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7

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 $...


3

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\...


3

I write down the solution for the Heston model. You can directly generalise the result. Let $f=f(t,s,v)\in C^{1,2,2}(\mathbb{R}_+^3)$ be a real-valued function (portfolio value) and consider the two-dimensional stochastic process $(S_t,v_t)$ with \begin{align*} \mathrm{d}S_t&=(r-q) S_t \mathrm{d}t+\sqrt{v_t} S_t \mathrm{d}W_{1,t}, \\ \mathrm{d}v_t&=\...


5

It appears that you need to read some books such as Stochastic Differential Equations. For such type of equations, you need to use something called integrating factor such as the function $e^{-\mu t}$ here. Note that \begin{align*} d\big(e^{-\mu t} X_t \big) &= X_t d\big(e^{-\mu t}\big) + e^{-\mu t} dX_t\\ &=-\mu e^{-\mu t} X_t dt + e^{-\mu t} (\mu ...


2

I came across this thread while searching for a similar topic. In Nualart's book (Intoduction to Malliavin Calculus), it is asked to show that $\int_0^t B_s ds$ is gaussien and it is asked to compute its mean and variance. This exerice should rely only on basic brownian motion properties, in particular, no Itô calculus should be used (Itô calculus is ...


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}} + \...


1

$f'(X_t, t)$ refers to $\frac{\partial f}{\partial x} (X_t, t)$. If you make this change in notation, along with correcting the typo pointed out by @Alex C, the two versions of the Ito's lemma will match. Also, the notation used in Mark Joshi's book is not standard; your confusion in this scenario is natural.


1

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.


4

An integral with respect to a stochastic process is the theme of stochastic calculus for which you ought to get an introductory textbook as it is the key to financial models. A Brownian motion $(W_t)$ is the easiest integrand and typically the first example one encounters. Then, $\int_t^T 1\mathrm{d}W_s=W_T-W_t=W_{T-t}=\sqrt{T-t} Z$ where $Z\sim N(0,1)$. ...


1

Consider sequences of n flips with p=1/2 for simplicity. The sets defined based only on the first flip obviously divide the space in half, so each has positive probability (1/2) no matter what n is. But consider the outcome A (a singleton set) defined by seeing n heads. If n=2, P(A)=.25. More generally, P(A)=0.5^n, which goes to zero as n goes to infinity. ...


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