# Tag Info

Accepted

### Difference between weak ( or martingale ) and strong solutions to SDEs

The main difference between weak and strong solutions is indeed that for strong solutions we are given a Brownian motion on a given probability space whereas for weak solutions we are free to choose ...
• 120k
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### Solution to General Linear SDE

Here is the complete solution to the problem including some special cases for an easy start. With analogy to the integrating factor method from ODEs it seems natural to rearrange \begin{align*} \...
• 3,293
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### Why do we study Cameron-Martin Space and what is the motivation behind it?

The infinite-finite dimension dichotomy , and Fourier coefficients When we study (multivariate) normal random variables on $\mathbb R^d$, we often have equivalent ways of understanding them. One ...
Accepted

### Initial Distribution of Stochastic Differential Equations

Part 1: Weak existence In order to construct and study weak solutions to SDEs it is often very useful to take advantage of the close connection between SDEs and martingale problems. For simplicity of ...
• 120k
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### Stochastic Differential Equation solution for Geometric Brownian Motion

Your integration step is wrong. $Z$ here refers to Brownian motion, and thus you need to apply Ito Integration. One way I like to see the extra drift is needed, is that we take derivative of the ...
• 7,792

• 1,006
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### Why is Brownian Motion so Big in the Theory of Stochastic Differential Equations?

Actually it is a massive loss of generality. Here are two reasons why people use models based on Brownian motion: the central limit theorem. The CLT is saying that Gaussian distributions appear ...
• 120k
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### Must the strong solutions of an SDE be unique? Can there be two different strong solutions?

Example 1: Note that stochastic differential equations (SDEs) are a generalization of ordinary differential equations (ODEs); in particular, any ODE which does not have a unique solution does not have ...
• 120k
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### In calculus, if $\frac{dy}{dx}$ is not in fact a fraction, is the equation below for geometric brownian motion technically incorrect?

As suggested in the comments, the paths of $W_t$ (and also $S_t$) are almost surely nowhere differentiable, so the notation $\frac{dW_t}{d t}$ is meaningless in this context. The SDE as written is ...
• 12.7k
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### SDE Solution: Hull-White extension of Vasicek model

The SDE can be solved similarly as in the Vasicek model. Define $F(t,r(t)) = e^{\alpha t}r(t)$, then \begin{cases} \displaystyle \frac{\partial F}{\partial t} &= \alpha e^{\alpha t} r(t) \\ \...
• 1,410
Accepted

### Solution of SDE $dX_t = \mu(t)X_tdt + \sigma X_t dW_t$

The reason $\ln(x)$ is chosen is this: (Note: where I say $B_{t}$ I mean Brownian motion, which you denoted in your question as $W_{t}$) The SDE you provided is one of the few we can explicitly ...
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### Stochastic Differential Equation solution for Geometric Brownian Motion

I believe the answer by @Yujie Zha can be simplified substantially. Thanks to @Dr. Lutz Lehmann for providing a link to this, my solution is the same as the solution on page 15, but with more ...
• 619
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• 120k

### Show that $dX_t=\frac{X_t}{1-t}dt+dW_t$ can be written as $X_t=(1-t)\int_{0}^{t}\frac{1}{1-s}dW_s$

Let $Y_{t} = \int_{0}^{t} \frac{1}{1-s}\, dW_{s}$. Next take a look at $$X_{t} = (1-t) \int_{0}^{t} \frac{1}{1-s}\, dW_{s} = (1-t)Y_{t}$$ and differentiate using It^o's lemma \begin{align*} dX_{t} &...
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### Computing expected value of a function along a Brownian Path

Write $$E[\exp\big(t + \frac{1}{2}W(t)\big)]=e^{t} \cdot E[\exp\big( \frac{1}{2}W(t)\big)]$$ at this point remember the fact that $W(t)$ is a centered Gaussian r.v. with variance equal to $t$, and use ...
• 3,327

### A question about SDE and geometric Brownian motion.

I find notation such as $\frac{dN_t}{N_t}=rdt+\alpha dB_t$ incredibly unfortunate, it leads to the type of confusion you have encountered (and that I used to encounter myself). That is why I advise ...
• 1,219
### Expected value of $S_t$ where $dS_t=(\mu S_t+a)dt + (\sigma S_t +b)dW_t$
Following @KurtG's comment, you may observe that $\beta_t M_t = \int_0^t \beta_s dW_s$, being the Itô integral of an $L^2(\Omega \times [0,t])$ process, is a martingale, so that $E(\beta_t M_t) = 0$. ...