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3

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 $l_n=r+g+t\cdot n$ Then $$X_n=\frac{r_n}{l_n}\text{ and } r_{n+1}=r_n1\{U_{n+1}>X_n\}+(r_n+t)1\{U_{n+1}\le X_n\}$$ Denote ...



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