# Tag Info

12

In terms of the sample space of events $Ω$, an event $E$ happens almost surely if $P(E) = 1$, whereas an event happens surely if $E=Ω$. An example: suppose we are (independently) flipping a (fair) coin infinitely many times. The event $$\{ \text{I will get heads infinitely often}\}$$ is an almost sure event (because it is possible get only a finite ...

8

There is a difference between "almost surely" and "surely." Consider choosing a real number uniformly at random from the interval $[0,1]$. The event "$1/2$ will not be chosen" has probability $1$, but is not impossible. I recommend reading the relevant Wikipedia article, which I found very clarifying when I was learning probability.

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Definitely incorrect. Let F = X+Y. Suppose X and Y are IID normal. Then E[X|F] and E[Y|F] are both linear in F, and hence perfectly correlated.

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So long as $\mu$ and $\nu$ are distinct, the infinite product measures will be mutually singular. Indeed, choose an event $B$ with $\mu(B)\not=\nu(B)$. Then $$G:=\{x=(x_1,x_2,\ldots)\in {\Bbb R}^\infty: \lim_n n^{-1}\sum_{k=1}^n 1_B(x_k)=\mu(B)\}$$ satisfies $\mu^\infty(G)=1$ but $\nu^\infty(G)=0$ by the strong law of large numbers. A definitive answer ...

5

At the moment $T_{k-1}$ we are on a boundary of non-visited and visited points. Now what Lawler$^*$ says is that there are two possibilities (both having probability $1/2$): We step into the non-visited area. Then we have $T_{k} - T_{k-1} = 1$. We step back into the visited area. Then $$T_{k} - T_{k-1} = 1+ \text{time to reach the boundary again} \\+ ... 5 Using that$$M_t := \exp(i \xi B_t + \frac{t}{2} \xi^2)$$is a martingale, one can show that for any \lambda>0 there exists \beta>0 such that$$\mathbb{E}e^{\beta T_{\lambda}}<\infty.$$By Markov's inequality, this implies in particular that$$\mathbb{P}(T_{\lambda} \geq t) \leq e^{-\beta t} \underbrace{\mathbb{E}(e^{\beta ...

5

Take the defintion (for $P(B) \neq 0$): $$P(A|B) := \frac{P(A\cap B)}{P(B)}$$ Just use that for every term and you will see that the equation holds. Edit I'll make it clearer and use your notation: $$p(x,y|z)=\frac{p(x,y,z)}{p(z)}$$ and $$p(y|x,z)=\left(\frac{p(y,x,z)}{p(x,z)}= \right)\frac{p(x,y,z)}{p(x,z)}.$$ And this holds by definition. There is no ...

5

Let $x=\frac{p}{q}$ and $y=\frac{q^{\prime}}{p^{\prime}}=\frac{1-q}{1-p}.\;\;\;$ We know that $x>y$ since $\ln x>\ln y$, and we want to show that $\color{blue}{(p+q)\ln x>(p^{\prime}+q^{\prime})\ln y}$. Since $p=xq$ and $q^{\prime}=p^{\prime}y,\;\;$ $p+q=q(x+1)$ and $p^{\prime}+q^{\prime}=p^{\prime}(y+1)$. Then $\displaystyle ... 5 Hint for the case$S<1$: Prove this by induction for a finite number of probabilities using that$(1-a)(1-b)\ge 1-a -b$and then proceed to the limit. Hint for the rest of proof: If$\sum_{n=1}^\infty p_n<\infty$, then there exists$N$such that$\sum_{n=N}^\infty p_n<1$. 4 Everything seems correct. But yes, there is both quicker and better way, based on the fact (which is easy to prove) that a random variable is symmetric iff its characteristic function is real-valued. Given this, the symmetry of$S_n$immediately follows from $$\varphi_{S_n} (t) = \prod_{k=1}^n \varphi_{X_k}(t).$$ It is better because it allows to prove ... 4 In general,$X \in \mathcal{V}$does not imply$\mathbb{E}(|X_t|)<\infty$. Just consider the process $$X_t(\omega) := \begin{cases} 0, & t < 1, \\ \frac{1}{\omega}, & t \geq 1 \end{cases}, \qquad \omega \in \Omega := (0,1).$$ on the measurable space$((0,1),\mathcal{B}((0,1)))$endowed with the Lebesgue measure. Obviously,$X \in ...

4

What you have is proof that: $\mathsf P(U=x, V=x)=0$ \begin{align} P(U =x, V=x) & = \int_{-\infty}^x \int_{-\infty}^x f(u, v)\operatorname du\operatorname dv - \lim_{t \rightarrow x} \lim_{s \rightarrow x}\int_{-\infty}^{t} \int_{-\infty}^{s} f(u, v)\operatorname du\operatorname dv \\[1ex] ~ & = 0 \end{align} Which is totally not what is ...

