# Tag Info

11

A direct proof: For any measurable set $A$, applying independence of the events $\{X\in A\}$ and $\{X\in A\}$ gives $$\mathbb P(X\in A)=\mathbb P(X\in A)\mathbb P(X\in A)\implies \mathbb P(x\in A)\in \{0,1\}.$$ Hence $\mathbb P$ is a dirac measure. Therefore $X$ is a.s. constant.

7

Hint: $$\exp(-I_{E_n}) = e^{-1} I_{E_n}+ I_{E_n^c}$$ implies $$\mathbb{E}(\exp(-I_{E_n})) =(e^{-1}-1) \mathbb{P}(E_n)+1.$$ Now use that $$1+x \leq e^x.$$

6

The intuition is that if $g(x) \geq h(x) ~\forall x \in \mathbb R$, then $E[g(X)] \geq E[h(X)]$ for any random variable $X$ (for which these expectations exist). This is what one would intuitively expect: since $g(X)$ is always at least as large as $h(X)$, the average value of $g(X)$ must be at least as large as the average value of $h(X)$. Now apply ...

6

Since $X_n$ and $X$ are both measurable (they are random variables after all) and the function $f(x,y) = \lvert x-y \rvert$ preserves measurability (due to its continuity), $\lvert X_n - X \rvert$ is measurable as well. Maybe it is better for your understanding if I argue in the following way. If $X_n$ and $X$ are measurable, then so is $X_n - X$. This is a ...

5

I think it's pretty hard to find a book which covers martingale theory; usually, books either give just an introduction or they focus on one particular aspect of martingale theory. I'll list some books which might be of interest and sketch (roughly) which parts they cover: David Williams: Probability with Martingales (Basic properties, optional stopping, ...

4

I will treat the case where M is a continuous semimartingale. Unfortunately it is generally not the Riemann Stieltjes integral. You know that the Stieltjes measure of g, is only defined if g has finite variation. However, as you know, many stochastic processes does not have sample paths with finite variation, and therefore such an integral does not exist. ...

4

We have that for $y\in(0,1)$ that $$P(Y \leq y) = P(X^3 \leq y)=P(X \leq y^{1/3}) =\int_0^{y^{1/3}} f_X(x)dx =y^{1/3}$$ and therefore $$f_Y(y)=\frac{d}{dy}P(Y \leq y)= \frac{1}{3y^{2/3}}$$ so nothing is wrong.

4

It's a simple calculation. First, $$\mathbb{E}[(X-\mathbb{E}[X])^2] = \mathbb{E}[X^2 + \mathbb{E}[X]^2 - 2X\mathbb{E}[X]].$$ Linearity of expectation implies $$\mathbb{E}[X^2 + \mathbb{E}[X]^2 - 2X\mathbb{E}[X]] = \mathbb{E}[X^2] + \mathbb{E}[X]^2 - 2\mathbb{E}[X]^2 = \mathbb{E}[X^2] - \mathbb{E}[X]^2,$$ using the fact that $\mathbb{E}[X]$ is a constant. ...

4

$$\mathbb{E}\left[\left(X-\mathbb{E}\left[X\right]\right)^2\right] =\mathbb{E}\left[X^2\right]-\mathbb{E}\left[X\right]^2 =\mathbb{E}\left[X\right]^2-\mathbb{E}\left[X\right]^2=0$$ thus $X-\mathbb{E}\left[X\right]=0$ a.e. . IMPORTANT EDIT : This requires $X$ to be square integrable.

4

For all $a$, we have $$a=E[a-X]\le E[(a-X)I_{X<a}]$$ where $I$ is the indicator function.Use Cauchy-Schwarz inequality we have $$a^2\le E[(a-X)^2]E[I^2_{X<a}]=E(a-X)^2P\{X<a\} =(Var(X)+a^2)P\{x<a\}$$ so $$P\{X\ge a\}=1-P\{X<a\}\le\dfrac{\sigma^2}{\sigma^2+a^2}$$

4

We have to consider three cases separately: $S_{k-1}(\omega)<0$: Then $S_k(\omega) \leq 0$ and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)| = -S_k(\omega)+ S_{k-1}(\omega) = - X_k(\omega).$$ $S_{k-1}(\omega)=0$: Then $|S_k(\omega)|=1$ and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)|=1.$$ $S_{k-1}(\omega)>0$: Then $S_k(\omega) \geq 0$ and therefore ...

