# Tag Info

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


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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