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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) ...


4

First, $P(E_n)=1/n$ simply by symmetry, because $E_n$ happens when $\max\{X_1,X_2,\ldots,X_n\}$ is realized with $X_n$. Since the common distribution is continuous there is almost surely no ex aequo and $\max\{X_1,X_2,\ldots,X_n\}$ has as much chances to be realized with each $X_k$ with $1\leqslant k\leqslant n$, hence the result. Likewise, the ...


4

The idea is that, since the series $\sum\limits_nP(X_n\ne-1)$ converges, Borel-Cantelli lemma (the simple one) indicates that, almost surely, there exists some finite $N$ such that $X_n=-1$ for every $n\geqslant N$. Every such sequence $(X_n)$ is such that $S_n/n\to-1$ hence $P(S_n/n\to-1)=1$. The strong law of large numbers does not apply, although $(X_n)$ ...


3

The transition rate matrix is another name for the generator matrix for a continuous time markov chain. A transition matrix is for a discrete time markov chain. $M$ would be for a continuous time markov chain (the underlying process for the queue is based on Poisson processes) while $T$ would be for a discrete time markov chain (the underlying process ...


3

Assuming you mean $\mathrm{E}(X|Y)$, then since the distribution of each die is identical and independent, if we are given that their sum is $y$, the expected value of each die would be the same: $y/2$. If you did mean $\mathrm{E}(X/Y)$, then in light of the preceding, this would be $\frac12$. Explicit Calculation of $\boldsymbol{\mathrm{E}(X|Y)}$ Given ...


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

Let $W_t$ be win today, $W_y$ be win yesterday, $L_t$ be lose today and $L_y$ be lose yesterday. We have $P(W_t)=P(W_t|W_y)\cdot P(W_y)+P(W_t|L_Y)\cdot P(L_Y)$. From the given information, this becomes $P(W_t)=.8P(W_y)+.4P(L_y)$. If the probabilities are to be stable (long-term), we should have $P(W_y)=P(W_t)$ and $P(L_y)=P(L_t)$. Thus ...


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

If Debra selects $N$ red cubes, then she must have selected $(3 - N)$ black cubes, so her winnings are given by: $$ W = 2N - 1(3 - N) = 3N - 3 $$ so $a = 3$ and $b = -3$.


2

CLT does not apply here 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 $1$ (...) ...


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

Using the result here one can see that $k\in C^\infty(\mathbb R)$. But if this is all we have, the statement is false: take $k(w)=\exp(w^4)$, and then the integral is infinite. You need some assumptions about the behavior of $k$ at infinity (or of $h$, from which $k$ was obtained). For example, if $k$ is bounded by $M$, then ...


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

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 $$\sum_{n\ge 0} p_{ij}^{(n)} \ge \sum_{n\ge k} p_{ij}^{(n)} = \sum_{n\ge 0} p_{ij}^{(n+k)}\ge p_{ij}^{(k)} \sum_{n \ge 0} p_{ii}^{(n)} = \infty$$


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

We show how to handle the problem for one value of $y$, say $y=9$. Given that $Y=9$, $X$ takes on values $3$ to $6$ with equal probabilities. Thus $$E(X|Y=9)=\frac{3+4+5+6}{4}.$$ One value of $y$ done, $10$ more to do. Remark: The symmetry argument of robjohn is much better.


2

This is the (much studied) Erdős–Rényi model of random graph. Exact results for fixed values of $n$ are seldom. Let $\epsilon\gt0$. If $1-p\gt(1+\epsilon)\ln n/n$ then the graph is connected with probability converging to $1$ when $n\to\infty$. If $1-p\lt(1-\epsilon)\ln n/n$ then the graph contains isolated vertices, and thus is disconnected, with ...


2

This post expands upon Did's post. As Did mentioned, there is no nice closed form for this probability for general $n$. However, this probability does satisfy a "nice" recurrence (nice in the eye of the beholder, I guess). Let $P(n,p)$ be the probability that $G(n,p)$ is connected, which is the value that you want. Then $P(n,p)$ satisfies the equation ...


2

Suppose the probability of head is $p$. Suppose you have thrown $n$ tosses and landed $a$ heads. And suppose you contemplate playing $m$ more tosses. The expected payoff of this is $$ \frac{a+pm}{n+m}. $$ This is greater than or equal to $\frac{a}{n}$ iff $$ (a+pm)n-(n+m)a=m(np-a)\geq 0. $$ So if $a>pn$, you shouldn't continue. If $a\leq np$, then you ...


2

\begin{align*} \operatorname{cov}(X+Y,X-Y)\equiv&\,\mathbb E[(X+Y)(X-Y)]-[\mathbb E(X+Y)][\mathbb E(X-Y)]=\mathbb E(X^2)-\mathbb E(Y^2)-0\times0\\ =&\,1-1=0.\end{align*} Hence, $X+Y$ and $X-Y$ are uncorrelated. It is not difficult to see that $X+Y$ and $X-Y$ are jointly normal. Now recall that two jointly random normal variables are independent if ...


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

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: ...


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

Since $e^{ax} \geq 1 + ax + \frac{1}{2}(ax)^2$ for $a > 0$, $$xe^{-ax} = \frac{x}{e^{ax}} \leq \frac{x}{1+ax+\frac{1}{2}(ax)^2},$$ and this last term goes to 0 as $x \rightarrow \infty$.


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

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

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

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

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 ...



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