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4

In general, the inequality $$\mathbb{E}\|X-Y\|^2 \leq \mathbb{E}\|X-Z\|^2 + \mathbb{E}\|Z-Y\|^2$$ does not hold. Simply consider arbitrary $X$, $Z := \frac{1}{2} X$ and $Y = \frac{1}{4}X$. Then $$\mathbb{E}\|X-Y\|^2 = \frac{9}{16} \mathbb{E}\|X\|^2$$ whereas $$\mathbb{E}\|X-Z\|^2 + \mathbb{E}\|Z-Y\|^2 = \left( \frac{1}{4} + \frac{1}{16} \right) ...


3

A fundamental theorem of Fukushima says that a sufficient condition for a Dirichlet form to be associated to a Markov process is that it be regular: in particular, this condition requires the state space to be locally compact, and doesn't cover infinite dimensional state spaces. However, Albeverio, Ma and Röckner have given a necessary and sufficient ...


3

Since the Markov chain is irreducible, it is possible to get from state $i$ to state $j$, so $p_{ij}^{(k)} > 0$ for some $k$. Then $p_{ij}^{(n+k)} \ge p_{ii}^{(n)} p_{ij}^{(k)}$, so $$\sum_{n\ge 0} p_{ij}^{(n)} \ge \sum_{n\ge k} p_{ij}^{(n)} = \sum_{n\ge 0} p_{ij}^{(n+k)}\ge p_{ij}^{(k)} \sum_{n \ge 0} p_{ii}^{(n)} = \infty$$


3

You only need $X$ is $\mathcal{F}_2$-measurable, which gives: $$ E[XY|\mathcal{F_2}]=X E[Y|\mathcal{F}_2]. $$ Now, using $\mathcal{F}_1\subset\mathcal{F}_2$ and iterated conditioning for the first equality below, we have $$ E[XY|\mathcal{F}_1]=E[E[XY|\mathcal{F_2}]|\mathcal{F}_1]=E[XE[Y|\mathcal{F_2}]|\mathcal{F}_1]. $$


3

CLT does not apply to the present setting because (the simplest version of) CLT assumes that $(X_n)$ is i.i.d. while here the distribution of $X_n$ depends on $n$. With Borel-Cantelli I obtained; $$\sum_n\Pr(X_n\neq0)<\infty\Longrightarrow\Pr(\limsup\limits_n\{X_n\neq0\})=0$$ What does it mean now? The set $\limsup\limits_n\{X_n=0\}$ has probability ...


3

First part: For every $n$, the simple Markov property at time $nK$ yields $$P_j(T_i\gt(n+1)K\mid T_i\gt nK,X_{nK}=k)=u(k),\qquad u(k)=P_k(T_i\gt K),$$ and $u(\ )\leqslant1-\varepsilon$ uniformly by hypothesis hence $$P_j(T_i\gt(n+1)K,X_{nK}=k)\leqslant(1-\varepsilon)P_j(T_i\gt nK,X_{nK}=k).$$ Summing these over $k$ yields ...


2

$S_N$ is a function of the random variable $N$ (as well as of the $X_i$), but $E[S_N]$ is not; it is a real number. $E[S_N\mid N]$ is a function of the random variable $N$ but not of the $X_i$. This random variable $E[S_N\mid N]$ takes on value $nE[X]$ whenever $N$ has value $n$. The right side of your $(1)$ is an application of the law of the ...


2

$${n\choose n-j-1}F(z_{k-1})^{n-j-1}{j+1\choose j}(F(z_k)-F(z_{k-1}))^j$$ Deciphering the formula: The first binomial factor is the number of ways of choosing $n-j-1$ elements from the whole sample (the ones that will be below $z_{k-1}$). The first power of $F$ is the probability that these elements are indeed all below $z_{k-1}$. The second binomial ...


2

To show $E[X|\mathcal{G}]$ equals some $Y$. You need to check: $Y$ is $\mathcal{G}$-measurable; $\int_A Y(\omega)dP(\omega)=\int_A X(\omega)dP(\omega)$ for all $A\in\mathcal{G}$. In this case, the claim is $Y=X$ works. (1) holds by assumption and (2) holds trivially.


