa e^{\kappa t} \int_0^t(\rho_s-\lambda_s)ds+\delta e^{\kappa t}\int_0^tdN_t$$
Hence after a jump occurs at $t$:
$$ \Lambda_t=\Lambda_{t-}+\delta e^{\kappa t}$$
Therefore by Ito's lemma for jump-diffusion processes:
$$ \begin{align}
d\Lambda_t & = \...
4
Let $W_t= -B_t$. Moreover, let $a= - \frac{r-\frac{1}{2}\sigma^2}{\sigma}$ and $b= -\frac{1}{\sigma}\ln \frac{S^*}{S_0}$. Then, as in this question,
\begin{align*}
\mathbb{P}\left(\tau \ge T \mid W_T\right)\pmb{1}_{\{W_T \le b-aT\}} &= \mathbb{P}\left(W_t+at \le b, t\in[0, T] \mid W_T\right)\pmb{1}_{\{W_T \le b-aT\}}\\
&=\Big[1-\exp\Big(-\frac{2}{T}b\...
4
Using Itô's Lemma, notice that:
$$d(tS_t)=tdS_t+S_tdt=dX_t+S_tdt$$
Hence:
$$X_t=tS_t-\int S_udu$$
Using independence of Brownian increments, $E(S_udW_u)=E(S_u)E(dW_u)=0$, and the chain rule for the 4th step:
$$\begin{align}
E(X_t)&=E\left(\int dX_u\right)
\\
&=\int uE(dS_u)
\\
&=\int u\mu E(S_u)du
\\
&=S_0\int u\mu e^{\mu u}du
\\
&=S_0\...
4
If you are happy to try the brute force approach, then here are the relevant formulae:
In ordinary calculus, you have the product rule for the differential of two variables:
$$d \left( x_1 x_2\right)=x_1 dx_2+x_2 dx_1$$
The general version of this for differential of products of n variables is:
$$d \prod_{i=1}^{n}{x_i}=\sum_{i=1}^{n}{ \prod_{j=1 j \ne i}^...
4
Just wanted to add to @StackG's great answer using a different approach. Please, double-check my solution as well because I'm not 100% sure.
Let $\sigma_t$ be sufficiently regular such that $\dot{\sigma}_t \stackrel{def}{=}\frac{d \sigma}{dt}$ is well defined. Then, Ito's lemma:
$$
d(\sigma_t W_t) = \dot{\sigma}_t W_t dt + \sigma_t dW_t
$$
which in integral ...
4
I finally got the idea behind the example. To illustrate it in a more general setting I will present a rigorous proof:
Let $x_k$ denote the salary and $b_k$ the number of persons that earn $x_k$ or more. Following the proposed power law by Taleb, we have:
$x_{k}:=x_02^k$ and $b_k:=b_0\left(\frac{1}{2}\right)^{ka}$, where $a\geq1$ and $k\in \mathbb{N}$.
The ...
4
Let $Y_t= e^{B_t}$ and $Z_t = B_{t}-t / 2$. Then,
\begin{align*}
dX_t &= Z_t dY_t + Y_t dZ_t + d\langle Y, Z\rangle_t\\
&=(B_{t}-t / 2)e^{B_t}\big( dB_t + 1/2\,dt \big) + e^{B_t}\big(dB_t -1/2\, dt\big) + e^{B_t} dt\\
&=e^{B_t}(B_t-t / 2+1)dB_t + e^{B_t}(B_t/2-t / 4 -1/2+1)dt\\
&=e^{B_t}(B_t-t / 2+1)d\big(B_t+1/2t\big).
\end{align*}
We ...
3
For the last question. We assume that
\begin{align*}
S_t = S_0 e^{(r-q-\frac{1}{2}\sigma^2)t + \sigma W_t},
\end{align*}
where $W$ is a standard Brownian motion, $r$ is the interest rate, $q$ is the dividend yield, and $\sigma$ is the volatility.
Then,
\begin{align*}
X_{u+a}-X_a &= (r-q-\frac{1}{2}\sigma^2)a + \sigma(W_{u+a}-W_u)\\
&\sim (r-q-\frac{...
3
We assume that $\gamma(s, t)$ is differentiable with respect to $t$. Then,
\begin{align*}
dx_t = \left(\int_0^t \frac{\partial\gamma(s, t)}{\partial t} dW_s \right)dt + \gamma(t, t) dW_t.
