# Tag Info

7

Then can you help me to show that $\alpha$ is in fact stopping time? Well, not very easily I am afraid, since $\alpha$ is not always a stopping time for the filtration $(\mathcal F_t)_{t\geqslant0}$ defined by $\mathcal F_t=\sigma(X_s;0\leqslant s\leqslant t)$. For a counterexample, consider some Bernoulli random variable $U$ such that $P(U=1)=\frac12$ ...

2

...we can't use Ito [l]emma because $\displaystyle\int_0^t W_sds$ is not in the form $X_t = \displaystyle\int_0 ^t \sigma_s dW_s + \int_0 ^t \mu_s ds$. Actually $X_t=\displaystyle\int_0^t W_sds$ is in the form $X_t = \displaystyle\int_0 ^t \sigma_s dW_s + \int_0 ^t \mu_s ds$ with, for every $t\geqslant0$, $$\sigma_t=0,\qquad \mu_t=W_t.$$

2

Well, in the first case, when you consider a Brownian motion $B_t$, the following is usually understood: You are considering a background probability triple $(\Omega,\mathcal{F},P)$, with $\Omega$ being some set, $\mathcal{F}$ being some $\sigma$-algebra and $P$ being some probability space, and you consider a measurable mapping $B:\Omega\to C[0,\infty)$ ...

1

I'm sorry, I was very stupid. We can just apply Ito's formula and get: \begin{align*} & \int_a^b f(t)dW_t=f(b)W_b-f(a)W_a-\int_a^b W_t f'(t)dt. \end{align*} This yields that indeed we can find the desired pathwise upper bound of $\int_a^b f(t)dW_t$ in terms of $||f||$, $||f'||$, $\sup_{t \in [a,b]}|W(t)|$.

1

In the second case it is not necessarily the case that $dX_t=\mu_t\,dt+\sigma_t\,dB_t$. But the second case should imply the first. What is $dX_t\cdot dX_t$ for the first case ?

1

You just have to be careful with the notation because $t$ can appear in two arguments. The forward rate is $$f(t,T)=f(0,T) + \int_{0}^t \alpha(s,T)ds+ \sigma W_t.$$ The short rate is $$r_t=f(t,t)=f(0,t) + \int_{0}^t \alpha(s,t)ds+ \sigma W_t = F(t,X_t),$$ where $F_X = 1$, $F_{XX} = 0$ and $$X_t = \int_{0}^t \alpha(s,t)ds+ \sigma W_t,\\dX_t = ... 1 If 2aT\geqslant1, then \exp(aW_T^2) is not integrable hence E(\exp(aW_T^2)\mid F_t) does not exist. From now on, assume that 2aT\lt1. Define Z by$$W_T-W_t=\sqrt{T-t}\cdot Z,$$then Z is standard normal and independent of F_t and$$W_T^2=W_t^2+2\sqrt{T-t}W_tZ+(T-t)Z^2,$$hence$$ E(\exp(aW_T^2)\mid F_t)=\exp(aW_t^2)\cdot ...

1

Here is a more general statement: Consider some Markov process $(Y_t)$ and define, for every $t$, $Z_t=F(t,Y_t)$, where $F$ is measurable and, for every $t$, $F(t,\ )$ has a measurable inverse. Then, $\sigma(Z_s;s\leqslant t)=\sigma(Y_s;s\leqslant t)$ for every $t$ and the process $(Z_t)$ is Markov. Can you prove this and apply it to your setting?

1

Let $(Z_t)_{t \geq 0}$ be a Markov process with generator $B$. Then we know from Dynkin's formula that $$u(Z_t)-u(Z_0)- \int_0^t Bu(Z_s) \, ds$$ is a martingale if $u$ is "nice" (e.g. bounded and in the domain of the generator). In particular, if $u$ satisfies $Bu=0$, then we see that $$\mathbb{E}^xu(Z_t) = \mathbb{E}^x u(Z_0) = u(x).$$ This means the ...

Only top voted, non community-wiki answers of a minimum length are eligible