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Set $$M_t := f(X_t) - \int_0^t Af(X_s) \, ds$$ and write $P_t f(x) := \mathbb{E}^x f(X_t)$ for the semigroup associated with $(X_t)_{t \geq 0}$. Using the definition of the generator, it is not difficult to see that $$\frac{d}{dt} (\mathbb{E}^x f(X_t)) = \frac{d}{dt} P_t f(x) = P_t Af(x).$$ By the fundamental theorem of calculus, this implies $$P_t ... 2 Take a standard brownian motion B_n and p=4.$$\mathbb{E} \left[ (B_m - B_n)^4 \right] =\mathbb{E} \left[ B_{m - n}^4 \right] = 3 (m-n)^2 \neq 3 ( m^2 - n^2 ) = \mathbb{E} \left[ B_m^4 - B_n^4 \right]$$2 Recall that for any function f \geq 0 and any measure \mu we have$$\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$$\mathbb{P}-almost ... 2 Define an array as a sequence of random variables on a probability space (\Omega, \mathcal{F},P). Introduce doubly infinite arrays of random variables X_{n,j},~\mathcal{F}_{n,j} for j,n\geq 1, and set sub \sigma -algebras of a \sigma -algebra. Adapting the array to the filtration, then \{X_{n,j}\} is a MDA if the relation ... 2$$X_{n+1}=(Z_1+\cdots+Z_n+Z_{n+1})^2=X_n+2Z_{n+1}(Z_1+\cdots+Z_n)+Z_{n+1}^2.$$Now use the fact that expectation is linear, and that Z_{n+1} is independent of F_n, along with the fact that E[Z_{n+1}]=0, E[Z_{n+1}^2]>0. 1 Denote S_n = Z_1 + \cdots + Z_n. S_n is easily shown to be a martingale. Now$$ \mathbb E[X_{n+1} \mid \mathcal F_n] =\mathbb E[S_{n+1}^2 \mid \mathcal F_n] \geq (\mathbb E[S_{n+1} \mid \mathcal F_n])^2 = S_n^2 = X_n where the inequality follows from Jensen's inequality. 1 I believe you are confusing the mean and mode of your distribution, which is asymetric. S_t follows a lognormal distribution with parameters \mu = -\frac{1}{2} t and \sigma^2 = t. So, E [S_t] = e^{\mu + \frac{1}{2} \sigma^2} = e^{-\frac{1}{2}t + \frac{1}{2} t} = e^0 = 1, but the mode of the distribution is e^{\mu - \sigma^2} = e^{-\frac{3}{2}t}. ... 1 You do have a martingale! \begin{align*} \mathbb{E} [Z_t | \mathcal{F}_{t-1}] &= \mathbb{E} [X_t | \mathcal{F}_{t-1}] + \mathbb{E}[Z_{t-1}|\mathcal{F}_{t-1}] \\ &= 0 + Z_{t-1} \end{align*} 1 Since (X_t) is a Markov process, \mathbb{E}(g(X_t)\mid {\cal F}_s)=P_{s,t}g(X_s) where P_{s,t} is the transition kernel. Therefore \mathbb{E}(g(X_t)\mid {\cal F}_s) is measurable with respect to \sigma(X_s)=\sigma(Z_s). 1 Actually, you may want to account for the underlying random process... Let \{U_n\}_{n=0}^\infty be a sequence of i.i.d. uniform random variables on [0,1]. Let r_n denote the number of red balls at time n and l_n=r+g+t\cdot n ThenX_n=\frac{r_n}{l_n}\text{ and } r_{n+1}=r_n1\{U_{n+1}>X_n\}+(r_n+t)1\{U_{n+1}\le X_n\} Denote ...