15
A function $f : \mathbb R^n\backslash\{0\} →\mathbb R$ is called (positive) homogeneous of degree $k$ if
$$f(\lambda \mathbf x) = \lambda^k f(\mathbf x) \,$$
for all $\lambda > 0$. Here $k$ can be any complex number. The homogeneous functions are characterized by
Euler's Homogeneous Function Theorem. Suppose that the function $f : \mathbb R^n \...
13
Edit: Freddy's answer is good -- we wrote concurrently. He rightly points out that QF is a broad field, and that it is among other things a community. Here, I describe a practical, down-to-earth path for getting your feet wet in one key piece of it -- software and model development for derivatives analysis, starting with vanilla options. Your best bet ...
13
At the top of this list I still recommend you to seek employment in order to learn from others in QF space. Could you possible work in a quant team within an investment bank where you currently reside?
Start to reach out to the quant finance community so you are connected once you decide to locate to where you can practice this discipline.reach out to alumni,...
11
My understanding is because the Ito's integration definition keeps the martingale property.
With Brownian motion $W(t, \omega)$ defined, to define stochastic integration in a Riemann–Stieltjes style:
$$\int_0^t f(t, \omega) d W(t, \omega) = \lim_{\| \Delta_n\| \to 0 } \sum_{i=1}^{n} f(\tau_i,\omega) \left ( W(t_i, \omega) - W(t_{i-1}, \omega) \right ) $$
, ...
9
I'd say to read Prof. Shreve's well-known two-volume textbook Stochastic Calculus for Finance I and II.
9
In fact Ito and Stratonovich calculus are both mathematically equivalent. In the following paper you can e.g. see that both derivations lead to the same result, i.e. the Black-Scholes equation:
Black-Scholes option pricing within Ito and Stratonovich conventions by J. Perello, J. M. Porra, M. Montero and J. Masoliver
From the abstract:
Options financial ...
8
As Sanjay said, you can apply Itô's Lemma to $f(t,x)=x^2$ and obtain
\begin{align*}
\mathrm{d} S^2_t=\left(2\mu S_t^2+\sigma^2S_t^2\right)\mathrm{d}t+\left(2\sigma S_t^2\right)\mathrm{d}W_t.
\end{align*}
Thus, $(S_t^2)$ is again a geometric Brownian motion and hence, for each time point $t$ log-normally distributed with drift $2\mu+\sigma^2$ and volatility $...
7
The other answers are useful and sensible. I have worked full time in equity research for nearly two decades, so very much a "qualitative" rather than a quantitative approach. However, all the firms for which I have worked had quants and because of my casual interest in the area I've spent a lot of time talking to quant teams over the years, often over a ...
7
I think one of the best (and very current) articles about how to break into QF (for any kind of background) is:
"On becoming a Quant" by Mark Joshi
For your special background in mathematics see this excerpt from section 9:
The main challenge for a pure mathematician is to be able to get one’s
hands dirty and learning to be more focussed on getting ...
7
At the first glance, what you are asking for is a model admitting arbitrage, so there is a zero chance of losing money and positive chance of yielding profits. Well, many equilibrium models start with assuming arbitrage is not possible (otherwise it would be trivial wouldn't it).
But, in my opinion, what you actually seek is the Efficient Markets Hypothesis....
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}\, ...
6
To price financial instruments such as options, bonds and stocks must be priced so as to be "arbitrage free". The concept of arbitrage can be made precise by one of the fundamental ideas of quantitative finance, the so called Arbitrage Theorem.
Put differently the Arbitrage Theorem provides a very elegant and general method for pricing derivative ...
6
Under the risk-neutral measure the discounted (under some numéraire) price process is a martingale. If we have a bank account with dynamics $dB_t = r B_t dt$ then the discounted asset $X_t = \frac{S_t}{B_t}$ will have the dynamics
\begin{equation}
dX_t = \frac{dS_t}{B_t}- \frac{S_t dB_t}{B_t^2} = (\mu - r S_t) \frac{1}{B_t} dt + \frac{\sigma}{B_t} dW_t
\end{...
5
A detailed description of the Hurst Exponent can be found here. A further (rather short search of Google) turned up this site claiming to provide an Excel Workbook with, among other things, Hurst Exponent estimation.
5
No, a sum of two GARCH processes is generally not a GARCH process.
(I am not even sure whether there exists a nontrivial special case where the opposite holds.)
By GARCH I mean the classic definition of GARCH due to Bollerslev (1986), not an arbitrary variation like EGARCH, IGARCH, FIGARCH or whatever else.
Let me provide an example. Take two ...
5
Sorry, but despite being used as a popular example in machine learning, no one has ever achieved a stock market prediction.
It does not work for several reasons (check random walk by Fama and quite a bit of others, rational decision making fallacy, wrong assumptions ...), but the most compelling one is that if it would work, someone would be able to become ...
5
One possibility worth exploring is to use the support vector machine learning tool on the Metatrader 5 platform. Firstly, if you're not familiar with it, Metatrader 5 is a platform developed for users to implement algorithmic trading in forex and CFD markets (I'm not sure if the platform can be extended to stocks and other markets). It is typically used for ...
