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12

Start with $x_1=0$ and then just let $x_{n+1}=1$ whenever $f(n)/n\leq p$ and $x_{n+1}=0$ whenever $f(n)/n> p$. It's easy to prove by induction that we will always have $\left|\frac{f(n)}n-p\right|\leq\frac 1n$ and so $f(n)/n\rightarrow p$.

10

It turns out the numerical findings about $\mathbb{E}[N_3]$ by David E is not a coincidence, it is exact! $$\mathbb{E}[N_3] = \frac{1}{1-\sin(1)}$$ Let $X_1, X_2, \ldots$ be a sequence of random variables. For each $n$, we will assume $X_n$ take value from the set $\langle n \rangle \stackrel{def}{=}\{ 1, 2, \ldots, n \}$ with uniform probability. After ...

9

This is false, when $\lim_{n \to \infty} p_n$ doesn't exist. For instance, consider $p_{2n-1} = \dfrac1{2^n}$ and $p_{2n} = 1-\dfrac1{2^n}$. This gives us $\sum_{n=1}^{2N} p_n = \sum_{n=1}^{2N} (1-p_n) = N$, whereas $$\sum_{n=1}^{2N} p_n(1-p_n) < \sum_{n=1}^N \dfrac1{2^{n-1}} < 2$$ However, the claim is true when $\lim_{n \to \infty} p_n$ exists. ...

8

Since $\mathbb E[X^2] = \mathbb E[X\mathbb E[Y|X]] = \mathbb E[\mathbb E[XY|X]] = \mathbb E[XY]= \mathbb E[\mathbb E[XY|Y]] = \mathbb E[Y\mathbb E[X|Y]] = \mathbb E[Y^2],$ observe that $$\mathbb E[(X-Y)^2]=\mathbb E[X^2+Y^2-2XY]=0,$$ which implies that $X$ and $Y$ differ on a set of measure zero. For the weaker condition where $X$ and $Y$ are ...

7

Note that $$\bigcup_{n=1}^\infty A_n \supset \bigcup_{n=1}^\infty A_n \cap \bigcup_{n=2}^\infty A_n \cap \ldots \cap \bigcup_{n=k}^\infty A_n \cap \ldots = \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_n = \limsup_{n\to\infty} A_n$$ And use $A\subset B \Rightarrow P(A) \le P(B)$

6

Reprenting the health of each Patron by $y$, and the quantity of interest by $N_y$, I get: \begin{align*} \Pr[ N_3 = 1 ] &= 0 \\ \Pr[ N_3 = 2 ] &= 0 \\ \Pr[ N_3 = 3 ] &= 1/6 \\ \Pr[ N_3 = 4 ] &= 1/6 \\ \Pr[ N_3 = 5 ] &= 19/120 \\ \Pr[ N_3 = 6 ] &= 47/360 \\ \Pr[ N_3 = 7 ] &= 173/1680 \\ \Pr[ N_3 = 8 ] &= 131/1680 \\ \Pr[ N_3 = ...

6

This is false. Let $$p_i=\begin{cases} \frac{1}{2^i}&i\text{ odd}\\1-\frac{1}{2^i}&i\text{ even}\end{cases}$$ Then $p_i(1-p_i)<\frac{1}{2^i}$, so $\sum p_i(1-p_i)$ converges.

6

The claim is not correct as the following counterexample shows: Consider $((0,1],\mathcal{B}(0,1])$ endowed with the Lebesgue measure. Define a sequence of random variables by \begin{align*} Y_1(\omega) &:= 1-1_{(1/2,1]}(\omega) \\ Y_2(\omega) &:= 1 - 1_{(0,1/2]}(\omega) \\ Y_3(\omega) &:= 1-1_{(3/4,1]}(\omega) \\ Y_4(\omega) &:= ... 5 It can be shown that nonnegative random variables X and Y have the same distribution so long as \mathbb{E}[X^\alpha]=\mathbb{E}[Y^\alpha] is finite for all \alpha\in(a,b], any 0\le a <b. Setting U=\log X and V=\log Y, define the functions f(\alpha)=\mathbb{E}[1_{\{X > 0\}}e^{\alpha U}],\ g(\alpha)=\mathbb{E}[1_{\{Y > 0\}}e^{\alpha ...

