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3

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 follows $$Y_n=\begin{cases}1,& X_n=6\\0,&X_n\neq6\end{cases}$$ 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 ...


1

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 ...


1

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 ...


1

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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