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First we show that that if $X\colon \Omega \rightarrow \mathbb{R}^n$ is normal $\mathcal{N}(m, C)$, $Y\colon \Omega \rightarrow \mathbb{R}^n$ is normal $\mathcal{N}(m^{\prime}, C^{\prime})$ and they are independent then $X+Y$ is normal $\mathcal{N}(m_1+m_2, C_1+C_2)$. Here $m$, $m^{\prime} \in \mathbb{R}^n$ and $C=[c_{jk}]$, $C^{\prime}=[c_{jk}^{\prime}]$ ...

2

We have $dX_t = dW_t - \frac{a}{2}dt$ and $d[X]_t = d[W]_t = dt$ since the deterministic piece $-at/2$ doesn't contribute to the quadratic variation. So, $$f(X_t) = f(X_0)+\int_0^t f'(X_s)\, dW_s - \frac{a}{2} \int_0^t f'(X_s) \,ds + \frac{1}{2} \int_0^t f''(X_s) \, ds \\ = f(X_0)+\int_0^t f'(X_s)\, dW_s + \frac{1}{2} \int_0^t \big[f''(X_s) - af'(X_s) ... 2 This is in Feller volume 2, and on page 2 of this paper. 2 For any \omega \in \Omega, we can construct a sequence (\Pi_n)_n=(\Pi_n(\omega))_n of partitions of [0,T] such that the mesh size |\Pi_n| tends to 0 as n \to \infty and$$\lim_{n \to \infty} \sum_{t_j \in \Pi_n} (B_{t_j}(\omega)-B_{t_{j-1}}(\omega))^2 = \infty.$$This means in particular that$$[B,B](T)(\omega) := \sup_{\Pi} \sum_{t_j \in \Pi} ...

2

There are several definitions for the quadratic variation: For a "nice" process $(X_t)_{t \geq 0}$ ("nice" means semimartingale), the quadratic variation is defined by $$[X]_t := X_t^2-X_0^2 -2 \int_0^t X_{s-} \, dX_s.$$ For $X_t := B_t^2$, we know from Itô's formula that $$d(B_s^2)= 2 B_s \, dB_2+ \, ds$$ and $$B_t^4 = 4 \int_0^t B_s^3 \, dB_s + 6 ... 2 For every t\geqslant0, (B_{t+s}^2)_{s\geqslant0} is distributed as (X_s)_{s\geqslant0}, where, for every s\geqslant0,$$X_s=B_t^2+2\sqrt{B_t^2}\cdot W_s+W_s^2,$$where (W_s)_{s\geqslant0} is a Brownian motion independent of (B_u)_{0\leqslant u\leqslant t}. Thus, indeed, (B^2_t)_{t\geqslant0} is a Markov process. The discrete analogue of this ... 2 This notion is called tightness of a sequence of measures. You can apply it in probability theory with the sequence P\#{X_n} of the image probabilities under the action of X_n:$$ P\#{X_n}(A) = P(X_n\in A) $$You transfer the topology issues to the (metric, often polish) space \mathcal X where$$ X:\Omega \to \mathcal X $$Note also that in the ... 2 Surely not an exam proof, what is going to come. We will use Theorem 5.4.1 from Ethier and Kurtz, Markov Processes, Wiley, 1986 (EK86); the important part is quoted below. Let us first write down the generator A_n of the process X^{(n)} $$A_n f(x) = \sin (nx) f'(x) + \frac{1}{2} f''(x) , x \in \mathbb{R}.$$ The expected ... 2 Let X_t=t\,B_{1/t}. Every linear combination of the random variables X_t is a linear combination of the random variables B_t hence it is normal. Thus, the process (X_t) is gaussian, in particular every increment X_t-X_s is gaussian. More generally, let Y_t=a(t)B_{c(t)}+b(t). For every functions (a,b,c), the process (Y_t) is gaussian. 1 As D is open and y \in D, we have$$r := d(y,\partial D) := \inf\{z \in \partial D; |z-y| \}>0.$$On the other hand, the boundedness of D implies that we can choose R>0 such that |z| \leq R for any z \in \partial D. Consequently,$$\log r \leq \log|B(T)(\omega)-y| \leq \log (R+|y|)$$holds for any \omega \in \Omega. Taking expectation ... 1 Key-facts: The function x\mapsto\|x\|^{-1} is harmonic on \mathbb R^3\setminus\{0\} and the distribution of \sqrt{t}\underline{X} is radially symmetric. Assume that \underline{X}_0 and \underline{X} are independent (otherwise anything can happen) and note that the result is a consequence of the fact that, for every random variable B whose ... 1 As mentionned by saz from Itô's lemma applied to Y_t=B^2_t you get : dY_t=finite_variation_ter.dt + 2.B_tdB_t so the quadratic variation is (using Itô's isometry) is equal to :$$<Y>_t=4\int_0^t B_s^2ds$$Please note that this is a stochastic process. Best regards 1 Hint: under the risk neutral probability, the prices of the securities having an L^2 payoff are martingales. So you probably need to compute the Ito-differential of$$ (t,\omega)\to S^3_t e^{(2r+3σ^2)(T−t)} $$1 As the process (W(t))_{t\ge 0} has the same distribution than (cW(\frac t{c^2}))_{t\ge 0}, you get with c = \sqrt t:$$ P( W(t) > 0 ;\ \ W(2t) > 0) = P\left( cW(\frac t{c^2}) > 0 ;\ \ cW\left(2 \frac t{c^2}\right) > 0\right)\\ = P\left( W\left(\frac t{c^2}\right) > 0 ;\ \ W\left(2 \frac t{c^2}\right) > 0\right)= P( W(1) > 0 ; ...

1

Let $(P_t)_{t\ge0}$ denote a regular conditional distribution of $W$ given $T$. This means that each $P_t$ is a probability measure on $C[0,\infty)$, the set of continuous functions from $[0,\infty)$ to $\mathbb{R}$ endowed with the $\sigma$-algebra induced by the coordinate projections, such that $$P(W\in A,T \in B) = \int_B P_t(A) d Q(t)$$ where $Q$ is ...

1

It suffices to prove that $\int_0^t \phi_s^2 \, ds$ is finite for all $t \in [0,\infty)$. This allows us to define the stochastic integral $$\int_0^t \phi(s) \, dB_s$$ for any $t \in [0,\infty)$ (as a local integral). This means that your proof is correct. In fact, the very argumentation shows that $\Lambda_{\text{loc}}^2$ contains all functions $f$ ...

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