\end{align*}
3
You can show that "the implied variance of an ATM short maturity option is equal to the expectation under the risk neutral measure of the integrated variance over the life of the option." As you move away from the assumptions: ie not ATM, longer maturity, risk neutral measure far from true, then the forecasting power diminishes. (Google 'stochastic ...
3
For the first question, since by definition,
\begin{align*}
\varepsilon_t^{if} = e^{i \int_0^{t}f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{dM_s}{\sqrt{\xi}} + \frac{1}{2}\int_0^t f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{d\langle M\rangle_s}{\xi}},
\end{align*}
then,
\begin{align*}
d\varepsilon_t^{if} = i \varepsilon_t^{if} f\Big(\frac{1}{\xi}\...
3
Questions:
1=> Does anyone have a suggestion to determine a trend correctly.
My answer is in general and an opinion. Hong Kong Stock Exchange is third largest market behind Tokyo and Shanghai and most volatile market in the world. It is related to Singapore, Shanghai and Shenzhen, Korea, Taiwan and other famous Asian markets.
For rough overall trends, ...
3
We assume that the price at time $t$ of a zero-coupon bond, with maturity $u$ and unit face value, is of the form
\begin{align*}
f(u-t, r_t, x_t) = E\left(e^{-\int_t^u r_s ds}\mid \mathcal{F}_t\right).
\end{align*}
Note that
\begin{align*}
M(t, r_t, x_t) &\equiv f(u-t, r_t, x_t) e^{-\int_0^t r_s ds} \\
&=E\left(e^{-\int_0^u r_s ds} \mid \mathcal{F}_t ...
3
Note that, for $t>s>0$,
\begin{align*}
X_t-X_s &= \frac{1}{t}\int_0^t udW_u - \frac{1}{s}\int_0^s udW_u\\
&=\frac{1}{t}\bigg(\int_s^t u dW_u + \int_0^s udW_u \bigg)- \frac{1}{s}\int_0^s udW_u\\
&=\frac{1}{t} \int_s^t u dW_u + \Big(\frac{1}{t} -\frac{1}{s}\Big)\int_0^s udW_u\\
&=\frac{1}{t} \int_s^t u dW_u - \frac{t-s}{t} X_s.
\end{align*...
3
I am not providing a full proof but a reference for you to read up the details. The key step is mentioned below.
Most models used in finance are Markovian which is kind of in line with the efficient market hypothesis. The key step of of seeing that the Heston process is Markovian is the following theorem.
Let $f$ be a bounded Borel function from $\mathbb{...
3
The SDE you are describing is called the Geometric Brownian Motion. In the end its just a model, which underlies certain assumptions, which are usually not met in the real world scenarios. There are many further extensions and variation of SDEs for modelling prices f.e. including a jump component (jump diffusion models), mean reversion (f.e. Ornstein-...
3
Take the analogy of equations modelling something in physics.
Just because you write down an equation, it does not mean it has to be connected to anything in reality. It only do so to the extent you have adapted the equation and it's parameters to fit reality.
In finance things are a bit more complicated when it comes to the predicting power though.
...
2
Note that $X$ is a continuous martingale. Moreover, the quadratic variation is given by
\begin{align*}
\langle X_t, \, X_t\rangle = \int_0^t |\sigma_u|^2 du = c^2 t.
\end{align*}
That is,
\begin{align*}
\langle X_t/c, \, X_t/c\rangle = t.
\end{align*}
From Levy's characterization, $X/c$ is by law a Brownian motion, which we denote by $\beta$. Then, by law,
\...
2
It appears that we need only to observe the following:
\begin{align*}
\lim_{\lambda\rightarrow 0}\frac{1}{\lambda}\int_0^{\lambda t}\sigma^2_u du &= \lim_{\lambda\rightarrow 0}\int_0^{ t}\sigma^2_{\lambda u} du\\
&= \int_0^{ t}\sigma^2_{0} du \\
&=\sigma^2_{0} t.
\end{align*}
2
We write the differential form of Ito formula for simplification. Actually, the differential form for Ito formula
$$
dF(W(t)) = 2W(t)dW(t) + dt
$$
means the integral form for Ito formula,
$$
\int{dF} = \int{2W(t)dW(t)} + \int{dt}
$$
which make sense in mathemaitcs.
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