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What your exercise asks you to prove, is a generalization of the Markov Inequality which can be done as follows \begin{align*}M&=E[\phi(|X|)]=\int_{-\infty}^{+\infty}\phi(|x|)f_X(x)dx\\&=\int_{|X|< c}^{}\phi(|x|)f_X(x)dx+\int_{|X|\ge c}^{}\phi(|x|)f_X(x)dx\\&\ge \int_{|X|\ge c}^{}\phi(|x|)f_X(x)dx\\&\overset{*}\ge\phi(c)\int_{|X|\ge ... 3 Hint: Define the random variable with Y_n \in \{0,1\} (indicator variable with 2 states) as followsY_n=\begin{cases}1,& X_n=6\\0,&X_n\neq6\end{cases}$$with transition matrix$$\mathbf P_{(1)}=\begin{array}{r|cc|r}&0&1&\\\hline0&\frac{4}{5}&\frac{1}{5}\\1&1&0 \end{array}$$Initially Y_0=1. You want to find the ... 3 The (simple) Markov property is used at the first equal sign after 1\geqslant\cdots, to assert that, for each time k, the probability P(X_k=j,\text{something happens after time k}) is also P(X_k=j) times P_j(\text{same thing shifted by time k happens}). Re your second question, introduce \alpha=\inf\limits_{0\leqslant\ell\leqslant ... 2 For every t\geqslant0, (B_{t+s}^2)_{s\geqslant0} is distributed as (X_s)_{s\geqslant0}, where, for every s\geqslant0,$$X_s=B_t^2+2\sqrt{B_t^2}\cdot W_s+W_s^2,$$where (W_s)_{s\geqslant0} is a Brownian motion independent of (B_u)_{0\leqslant u\leqslant t}. Thus, indeed, (B^2_t)_{t\geqslant0} is a Markov process. The discrete analogue of this ... 2$$\mathbb E\phi(|X|) \geq \mathbb E\phi(|X|)1_{\{|X| >c\}} \geq \phi(c)\mathbb E1_{\{|X|>c\}}$$1 Here is a simple case showing that (\Omega_s) is irrelevant but that (r(t)) is usually not Markov. Assume that \sigma(s,t,\nu)=t-s for every s\leqslant t and every \nu, then elementary computations yield$$r(t)=F(0,t)+\tfrac18t^4+X_t,\qquad X_t=\int_0^tB_s\mathrm ds, hence the process $(r(t))$ is Markov if and only if the process $(X_t)$ is. But ...

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In this specific example, one can argue as follows. The event $\{U_n=u_n\}$ contains all the information one needs to calculate the probability distribution of $U_{n+1}$, since all one needs is the maximal value up to time $n$ (i.e. the value of $U_n$), which will be compared with the new dice outcome at time $n+1$. Having information about all the past ...

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This is essentially asking "Prove that the probability of the $n+1^{th}$ event depends only on the outcome of the $n^{th}$ event, and not on the ones prior to that." In this context, it is clear that the probability of $U_{n+1}$ being $u_{n+1}$ definitely depends on what $U_n$ was - clearly, $U_{n+1}\geq U_n$, however, knowing what $U_{n-1}$ was provides no ...

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We know the continuous time Markov chain is also a composition of an imbedded discrete time Markov chain $Y$ with a P(o)isson process $N$ such that $X(t)=Y(N(t))$. Do we know that? Actually a DTMC composed with a homogenous Poisson process jumps after exponentially distributed times whose common parameter is the parameter of the Poisson process while ...

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