4

It doesn't really make sense, as pointed out by @drhab. But you could interpret the problem as follows: let $p(R)$ be the probability that $x^2+2bx+c$ has real roots when $b$ and $c$ are chosen randomly and independently from a uniform distribution on $[-R,R]$; then find $\lim_{R\to\infty}p(R)$. In this case we have $$p(R)=P(c\le ... 4 Hint: S_n - S_{4n}/2 is the sum of 4n independent random variables, of which n have the distribution of X_1/2 and 3n have the distribution of -X_1/2. Apply Lindeberg to the sequence Y_j where Y_j has the distribution of -X_1/2 if j \equiv 1 \bmod 4 and X_1/2 otherwise. On second thought, you don't really need Lindeberg if you ... 4 Brownian motion, Solution I Since W_t \sim N(0,t), we have$$\mathbb{E}(|W_t| 1_{\{|W_t|>K\}}) = \frac{2}{\sqrt{2\pi t}} \int_K^{\infty} x \exp \left( -\frac{x^2}{2t} \right) \, dx = \sqrt{\frac{2}{\pi}} \sqrt{t} \exp \left(-\frac{K^2}{2t} \right)$$and therefore$$\sup_{t \geq 0} \mathbb{E}(|W_t| 1_{\{|W_t|>K\}})=\infty.$$Brownian motion, ... 4 There are various forms for the remainder term of a finite Taylor expansion. One of them is$$f(x)=\sum_{k=0}^nf^{(k)}(a){(x-a)^k\over k!} +\int_a^xf^{(n+1)}(t){(x-t)^n\over n!}\ dt\ ,\tag{1}$$whereby f is assumed sufficiently differentiable in the neighborhood of x=a. This is of course valid as well if f is complex-valued. For the proof of (1) ... 4 Probability is not just defined on a set. It's also defined on a \sigma-algebra of subsets. Namely, a probability structure is made of three parts: The underlying set, say X. The sets to which you can assign probability. The function which assigns probability. There are several axioms which we require of these structures, such as the sets to which ... 4 If you wonder why so far only the "trivial" example of a Dirac measure has been mentioned, there is actually a good reason for this. The young Ulam wondered whether there is an uncountable set X for which there is a probability measure$$ \mu \colon \mathcal P(X) \rightarrow [0,1] $$that vanishes on singletons (i.e. \mu(\{x\})= 0 for all x \in X). He ... 4 An alternative intuition is to see it as 4 groups of 13 with 4 friends all in different groups. The first can be in any group, the 2nd has 39 permissible slots out of 51, and so on. Thus Pr = \frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49} 4 A Dynkin system is a \sigma-algebra iff it is stable under finite intersections. Is your family stable under finite intersections? 3 That "Wikipedia" definition is for the special case where the covariance matrix is \sigma^2 I. In that case they are equivalent. A measure \mu on \mathbb R^n has density \rho with respect to Lebesgue measure \lambda^n iff \mu(A) = \int_A \rho(x) \; d\lambda^n(x) for all Lebesgue measurable sets A. 3 Because most of the time we identify functions which are almost surely equal. The whole theory of Lebesgue spaces carries this identifications, and for good reasons; there are a lot of properties that hold; for example we have X = Y \ a.s. \implies E(X) = E(Y) In general we don't care too much about sets of measure zero; many theorem only prove things ... 3 For any cardinal \kappa, you can define the product measure on 2^{\kappa} where 0, 1 are assigned one half measure each. This measure is defined on the sigma algebra generated by clopen subsets of 2^{\kappa}. Let us call the corresponding measure algebra \text{Random}(\kappa). The class of measure algebras thus obtained forms the building block for ... 3 The solution is arguably simpler if you distinguish between permutations within the hands. There are 4! ways of distributing the aces over the 4 players, and 13^4 possible positions for the aces in the hands. That leaves 48! possibilities for filling the remaining 48 slots. The size of the sample space in this perspective is just 52!, so the ... 3 By the fundamental theorem of calculus,$$ e^{ix} = 1 + (ix) \int_{0}^{1} e^{iux}\, du, $$which is the case r = 1 of$$ e^{ix} = \sum_{k=0}^{r-1} \left[\frac{(ix)^{k}}{k!}\right] + \frac{(ix)^{r}}{(r - 1)!} \int_{0}^{1}(1 - u)^{r-1} e^{iux}\, du. \tag*{$P(r)$} $$(N.B. (1 - u)^{r-1} in the integrand, not (1 - u)^{r}.) Assume inductively that P(r) ... 3 The RHS is direct to evaluate. The variance of B_s is s so:$$\int_0^t E[B_s^2] ds = \int_0^t s ds = \frac{t^2}{2}$$You are right about use Ito's Lemma for the LHS. By Ito:$$B_t^2 = t + 2\int_0^t B_s dB_s$$So \begin{eqnarray*} E[(\int_0^t B_s dB_s)^2] &=& E\left[\left(\frac{B_t^2 - t}{2}\right)^2 \right] \\ &=& ... 3 You don't need to use Lindeberg. Note that$$S_{4n} = S_n + \sum_{i=n+1}^{4n} X_i,$$and that the two terms are independent. So you can separate X_n \equiv S_n/\sqrt{n}-S_{4n}/\sqrt{4n} into two independent parts and apply central limit theorem to both. This shows that$$ X_n \Rightarrow \frac 1 2 Z_1 + \frac {\sqrt{3}}{2} Z_2 \sim N(0, \sigma^2), $$... 3 Using the martingale property,$$ \begin{align} \text{E}(W_n) &= \text{E}[\text{E}(W_n | W_{n - 1})] \\ &= \text{E}(W_{n - 1}) \\ &= ... \\ &= \text{E}(W_1) , \end{align} $$and so you have a uniform bound on both \text{Var}(W_n) and \text{E}(W_n)^2. 3 The paper "On The Bound of Proximity of the Binomial Distribution to the Normal One" - Nagaev, Chebotarev (2010) has improvements to C specifically for the Binomial Distribution. Theorem 2 on page 3 gathers together the results in the paper to show that C can be taken to be .4215 in general. The paper notes that an Esseen (1956) paper demonstrates that ... 3 Hints: Since (W_t)_{t \geq 0} is a Brownian motion, it has independent normally distributed increments; in particular, W_3-W_2, W_2-W_1,W_1 are independent random variables which are Gaussian with mean 0 and variance 1. This implies that (W_1, W_2-W_1,W_3-W_2) is (jointly) Gaussian with mean \mu=0 and covariance matrix$$C = \begin{pmatrix} 1 ...

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