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$(x,y)$ is uniform so it's density is constant, say $c$. To determine $c$, note that $F(1,1)=1$ or $$\int_0^1\int_0^1cdxdy=1\Rightarrow c=1$$ It's not clear what do you mean by $P\{Y\ge x|Y\ge 1/2\}$. If $x$ is a constant then $$P\{Y\ge x|Y\ge 1/2\}=2(1-x)\wedge 1$$ because the marginal distribution of $Y$ is uniform over $[0,1]$. If you actually ...

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To show independence you just need to show that $\Bbb{P} (B_t- B_s \in A_0, B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k) = \Bbb{P} (B_t- B_s \in A_0)\Bbb{P} ( B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k)$ for every $r_1, \ldots r_k \leq s$ and $A_1, \ldots, A_k \in \mathcal{B}$. This is because $\sigma(\{B_r, r \leq s\})$ is generated ...

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Recall that for any function $f \geq 0$ and any measure $\mu$ we have $$\int f \, d\mu = 0 \implies f=0 \quad \mu\text{-a.s.}$$ Therefore, as $(\langle X \rangle_t)_{t \geq 0}$ is an increasing process, $$\mathbb{E} \left( \int_0^t e^{X_s} \, d\langle X \rangle_s \right)=0$$ implies $$\int_0^t e^{X_s} \, d \langle X \rangle_s =0$$ $\mathbb{P}$-almost ...

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Consider the first projection map $\pi_1: \Bbb R \times \Omega \to \Bbb R$. This map is measurable by the definition of the product sigma algebra $\mathcal B \otimes \mathcal F$. Similarly, the second projection map $\pi_2: \Bbb R \times \Omega \to \Omega$ is measurable. Thus $X \circ \pi_2$ is measurable as a map from $\Bbb R \times \Omega \to \Bbb R$, ...

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This might help you to handle measurability on product spaces To get the measurability consider first specific functions $$f(x,y)= 1_A(x)1_B(y)$$ $x \mapsto \int_Y f(x,y)\, d\nu(y)= \nu(B)1_A(x)$ This is a measurable function of the first variable. Now consider $\mathcal{C}= \big\{D \in X \times Y; x \mapsto \int_Y1_D(x,y)\, d\nu(y) \text{is ... 1 Integration by parts is the right idea, but let's start with the left side of$\displaystyle\int_{0}^{\infty}(1-F(x))\,dx - \int_{-\infty}^{0}F(x)\,dx = E(X)$instead of the right side. We have$\displaystyle\int_{0}^{\infty}(1-F(x))\,dx = \underbrace{\left[x(1-F(x))\right]_{0}^{\infty}}_{0}-\int_{0}^{\infty}-xF'(x)\,dx = \int_{0}^{\infty}xf(x)\,dx\$, and ...

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