# Tag Info

3

Intuitive argument: by symmetry we should have $P(X_1 \ge 0 \mid W_1 = 0) = \frac{1}{2}$. However, $P(f(1, W_1) \ge 0 \mid W_1 = 0)$ is either $0$ or $1$ depending on whether $f(1,0) \ge 0$. Of course this is not really a proof because I conditioned on an event of probability 0. So let's try to use the same idea in an actual proof. Consider the ...

2

Consider a function $f: [0,T] \times \Omega \to \mathbb{R}$ which is also in your "good conditions" space, denoted $\mathcal{V}$. Consider the operator $I : \mathcal{V} \to L^2(\Omega, \mathcal{F}_T, P)$ given by $$I_T(f) := \int_0^T f(s,\omega) \, dB_t(\omega).$$ Ito's isometry tells us $$E\left( \left( \int_0^T f(s) \, dB_t \right)^2 \right) = \int_0^T ... 1 It seems like you need some further clarification. What you want to do is to fix a point x and a time t, and consider them as known. Now your solution is$$ f(t,x) = E[ h(X_T) | x,t]where X_u is a stochastic process satisfying \begin{aligned} &\mathrm{d} X_u = 2u \, \mathrm{d}u + u^2\, \mathrm{d}W_u \quad \text{when} \quad t ... 1 You don't have the right argument, the real reason why you can get the third line is because first B and Z are orthogonal so that\mathbb{E}^{\mathbb{Q}^{t} }\bigg[\bigg(\int_0^t\sigma_r^t(s)dB_s^t \bigg)\cdot\bigg(\int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{Z^t}\mathbb{1}_{S_t > K}]dZ_s^t \bigg)\bigg] =0 and because : ...

1

As you correctly write, the integral $\int_0^t b(X_s) dX_s$ is well defined (as an extended Itô integral) provided that $\int_0^t b(X_s)^2 ds<\infty$ almost surely. So the question is: When is the integral $\int_0^t f(X_s) ds$ finite almost surely? The answer to the latter question is well known: this is the case if$^*$ \$f\in ...

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