2

In general, it is not true even for non-negative sequences of random variables that the almost sure convergence to $0$ is equivalent to the convergence of the expectation to $0$. (and this example shows is for an appropriate choice of $a$ and $b$) For the convergence in $\mathbb L^r$, follow the approach you suggested. For the almost sure convergence, you ...


2

In general $$ \eqalign{P(X + Y \le z) &= \iint_{\{(x,y): x+y \le z\}} dx\; dy\; f_{XY}(x,y)\cr &= \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy \; f_{XY}(x,y) }$$ If $X$ and $Y$ are independent, $f_{XY}(x,y) = f_X(x) f_Y(y)$ so this becomes $$ \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy\;f_X(x) f_Y(y) = \int_{-\infty}^\infty dx\; ...


2

Recall that if $X \sim \mathrm{Binomial}(n,p)$, then $$\Pr[X = x] = \binom{n}{x} p^x (1-p)^{n-x}, \quad x = 0, 1, 2, \ldots, n.$$ Then use this to calculate the ratio $$\frac{\Pr[X = x+1]}{\Pr[X = x]},$$ being careful to cancel like terms.


2

I believe what's meant is that the sequence of functions $L_n$ converges to a function $L$ with the property that $L(x) \le f(x)$ for every $x$, and similarly for the $U_n$. That's certainly true, and seems to make sense in the context given. As @Nate notes below, this convergence is probably meant to be understood pointwise.


2

More generally... one may want to keep in mind that Markov chains (in discrete time) and Markov processes (in continuous time) are different (although related) objects. 1. A Markov chain $(X_n)$, indexed by $n$ integer, is described by a transition matrix $P$, such that, for every states $(x,y)$ and every $n$, $$\Pr(X_{n+1}=y\mid X_n=x)=P_{xy}.$$ Thus: ...


1

The order of draws does not matter at all. You have $\frac 1{12}$ chance of getting the first pick each year. Before the first draw, you have $\frac 1{12^2}$ chance of getting the first pick the next two years in a row. However, you would probably remark on it if the same person got the first pick twice. That has a chance of $\frac 1{12}$ as somebody has ...


1

Let us write $x$, $\hat x$ and $u$ for your $x_t$, $\hat x^t$ (or is it $\hat x$?) and $u_0^t$ (or is it $u_t^0$?), respectively. By definition of conditional expectation, $x-\hat x$ is orthogonal to the space $L^2(u)$ of square integrable random variables measurable with respect to $u$. Since $\hat x$ is in $L^2(u)$, $x-\hat x$ and $\hat x-E(x)$ are ...


1

Unless otherwise specified, statements about convergence, inequalities, etc, of functions are usually meant to be interpreted pointwise. So "$L_n \uparrow L$" means $L_n(x) \uparrow L(x)$ for every $x$, and "$L \le f$" means $L(x) \le f(x)$ for every $x$.


1

In general, I would try finding the CDF. $$ F_K(k) = \iiint_{\{(x,y,z): f_c(x,y)+f_c(y,z) + f_c(x,z)\le k\}} f_X(x) f_Y(y) f_Z(z)\; dx\; dy\; dz$$


1

If there have been four claims at most, you need to consider values of $N$ from $\color{blue}{0}$ to $4$ inclusive, hence the total probability is $$P(N=0)+P(N=1)+P(N=2)+P(N=3)+P(N=4)=\frac{5}{6}$$ For the probability of at least one outcome, the values of $N$ from $1$ to $4$ need to be considered ...


1

The area (probability) up to $0$ is $(1)(0.2)$, which is $0.2$. To get to area $0.5$, we need another $0.3$. So if $m$ is the median then $$\int_0^m (0.2+1.2x)\,dx=0.3.$$ The rest is calculation. We could also solve the problem geometrically. For the mode, graph the density function. It reaches its maximum at $1$.


