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1

$i \times \sigma \times W$ is a solution of your equation. Its variance at time $t$ is equal to $\sigma^2 \times t$ which is positive. Please check this page for more details about how to compute variance for complex random variables: Wikipedia: complex random variables The variance is always a nonnegative real number. It is equal to the sum of the ...


5

Note merely that $B_t=B_s+(B_t-B_s)$ which is the sum of independent normally distributed random variables. In particular, $B_s$ is $\mathbb{F}_s$-measurable and $B_{t-s}$ is independent of $\mathbb{F}_s$. Thus, \begin{align*} \mathbb{E}_s[Z_t] &= \mathbb{E}_s\left[\exp\left(-\frac{1}{2}\theta^2t+\theta B_t\right)\right] \\ &= \mathbb{E}_s\left[\exp\...


2

Actually this is just the Black-Scholes SDE with zero drift and $-\frac{1}{2}\sigma^2$ volatility. If you plug that into the well known solution, you get $S_t=S_0e^{\frac{1}{8}\sigma^4t-\frac{1}{2}\sigma^2 W_t}$ but let's calculate it with Ito's formula. Choose $f(x)=\log(x)$, then we have $f'(x)=\frac{1}{x}$ and $f''(x)=-\frac{1}{x^2}$. Inserting in Ito's ...


1

Let $(B_t)$ be a standard Brownian motion. Then, $B_t\sim N(0,t)$ and $B_t-B_s\sim N(0,t-s)$. Informally, you can say $\mathrm{d}B_t\sim N(0,\mathrm{d}t)$ where $\mathrm{d}B_t=B_{t+\mathrm{d}t}-B_t$ is an infinitesimal increment.


1

Let $Y = \log X$, then: $$\begin{align} Y &= Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t \\ EY_t &=Y_0 + (\mu-\frac{\sigma^2}{2})t \\ EY_tEY_s &= Y_0^2 + Y_0 (\mu-\frac{\sigma^2}{2}) (t+s) + (\mu-\frac{\sigma^2}{2})^2 t s \\ E(Y_tY_s) &= E\left((Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t) (Y_0 + (\mu-\frac{\sigma^2}{2})s + \sigma W_s)\...


0

Thanks a lot for the above great answer:). Here I also added the link for Levy's Characterization of Brownian Motion: http://individual.utoronto.ca/normand/Documents/MATH5501/Project-3/Levy_characterization_of_Brownian_motion.pdf


7

I've seen that Gordon answer is more concise and to the point. Take this as a complementary answer. This is a general approach that will work for all this type of linear SDEs, not just this one. Assume we have the following linear SDE $$dX_t = (F_t X_t +f_t)dt + (G_t X_t +g_t)dB_t \tag*{(1)}$$ where $F, G, f$ and $g$ are Borel measurable bounded ...


10

Let \begin{align*} Y_t = e^{(a+\frac{c^2}{2})t-cW_t}. \end{align*} Then \begin{align*} dY_t = Y_t\left[\big(a+c^2\big)dt -c dW_t \right]. \end{align*} Moreover, \begin{align*} d(X_tY_t) &= Y_t dX_t + X_t dY_t + d\langle X, Y\rangle_t\\ &=abY_tdt. \end{align*} That is, \begin{align*} X_t = Y_t^{-1}\left(X_0 + ab\int_0^t Y_sds\right). \end{align*}


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