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5

The maximum $M_n$ of interest is such that $$(1-2/\sqrt{n})K_n\leqslant M_n\leqslant1,\quad\text{where}\quad K_n=\max\{X_i\mid 1\leqslant i\leqslant\sqrt{n}\}.$$ By independence, $P(K_n\leqslant x)=P(X_1\leqslant x)^{\sqrt{n}}=x^{\sqrt{n}}\to0$ when $n\to\infty$, for every $x$ in $(0,1)$. Hence $K_n\to1$ in probability, $(1-2/\sqrt{n})K_n\to1$ in ...


5

Knowing that the 6th coin flip is still 50-50 and completely independent from the last 5 should make the entire intro of your question void. You should ask - is it worth borrowing \$100 in order to have a 50-50 chance of winning an additional \$100. Everything else is just just noise. To answer this just think about the simple goal of maximising the ...


5

For points $P$ and $Q$ in the plane, let $\Gamma(P,Q)$ denote the circle with segment $PQ$ as its diameter. We first prove a lemma. Lemma: Let $\Gamma$ be a circle, $P$ a point inside $\Gamma$, and $P'$ the reflection of $P$ about the center of $\Gamma$. Denote by $\mathcal{L}$ the locus of points $Q$ inside $\Gamma$ such that $\Gamma(P,Q)$ is internally ...


4

This answer is under the assumption that these are fair dice. You have $P(x_2 - x_1 = k) = (6-|k|)/36$, for $-5 \le k \le 5$. (This is just a short way of writing what you'd get out of counting case by case.) Now $$P((x_2 - x_1) = (x_4 - x_3)) = \sum_{k=-5}^5 P((x_2 - x_1) = k, (x_4 - x_3) = k)$$ and since $x_1, x_2$ are independent of $x_3, x_4$, this ...


4

Each $A_n$ is a subset of $\Omega$. In the previous axiom we only needed to refer to one arbitrary set, so we just called it $A$. Now we want a collection of countably many arbitrary subsets $A_1,A_2, A_3,A_4,A_5,...$ so we need to label them somehow, and we do that via subscripts running over the natural numbers. Now remember, a $\sigma$-algebra is a ...


3

I have some partial results; perhaps someone can continue/correct my work. \begin{align*} &\phantom{{}={}}P\left(1-\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}>\epsilon\right)\\ &=P\left(1-\epsilon>\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}\right)\\ &=\prod_{i=1}^{n/2} P\left(1-\epsilon ...


3

It means that they satisfy $A_n \subset A_{n+1}$. The limit is not a limit in the $\epsilon$-$\delta$ sense. It just means $\cup_{n=1}^\infty A_n$. For example, take $A_n = [0,n]$, then $\cup_{n=1}^\infty A_n = [0, \infty)$. From a probability perspective, there is a real limit associated with these nested sets. Suppose $p$ is the probability measure. ...


3

If we replace uniforms on $(0,1)$ by uniforms on $(0,w)$, the resulting random variable $Y$ has the same distribution as $wX$, where $X$ has Irwin-Hall distribution. In particular, $\Pr(Y\le y)=\Pr(wX\le y)=\Pr(X\le \frac{y}{w})$. It follows that if $f_X$ is the density function of the Irwin-Hall, then $Y$ has density $f_Y(y)=\frac{1}{w}f_X(y/w)$. In a ...


3

If $D$ is a domain in $\mathbb{R}^n$, and $x_0 \in D$ is a point, then the distribution of Brownian motion started at $x_0$ stopped at the time when it leaves $D$ is the harmonic measure of $\partial D$ with base point $x_0$. In the plane you can often use conformal invariance to calculate harmonic measure. Since you know that the harmonic measure of the ...


3

It is known as the chain rule. A justification can be seen, assuming of course $\Pr[C] > 0$ and $\Pr[A\cap C] > 0$, as $$\begin{align} \Pr[A\cap B \mid C ] &= \frac{\Pr[A\cap B \cap C ]}{\Pr[C]} = \frac{\Pr[A\cap B \cap C ]}{\Pr[A\cap C]}\cdot \frac{\Pr[A \cap C ]}{\Pr[C]} \\ &= \Pr[B\mid A\cap C ]\cdot \Pr[A \mid C ] \end{align}$$ where the ...


2

Hint: Note that $ A\cup \emptyset = A$, thus: $P(A\cup \emptyset) = P(A)+P(\emptyset)+P(A\cap\emptyset) = P(A)$, now $P(A)>0 \implies ?$


2

Since $P(X\in A\mid Y) = P(Z\in A\mid Y)$ a.s. we have $E[P(X\in A\mid Y)\cdot 1_{\{Y\in B\}}] = E[P(Z\in A\mid Y)\cdot 1_{\{Y\in B\}}]$. Using the tower property of the conditional expectation and $\sigma(Y)$-measurability of the indicators, you now get the desired result.


2

Hint: Set $Y_k = B_{t_k + s} - B_{t_{k-1}+s}$ where $t_0 = 0$. Note that $X_k = Y_1 + \dots + Y_k$, so you can express $f(X_1, \dots, X_n)$ as a Borel function of $Y_1, \dots, Y_n$. Now observe that $\{\mathcal{F}_s, \sigma(Y_1), \dots, \sigma(Y_n)\}$ are mutually independent $\sigma$-fields. Use a Dynkin lemma argument to conclude that $\mathcal{F}_s$ ...


