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 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{...


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}^...


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 ...


2

Thought to add this as a comment, but it appears too long. Your question does not appear complete, that is, the rationale for using two Brownian motions is not clear. Note that \begin{align*} dX_t &= \mu dt + \sigma_1 dW^1_t + \sigma_2 dW^2_t \\ &=\mu dt + \sqrt{\sigma_1^2 + \sigma_2^2 + 2 \rho \sigma_1\sigma_2}\frac{\sigma_1 dW^1_t + \sigma_2 dW^...


2

The expectation follows from Fubini since $\mathbb{E}\left[\int_0^t B_s^2 \mathrm{d}s\right] = \int_0^t \mathbb{E}[B_s^2] \mathrm{d}s= \int_0^t s\mathrm{d}s = \frac{1}{2}t^2$. The variance follows from Ito's Isometry and is answered here.


2

You have $$dZ_t = df\left(S_t, B_t, X_t\right) = \frac{\partial f}{\partial s}dS_t + \frac{\partial f}{\partial b}dB_t + \frac{\partial f}{\partial x}dX_t + \frac{1}{2}\left[\frac{\partial^2 f}{\partial s^2} d\langle S\rangle_t + 2\frac{\partial^2 f}{\partial s\partial x} d\langle S, X\rangle_t \right]$$ since $d\langle B\rangle_t$, $d\langle B, S\rangle_t$, ...


2

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*...


2

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

The GBM model is liked by practitioners for the modelling of stock prices for the following reasons: (i) The solution is log-normal, so the stock price distribution varies between zero and infinity: which is what we would expect from a real-world stock price. (ii) The model has independent increments, which means the future distribution of the stock only ...


2

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-...


2

In this (extremely technical) paper by Duffie et al it is shown that a Markov process is infinitely decomposable if and only if it is a regular affine process. So their results establishes a correspondence between Markov processes and regular affine processes. Okay, that (the paper) is too technical for me, but if I look at the characteristic function of ...


1

A stochastic process with independent increments is a markov process. the proof is available in the following document: (Lemma 1.1) http://statweb.stanford.edu/~adembo/math-136/Markov_note.pdf


1

\begin{align} S(T) - S(t) := \int_t^T dS(s) = r(T-t) + \sigma (W(T) - W(t)) \end{align} Since $W(T) - W(t) \sim N(0, T-t)$, we have $S(T) \sim N\left(S(t) + r(T-t),\sigma^2(T-t) \right)$.


1

The first step is to include jumps in the stock price. Then, you can easily add jumps into the variance process. If you only consider seldom, large jumps, you may want to use a jump-diffusion like the models from Merton (1976) and Kou (2002). The former uses a log-normal distribution for the jump size whilst Kou employs a double exponential distribution. A ...


1

Your process for $(S_t)$ is a geoemtric Brownian motion and since $S_t=S_0 e^{\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t}$, we have \begin{align*} \ln(S_t) &= \ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t \\ &\sim N\left(\ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)t,\sigma^2 t\right). \end{align*} Thus, \begin{align*} X_t &= e^{\...


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