4

There is a standard lower estimate for $1-\Phi(x) = P(N(0,1)\ge x)$: for all $x>0$, $$1-\Phi(x)> \frac{x}{x^2+1}\varphi(x) = \frac{x}{x^2+1}\frac1{\sqrt{2\pi}}e^{-x^2/2}.$$ You can find the proof e.g. here. So for $x\ge 1$ $$1-\Phi(x)> \frac{1}{x+1/x}\varphi(x)\ge \frac{1}{2x}\varphi(x),$$ consequently, the desired inequality holds with $K_2 = ... 4 I'll first restate what I understand to be the problem. You have a full binary tree of depth$k$with$2^k$leaves (in your case$k=10$), with exactly$n$leaves uniformly randomly selected. You are looking for the expected number of evaluated nodes, where a node is regarded as evaluated if there is at least one selected leaf among the descendants of its ... 4 Hint: if$P(B_1) = 0$and$P(B_2) = 0$, then$P(B_1\cup B_2) = 0$. 4 No, not necessarily. Consider the following well known example from Bernstein (1928). Suppose a box contains four tickets labelled 112, 121, 211, 222. Choose one ticket at random and consider the events $$A = \{1 \text{ occurs in the first place}\},$$ $$B = \{1 \text{ occurs in the second place}\},$$ $$C= \{ 1 \text{ occurs in the third place}\}.$$ Here ... 4 Consider$f=\max\limits_{n\in\mathbb{N}}n\mathbf{1}_{(0,1/n]}$, which is given by $$f(x)=\begin{array}{}n&\text{if }x\in\left(\frac1{n+1},\frac1n\right]\end{array}$$ This would be the smallest candidate for a dominating function. However, notice that $$\int_{\frac1{n+1}}^{\frac1n}f(x)\,\mathrm{d}x=\frac1{n+1}$$ Thus, $$\int_0^1f(x)\,\mathrm{d}x$$ ... 4 Note that the sequence of random variables$X_n = \frac{1}{n} I\{Y < n\} Y$is bounded by$1$and converges pointwise to$0$. The assertion now follows from the dominated convergence theorem. 4 The$L^p$martingale convergence theorem holds also true for non-negative submartingales. The proof relies on Doob's maximal inequality: Let$(X_j)_{j \in \mathbb{N}}$be a non-negative submartingale (or a martingale). Then$X_n^* := \sup_{j \leq n} |X_j|$satisfies $$\|X_n^*\|_p \leq \frac{p}{p-1} \|X_n\|_p$$ for any$p>1$. Moreover, for ... 3$\mathcal M$is not only closed --- it's actually compact. And the continuous image of a compact set is compact. 3 Here is a direct proof (which doesn't require any additional theorems such as the dominated convergence theorem): Fix$\epsilon>0$. Since$A_n := \{n \leq Y < \infty\}$is a sequence of decreasing sets satisfying$\bigcap_{n \in \mathbb{N}} A_n = \emptyset$, the continuity of the (probability) measure$\mathbb{P}$gives $$\lim_{n \to \infty} ... 3 Yes, it makes sense, but careful. Your set A depends on X, but its measure can be bounded independently of X using Markov's inequality and the fact that \mathcal X is bounded in \mathbb L^1. So you fix \varepsilon>0, and choose a K such that for each X\in\mathcal X, \mathbb P\{|X| \geqslant K\} is smaller than \delta, where \delta ... 3 No, unless the random variables are constant this doesn't need to be true. The distribution of a random variable bears very little information about the underlying probability space on which the random variable is defined. The random variables don't even need to be defined on the same probability space in order to have the same CDF. So see an example, ... 3 I'm not too familiar with Brownian motions, but I think the problem is basically the same as in the case of individual random variables as in muaddib's answer: "Take the process X where X_0, X_1 are iid positive random variables and X_2 is a normal random variable with variance X_0+X_1 and mean zero." Here X_2 could be said to have "random ... 3 Let \Omega = \{\omega_1,\omega_2\}, P(\{\omega_1\}) = P(\{\omega_2\}) = 1/2;$$ \xi_n(\omega_i) = (-1)^{n+i}, i = 1,2, n\ge 1. $$Then \xi_n converge to a symmetric Bernoulli random variable (they just share this law), but there is no almost sure convergence. (For both \omega the values alternate.) 3 Assuming your probability measure is uniform on [0,1/m], then it is absolutely continuous with respect to the Lebesgue measure on [0,1], and admits a density with respect to this measure in the form$$ f(t) = \begin{cases} m & \mbox{ if }t\in[0,1/m], \\ 0 & \mbox{ otherwise}. \end{cases}$$Then the expectation of X_n with respect to P_m can ... 3 Using the inequalities obtained here yields c = \frac{1}{\sqrt{2\pi}}. 3 Since the augmented filtration is right-continuous, we may assume that (M_t)_{t \geq 0} has càdlàg sample paths. Since (M_t)_{t \geq 0} is a local martingale, there is a sequence of stopping times (\tau_k) such that \tau_k \uparrow \infty and (M_{t \wedge \tau_k})_{t \geq 0} is a martingale. Set$$Y := M_{T \wedge \tau_k}$$for fixed k \in ... 3 No, this is not true. Consider g \equiv \infty and h(x) = \infty for x \in A, h(x) = 0 for x \notin A, where A is not measurable. Then h is not measurable, but$$ f = g+h \equiv \infty$$is measurable and$g$is also. In fact, you can choose$h : X \to [0,\infty]$arbitrary in the above. 3 Of course$\frac{X}{n}$isn't binomial distributed for$n > 1$, since$P(\frac{X}{n} = \frac{1}{n}) \ne 0\$ and the binomial distribution only assumes integer values. You argument shows nothing. Just because a random variable has a mean and a variance doesn't mean it is binomial distributed.

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