5
The general idea is to bootstrap the discount factors in the correct order, based on the data you have given. I'm going to make some assumptions that your bonds are paying annual coupons. The longest maturity is 2.5 years, meaning you need discount factors for 6M, 1.5Y and 2.5Y.
The 6M deposit has a rate of 5%, this tells you that you should use the 5% rate ...
5
Optimization is definitely important in Quantitative Finance, especially for portfolio optimization where we maximize utility of the return of a portfolio as linear weighted vector of asset returns subject to a desired risk level:
$$ \max_{w\in[0,1]^n} U(\mu_p(w),\sigma_p(w))\quad s.t. \sum_{i=1}^n w_i=1$$
where $w$ being the portfolio weights, and $U$ ...
5
I think the main difference even in this little example is the gain-loss asymmetry which is a known stylized fact: When you look at the big bump both time series posses your artificial one is perfectly symmetric whereas the real one takes longer for going up and then crashes in a relatively shorter time frame.
This is a known phenomenon in real financial ...
5
1) Gatheral expresses everything in forward terms: forward value of the spot and of the call.
Consider an asset $A$. You need to hold $A$ at time $T$ but since you don't need it now you don't want to buy it now. Instead you enter a forward contract with someone that says that at time $T$ you will pay the amount $K$ and get the asset in exchange. What ...
5
All the topics you've mentioned are wonderful and shouldn't be eschewed by reading some finance-oriented review book. I recommend these instead.
Linear algebra: Hoffman and Kunze and Halmos
Set theory: Halmos
Measure theory: Rudin and Tao
5
Swap
Just to be clear, (3.4c) leads to (3.5a) when we assume lognormal $R(\tau)$. Lognormal $R(\tau)$ means we can write
$$R(\tau) = R_0 e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau} Z}$$
with $Z$ normal, and I'm assuming a zero mean -- which I think is required. Then for (3.4c) we have for the expectation value:
$$ E\left[(R(\tau) - R_0)^2 \right] = ...
5
Here's how i'd have at it;
* I happen to know these are okay guesses.
** Let's assume it's just the potential energy, and that as the point of the "5000 years" part of the question.
The moon came from the earth - likely it's crust since the idea is that it was formed from an impact. The crust of the earth is less dense than the core, so the moon ...
5
Let define $\mathbb{Q}$ and $\mathbb{P}$ two equivalent probabilities on a filtered space $(\Omega,(\mathcal{F}_t)_{t\geq 0})$
Let define $Z_T=\frac{d\mathbb{Q}}{d\mathbb{P}}$ restricted to $\mathcal{F}_T$ measurable events.
It means that for $X_T$ being $\mathcal{F}_T$ measurable we have:
$$\mathbb{E}^{\mathbb{Q}}[X_T] = \mathbb{E}^{\mathbb{P}}\left[...
5
$\require{cancel}$
$$\text{PnL} = -[P(t+\delta t,S+\delta S)-P(t,S)] + rP(t,S)\delta t + \Delta(\delta S - rS \delta t + q S\delta t)$$
Assuming a pure diffusion, at the order 1 as $\delta t \to 0$
$$P(t+\delta,S+\delta S) = P(t,S) + \frac{\partial P}{\partial t}\delta t + \frac{\partial P}{\partial S}\delta S + \frac{1}{2}\frac{\partial^2P}{\partial S^2}(\...
5
Aside from the independence requirement for the increments, that is, the independence of $X_{s+t}-X_s$ and $\mathcal{F}_s$, you can check whether the increment $X_{s+t}-X_s$ has the distribution of $N(0, t)$. In fact, note that
\begin{align*}
X_{s+t}-X_s &= (\sqrt{s+t}-\sqrt{s}) Z\\
&\sim N\left(0,\, (\sqrt{s+t}-\sqrt{s})^2\right),
\end{align*}
which ...
5
It is actually rather simple.
Lets start with the fixed rate market. A can borrow at 5% while B can borrow at 7%. Simply said, A has a comparative advantage of 2% in the fixed rate market.
In the floating rate market, A borrows at LIBOR + 1% while B borrows at LIBOR + 2.5%. From here, I'm guessing you already know that A has the comparative advantage as ...
5
You are right about the dropped $\sim$, it's probably just a typo. Furthermore, remember that in stochastic calculus, you have to take into account second order derivatives, i.e.
$$d\left(\frac{1}{Y_t}\right) = -\frac{1}{Y_t^2}dY_t + \frac{1}{2}\frac{2}{Y_t^3}dY_t^2$$
which is the Taylor expansion up to second order. Then you substitute $dY_t$ in the right ...
5
As you mentioned, we have
$$\sum_{l=0}^{k}{p}=\frac{k(k+1)}{2}$$
You want to know
$$\sum_{k=0}^{n}{\sum_{l=0}^{k}{p}}=\sum_{k=0}^{n}{\frac{k(k+1)}{2}}=\frac{1}{2}\sum_{k=0}^{n}{k^2}+\frac{1}{2}\sum_{k=0}^{n}{k}$$
you know that
$$\sum_{k=0}^{n}{k^2}=\frac{n(n+1)(2n+1)}{6}$$
Therefore
$$\sum_{k=0}^{n}{\sum_{l=0}^{k}{p}}=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+...
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