# Tag Info

1

Continuity is clearly not enough. Take: $f (x) = x$; the underlying measure space for the process as $[0,1]$ with the Lebesgue measure; $X_s (\omega) = g(s, \omega)$ for some measurable function $g$. Then you are asking whether $$\int_0^1 \int_0^t g(s, \omega) \ \text{d}s \ \text{d} \omega = \int_0^t \int_0^1 g(s, \omega) \ \text{d}\omega \ \text{d} s,$$...

1

I think the process $X_t:=\int_0^t H_s\mathbb dW_s$ is not a martingale in general. It is under very weak assumptions for $H$ a local martingale and if you have for example $\mathbb E(X_t^2)=\mathbb E(\int_0^t H_s^2\mathbb ds)<\infty$ for all $t\geq 0$ it is also a martingale. Another sufficent condition would be $\forall t\geq0: \mathbb E(\sup_{s\leq t}|... 1 In Fubini's theorem we have two measures and two Lebesgue integrals. Here you have one Lebesgue integral (i.e the expectation) and an Ito integral. The Ito integral is not defined pathwise but as a$L^2$limit. Also note that$dW_t$is not a measure. Also, it's not always true that $$\mathbb{E}\left[\int_0 ^t H_udW_u\right]=0. \tag{1}$$ The equality (1) is ... 1 With your notations, we must calculate$\int_{\mathbb{R}^2} \chi_{D}(x,y) \lambda^2(x,y)$For all$x\in \mathbb{R}$, let$D_x=\{y\in\mathbb{R}| (x,y)\in D\}$We can write$\chi_{D}(x,y) = \chi_{[-r,r]}(x).\chi_{D_x}(y)$From Fubini$\int_{\mathbb{R}^2} \chi_{D}(x,y) \lambda^2(x,y) = \int_{\mathbb{R}}\left(\int_{\mathbb{R}} \chi_{[-r,r]}(x).\chi_{D_x}(y)d\...

1

The claim is false. Here is a simple counter-example: Consider $\mathbb R$. Then $\overline B_1 =[-1,1]$ and $\overline B_2 =[-2,2]$. It is easy to see that $$\overline B_2=\left\{x+y:(x,y)\in\overline B_1\times\overline B_1\right\}$$ However, take $A=[0,1]$. The  (\lambda |_1 * \lambda |_1)(A)= (\lambda \otimes \lambda)(\{(x,y) \in [-1,1]^2: 0 \leqslant ...

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