1

(a) Given the expressions of the pmfs for the binomial and geometric distributions: $$ \mathbb{P}\{X=0\} = \binom{n}{0} p^0(1-p)^n = (1-p)^n $$ while indeed \begin{align} \mathbb{P}\{Y > n\} &= 1 - \mathbb{P}\{Y \leq n\} = 1 - \sum_{k=1}^n (1-p)^{k-1}p \\ &= 1 - p\sum_{k=0}^{n-1} (1-p)^{k} = 1-p\frac{1-(1-p)^n}{1-(1-p)}\\ &= ...


1

The easiest way for a) is to note that the probability that $Y\gt n$ is the probability of $n$ failures in a row, which is $(1-p)^n$. However, the path you started on works also. For $\sum_1^{n-1}p(1-p)^{k}$ is the sum of a finite geometric series with first term $p$ and common ratio $1-p$. By the usual formula, this sum is $$p\frac{1-(1-p)^n}{1-(1-p)},$$ ...


1

Let $X=G\left(\dfrac 1 {1-\mathrm e^{-E}}\right)$. If $G$ is the inverse of $\dfrac1 {1-F}$, then the identity $X=\dfrac 1 {1-F\left(\dfrac 1 {1-\mathrm e^{-E}}\right)}$ written in the post is incorrect. Actually, $X=G\left(\dfrac 1 {1-\mathrm e^{-E}}\right)$ translates as $\dfrac1 {1-F(X)}=\dfrac 1 {1-\mathrm e^{-E}}$ hence $F(X)=\mathrm e^{-E}$ and, for ...


1

For your first hesitation: we can take $f$ to have values in $[-1,1]$ because we are already assuming $f$ is bounded. Thus, if $M = \sup_x |f(x)|$, $f/M$ does take values in $[-1,1]$. The constant $M$ does no harm because $E[f(X)/M) = \frac{1}{M}E[f(X)]$, so the general case of values in $[-M,M]$ follows immediately from the $[-1,1]$ case. For the question ...


1

$$\{X\leq 1\}=\{X<\frac{1}{2}\}\cup\{\frac{1}{2}\leq X\leq 1\}$$ and these sets are disjoint so that: $$P\{X\leq 1\}=P\{X<\frac{1}{2}\}+P\{\frac{1}{2}\leq X\leq 1\}$$ or equivalently: $$P\{\frac{1}{2}\leq X\leq 1\}=P\{X\leq 1\}-P\{X<\frac{1}{2}\}$$


1

You should understand the difference between a probability density function and a cumulative distribution function. The cumulative distribution function, which in your case is $F(x)$, always gives the value for $P(X \leq x)$ So, $F(1)$ would give you $P(X\leq1)$ and $F(\frac{1}{2})$ would give you $P(X\leq\frac{1}{2})$. In order to find $P(\frac{1}{2} < ...


1

$$\begin{align}\require{cancel} \Pr[X=x]&={n\choose {x}}p^{x}(1-p)^{n-x}\\ \Pr[X=x+1]&={n\choose {x+1}}p^{x+1}(1-p)^{n-(x+1)}\\ &={n\choose x+1}\left(\dfrac p{1-p}\right) p^x(1-p)^{n-x}\\ &={n\choose x+1}\left(\dfrac p{1-p}\right)\dfrac{\Pr[X=x]}{n\choose x}\\ &=\dfrac {n\choose x+1}{n\choose x}\left(\dfrac p{1-p}\right)\Pr[X=x]\\ ...


1

First, you wrote that if $X_n$ is not constant a.s., then $\forall c\in\mathbb{R},\epsilon>0; P(|X_n-c| > \epsilon) > 0$. This is not true. Change the $\forall$ to an $\exists$ and it will be true. Here's one approach: $\implies)$ So $X_n$ is not constant a.s. That means there exist $c\in\mathbb{R}$ and $\epsilon>0$ such that ...


1

Hint It is not generally true that $\sum P(A_{nm})=+\infty$. But it must be true for one of the $m$ following sequences: \begin{equation*} (B^{i}_{n})_{n\in\mathbb{N}}=(A_{nm+i})_{n\in\mathbb{N}}, \quad i=0,1..,m-1 \end{equation*}


1

Your function is the probability density function of a random variable distributed uniformly on the interval $[1,2]$. As well as having an integral across the real numbers of $1$, a probability density function also needs to be non-negative. Your function has this property.



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