5

Interesting question! If the curve is very very thick, we can get non-zero probability. For example there is a curve that goes through every point of the unit square. (Please see Wikipedia, Space Filling Curve.) But $y=\sin(1/x)$ is not thick enough. Remark: One should give at least an informal argument that our curve is not thick enough. Let $\epsilon\gt ... 5 Draw a Venn diagram to help you understand$P(A|B)=P(A\cap B)/P(B)$. Then use this to relate the quantities$P(A|B)$and$P(B|A)$algebraically. Let's discuss the first point. Suppose we have a finite sample space so we can count the number outcomes in each possible "event." To determine$P(A|B)$, we're essentially asking what the probability of getting an ... 5 Step$I$: Pick$6$balls and weigh them$3$on each side. Step$II$:$1$. If the balls balance each other, then weigh the remaining two balls and the heavier one will tilt the balance on its side.$2$. If the balls do not balance each other, then the heavier ball should be one of the three balls on the side the balance tilts. From these three balls, ... 5 No: If$X=\pm1$with equal probability and independently$Y=\pm2$also with equal probability, and you tell me the value of$X+Y$, then I can tell you the value of$X$(and of$Y$). So in this case$X$and$X+Y$are not independent. And in general they usually will not be. 5 This is always possible. The idea is that we'll build the sequence$x_k$so that$f(d)/d$is always the best rational approximation to$p$with denominator$d$. Note that, for any irrational number$p$and fixed denominator$d$, there is a unique best rational approximation to$p$with denominator$d$, namely the unique$n/d$satisfying the inequality $$... 4 If X is any random variable, and \mathcal{H} \subset \mathcal{A} is any sub-sigma-algebra, then by the definition of conditional expectation, it is always true that$$E[E[X|\mathcal{H}]] = \int_{\Omega} E[X|\mathcal{H}] dP = \int_{\Omega} X dP = E(X)$$Now just apply this when X=|Y| and \mathcal{H}=\mathcal{F}_t. As Did has pointed out, the equality ... 4 This problem can be numerically solved by using the Poissonization trick. This rather unknown but powerful trick (due to A.N. Kolmogorov) is discussed in Chapter 4 of the book Understanding Probability of Henk Tijms. This trick leads to the answer \int_{0}^{\infty}\big(1-(1-e^{-t/6}-(t/6)e^{-t/6})^6\big)dt\approx 24.134. 4 We have$$1 - \mu\left(\bigcap_{i = 1}^n A_i\right) = \mu\left(\bigcup_{i = 1}^n A_i^c\right) \le \sum_{i = 1}^n \mu(A_i^c) = \sum_{i = 1}^n [1 - \mu(A_i)] = n - \sum_{i = 1}^n \mu(A_i) < 1,$$and so$$\mu\left(\bigcap_{i = 1}^n A_i\right) > 0.$$4 Nobody is making assumptions in this matter; rather, someone is drawing conclusions. And those conclusions are not based simply on the marginal distributions. Conventionally, one has$$ X_1,\ldots,X_n \overset{\mathrm{i.i.d.}}\sim N(\theta,1),  \bar X = \frac{X_1+\cdots+X_n} n \sim N\left(\theta,\frac 1 n \right) $$This gives use the mean vector and ... 4 Use the Borel-Cantelli lemma. Essentially, for the series to converge almost surely, the probability X is bigger than 0 necessarily has to converge to zero, which means that P({X=0 i.o.}) = 1. Clearly that's not the case since P({X≠0}) > 0 and constant for all X. Therefore, the sum of the probabilities diverges. And, because the sum of the probabilities ... 4 Yes: every subset of the primes has asymptotic density zero, and there are 2^\omega=\mathfrak{c} infinite subsets of the primes. 3 Let$$F(x):=x^{13}\cdot(\frac{(-x)^0}{3!}+\frac{(-x)^1}{2!}+\frac{(-x)^2}{1!}+\frac{(-x)^3}{0!})^{13}$$Then$$4!^{13}\cdot\sum_{k=13}^{52} \frac{1}{k!}[x^k]F(x)=\frac{50972203946555791528902451677555189167087762981}{92024242230271040357108320801872044844750000000000} =0.000553899741\cdots$$is the required probability. Ignoring suits, the number of ... 3 Credit to Did for this:$$\left\{ \limsup_{n \to \infty}\frac{S_n}{\sqrt{n}} > M \right\}= \left\{ \limsup_{n \to \infty}\frac{S_n-S_m}{\sqrt{n}} > M \right\}$$for every m \geq 1. Now$$\left\{\frac{S_n-S_m}{\sqrt{n}} > M\right\} \in \sigma\left(\bigcup_{k=m}^n \mathcal{F_k}\right)$$and$$\left\{\limsup_{n \to \infty}\frac{S_n-S_m}{\sqrt{n}} ... 3 Given a probability space$(\Omega, \Sigma, \mathbb P)$and a real-valued measurable function$X$defined on$\Omega$, the symbols $$\int_\Omega X(\omega) \mathbb{dP(\omega)},\ \int_\Omega X(\omega) \mathbb{P(d\omega)}$$ are simply interchangeable notation and are defined to represent integration of the function$X$with respect to the measure$\mathbb P$, ... 3 Yes, you should first make sure that$\mathbb{E}[\vert XY\vert]$is finite. This follows from the inequality $$\vert XY\vert\leq \frac{X^2+Y^2}{2},$$ which follows from$(\vert X\vert -\vert Y\vert)^2\geq 0$. 3 You have three distinct boxes, and want to count ways to put 20 indistinct balls into them. The total count of solutions is, as you have calculated: $$\frac{22!}{2!20!} = 231$$ Now, to generate forbidden solutions you can choose 2 boxes, then put the same number of balls$n$into each of them, and the remainder into the other. Since the number$n$can vary ... 3 With a slight augmentation of your notation, define $$S_{n,a/\sqrt{n}}:=\sum\limits_{i=1}^{n}{\bf 1}_{X_i>a/\sqrt{n}}=\sum\limits_{i=1}^{n}(1-{\bf 1}_{X_i\leqslant a/\sqrt{n}})=n-\sum\limits_{i=1}^{n}{\bf 1}_{X_i\leqslant a/\sqrt{n}}\,,$$ where we have used the indicator functions$\ {\bf 1}_{X_i\leqslant a/\sqrt{n}}\ $. Now, we simply observe the ... 3 Have you tried integration by parts together with the knowledge that the antiderivative of$x\phi(x)$is$-\phi(x)$? The answer is readily computable via this approach. 3 To your specific query of what expressions like$\text dY_t$mean, I offer the following. If you are referring to such entities as they arise in, say, the study of Ito processes, such as $$\text dY_t = \mu_t\text dt + \sigma_t \text dW_t,\ \ Y_0=y$$ such entities don't mean anything except only as meaningful shorthand for equivalent statements in terms of ... 3 Hint: Condition on the result of the coin-flip. 3 Then we have that$P(A\cup B)=P(A)+P(B-A\cap B)>P(A)$, so that$P(B-A\cap B)>0$, so that$B\neq A\cap B$and$A\neq A\cup B\$, so in particular, the inclusion is proper.

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