2

Here's a different argument. Define $$X_t = \int_0^t f(s,\omega) dW_s \\ Y_t = X_t^4 \\ Z_t = X_t^2.$$ Apply the Ito formula to $Y_t$ and $Z_t$: $$Y_t = 4 \int_0^t f(s,\omega) X_s^3 dW_s + 6 \int_0^t f(s,\omega)^2 X_s^2 ds \\ Z_t = 2 \int_0^t f(s,\omega) X_s dW_s + \int_0^t f(s,\omega)^2 ds$$ Now substitute and take expectations, thereby canceling the ...


2

No, this is not true. We construct a counterexample in the plane. To make computations easy, we restrict ourselves to the unit square $U=[0,1]\times[0,1]$, but you will see that the idea extends to the entire plane easily. Define the set $N=[1/3,2/3]\times[1/3,2/3]$. Consider the probability density function $f(x,y)=\begin{cases} 0&(x,y)\not\in U,\\ ...


2

A simple approach (which avoids the task of checking the nonnegativity of the function in your question) is to compare the hitting times $\tau_1$ and $\tau_2$ with the hitting time $\theta$ of $\pm2\pi$ by the process $$dY=\mathrm dW-\mathrm{sgn}(Y)\,\mathrm dt,$$ starting from some $|y|\leqslant2\pi$. Since $|\sin|\leqslant1$, $\tau_1$ and $\tau_2$ are ...


2

The difference between the terms "probability measure" and "probability distribution" is in some ways more of a difference between terms rather than a difference between the things that the terms refer to. It's more about the way the terms are used. A probability distribution or a probability measure is a function assigning probabilities to measurable ...


2

My strategy. First notice that the expected return in this game in the long run is zero dollars. You win what you risk, and you would expect to win and lose with roughly equal frequency. If your bets are constant, you should come pretty close to break even in the long run. But that might mean going way into to debt for a while and then slowly winning your ...


2

Visualizing things makes the problem much simpler and easier to understand in such cases. Consider the value pairs $(n,m)$ is allowed to take on a 2D grid: $$ \begin{array}{ccccc} (1,1) & & & &\\ (2,1) & (2,2) & & & \\ (3,1) & (3,2) & (3,3) & & \\ (4,1) & (4,2) & (4,3) & (4,4) & \\ \vdots ...


2

Hint: $X_1 + \ldots + X_n < t$ iff there exist rational numbers $r_1, \ldots, r_n$ such that $r_1 + \ldots+ r_n < t$ and all $X_i \le r_i$.


2

For every measurable function $u$, $Y=u(X_1,\ldots,X_n)$ is $\sigma(X_1,\ldots,X_n)$-measurable since, for every Borel subset $B$, $[Y\in B]=[(X_1,\ldots,X_n)\in u^{-1}(B)]$ and $u^{-1}(B)$ is a Borel subset of $\mathbb R^n$.


2

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2

Due to the Jensen's Inequality, for a convex function $f$ and positive numbers $\alpha_i$ with unit summation, we have: $$f(\alpha_1x_1+\alpha_2x_2+\ldots+\alpha_nx_n)\leq\alpha_1f(x_1)+\alpha_2f(x_2)+\ldots\alpha_nf(x_n)$$ with equality if and only if $f$ is linear or $x_1=x_2=\ldots=x_n$ The important thing that you must pay attention to, is that there ...


2

Take $X$ to be Bernoulli with parameter $0$, i.e. $X=0$ a.s.; and $a = -1$. Then $$ a^2\Pr[\lvert X\rvert \geq a] = 1, \qquad \mathbb{E}[X^2] = 0 $$


2

This is called Simpson's paradox. The way it can happen is that women tended to be a larger proportion of the faculty in departments where fewer people get tenure. Imagine all but one or two of the women going into a department where one out of $20$ faculty members get tenure, and all but one or two of the the men going into a department that gives tenure ...


2

First, let us take a look at this question. One year Babe Ruth had a higher batting average than Lou Gehrig for the first half of the season and also for the second half of the season. But Lou Gehrig had a higher batting average for the entire season. How can that be? The batting average for a baseball player over a time period $T$ is$${{\#\text{ ...


2

The appropriate model has the five dice distinguishable. A more familiar example is tossing two coins. A model that treats two heads, two tails, and one of each as equally likely gives answers that do not match reality. There are thus $6^5$ equally likely outcomes. We now need to count the favourables. We divide into cases. (i) All dice show the same ...


1

Let $X_t$ be the number of busy servers in a $M/M/\infty$ queuing system. You want to compute $E(X_t|X_0=n)$. This can be done by considering the sum $$X_t=Y_t+Z_t,$$ where $\{Y_t|X_0=n\}\sim Binomial(n,e^{-\mu t})$ is the number of servers that did not complete service in the interval $(0,t]$ (out of the $n$ that were busy), and $Z_t\sim ...


1

A sequence of random vectors $$ \mathbf{X}_{n} = \begin{bmatrix} {(X_{n})}_{1} & \cdots & {(X_{n})}_{m} \end{bmatrix}^{T} $$ converges in probability to a constant vector $\mathbf{a}$ if $ {(X_{n})}_{i}$ converges in probability to $a_{i}$, $\forall i \in \lbrace{1, \dots, m }\rbrace$.


1

For any $i=1,\ldots,n$, we have, using the Total Law of Expectation, conditioning on the value of $U_i$, \begin{eqnarray*} && \\ E(U_i\vert \max\{U_1,..,U_n\}=t) &=& E(U_i\vert U_i=t\cap \max\{U_1,..,U_n\}=t)P(U_i=t\vert \max\{U_1,..,U_n\}=t) \\ && + E(U_i\vert U_i\neq t\cap \max\{U_1,..,U_n\}=t)P(U_i\neq t\vert \max\{U_1,..,U_n\}=t) ...



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