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7

Suppose that $X$ and $Y$ are independent random variables, both of which are equal to 1 with probability 1, and that $Z$ is independent of $X,Y$ and $Z=1$ with probability $\frac{1}{2}$, $Z=-1$ with probability $\frac{1}{2}$. Then $\mathbb{P}(XZ=-1,YZ=1)=\mathbb{P}(Z=-1,Z=1)=0$, while $\mathbb{P}(XZ=-1)\mathbb{P}(YZ=1)=\frac{1}{4}$. So $XZ$ and $YZ$ are not ...


6

$Z(t) = X(t) + Y(t)$ is a Poisson process with rate $x+y$. Given such a process, we can realize $X(t)$ and $Y(t)$ by assigning each occurrence of the $Z$ process independently to the $X$ or $Y$ process with probabilities $x/(x+y)$ and $y/(x+y)$ respectively. You're looking for the probability that $X(t) = m$ when $Y(t)$ hits $n$, i.e. that of the first ...


5

Let's say $f(x)$ is some function. The conditions that qualify it as a bona fide Probability Density Function (PDF) are: $f(x) \geq 0$ for all $x$. $\int\limits_{-\infty}^{\infty} f(x) \ dx = 1$. So you need only to verify that the function in question has these properties to conclude that it is a PDF for some random variable.


5

This is not true. Moreover, it seems pretty obvious to me that the fact like this cannot be true without any further assumption like the martingale property; however, the requirement that $N$ should be a stopping time (with respect to the filtration of $X$, I presume) makes it harder to find a counterexample. Let $\Omega$ be $[0,1]$ with the Lebesgue ...


5

The chance that any given number is lucky is $\frac1{10}$. Then, by linearity of expectation, the total expected number of lucky numbers is $10\left(\frac1{10}\right) = 1$.


4

Let me just address the question on $\mathbb{R}$. One way to ask this question is: Is there a family of $\aleph_1$-many subsets of $\mathbb{R}$, each of which is Lebesgue null, whose union is all of $\mathbb{R}$? (I'm interpreting the sets in this phrasing as the complements of the events in the question, to match with language used elsewhere.) It ...


4

There are $\binom{208}{100}$ equally likely ways to choose $100$ cards from $208$. Now we count the favourables. There are $\binom{104}{47}$ ways to choose $47$ red cards from the $104$ available. For each of these ways, there are $\binom{104}{53}$ ways to choose the accompanying $53$ black cards, for a total of $\binom{104}{47}\binom{104}{53}$. Finally, ...


4

Consider $\Omega = [-1, 1]$, $X(t) = t$, and set the codomain of $Y$ to $\{0, 1\}$, with $$ Y(t) = \begin{cases} 0 & t \le 0 \\ 1 & \text{else} \end{cases}. $$ Let $x = 0$ and $y = 0$ in your last paragraph. The first set is $\{0\}$ and the second is $[-1, 0]$, and their intersection is $\{0\}$. So the answer is "no" to that main question. It ...


4

The sufficiency is false. Indeed, let $X_1,X_2,\dots$ be i.i.d. uniformly distributed on $[-1,1]$. Then $E[X_n\mid X_1,X_2,\dots,X_{n-1}]=0$ in view of independence. However, the series $\sum_{n=1}^\infty X_n$ diverges, as $\limsup X_n =1$ almost surely. The necessity is true. It seems that there is no simple proof (at least I don't see one). Let me, before ...


4

The term "probability distribution function" is ambiguous. It could refer to a probability density function (PDF) or a cumulative distribution function (CDF), for instance. For discrete variables, it can refer to a probability mass function. A PDF yields non-negative values over its domain, and its integral over its domain is equal to $1$. Depending on ...


3

Informally, if we know that $X_1$ is small in absolute value, then $X_2$ cannot be small in absolute value. More formally, let $A$ be the event $|X_1|\le \frac{1}{2}$. A look at the sine curve shows that $\Pr(A)=\frac{1}{3}$. Similarly, let $B$ be the event $|X_2|\le \frac{1}{2}$. We have $\Pr(B)=\frac{1}{3}$. But since $X_1^2+X_2^2=1$, the event $A\cap ...


3

I find this doubtful, since we are effectively giving each subsequent observation ($x_n$) a higher weight than the previous one...not a recipe for convergence when the $x_i$ are random. In general: $$s_n = \sum_{i=1}^n a(a-1)^{n-i}x_i$$ Thus, if $a \in (0,1)$ $$\lim \limits_{n\to\infty} s_n = \lim \limits_{n\to\infty}\sum_{i=1}^n a(a-1)^{n-i}x_i$$ ...


3

No, we don't generally have $\mathbb E(Z) = \frac{n}{2k+1}$. For instance, for $n\le k$ we have $Z=0$ with probability $1$. To find the expected median for large $n$ for $k=1$, with $3$ bins, we can expand the multinomial coefficients around $m=\frac n3$ using Stirling's approximation: \begin{align} &\binom n{m-x,m-y,m+x+y}\\ ...


3

The first is certainly not correct. Your sample points could be only $10,0$ or something like that, with average $5$. The second one, you could have all sample points as $5$. The third one has to be correct, because if all sample points are strictly smaller than $5$, their average is strictly smaller than their supremum, which is $5$. Thus, the third one ...


3

For the first question, yes, many such subsets exist. For instance, let $A_p\subset\{0,1\}^\omega$ denote the set of sequences $(x_n)$ such that $\sum_{n=1}^N x_n/N$ converges to $p$ as $N\to\infty$. Then the sets $A_p$ are disjoint for different values of $p$, and $\mu_q(A_p)=1$ if $p=q$ and $0$ otherwise. In particular, every subset of $A_p$ is ...


3

Draw the part of the unit square $[0,1]^2$ that corresponds to this event. It's straightforward to find its area geometrically.


3

The total number of ways to arrange $14$ people is $14!$. The number of ways to arrange $14$ people with a specific person last is $13!$. So in a single experiment, the probability of that person being chosen last is $\frac{13!}{14!}=\frac{1}{14}$. Hence in $3$ experiments, the probability of that person being chosen last every time is ...


3

As requested in the comments: Although the desired claim sounds plausible, it isn't true. To see a simple counterexample, suppose we look at four people: Two women (with initials $A.C.$ and $B.D.$) and two men (with initials $A.D.$ and $B.C.$). We imagine that we are drawing uniformly randomly from this pool. Let $X$ be the event which determines ...


3

By symmetry. They must add up to $k$ for all $n$ boxes, and they're all the same, so they're $\frac kn$.


3

Since the Poisson process $(N_t)_{t \geq 0}$ has non-decreasing sample paths, we have $$N_{\tau} = \sup_{n \in \mathbb{N}} N_{\tau \wedge n}.$$ By the monotone convergence theorem, this implies that $$\mathbb{E}(N_{\tau}) = \sup_{n \in \mathbb{N}} \mathbb{E}(N_{n \wedge \tau}).$$


3

This law is assuming that you are "breaking up" the sample space for $Y$ based on the values of some other random variable $X$. In this context, both $Var(Y|X)$ and $E[Y|X]$ are random variables. Each realization assumes that we first draw $X$ from its unconditional distribution, then sample $Y$ from its conditional distribution given $X=x$. The first term ...


3

Let's look at a specific value, say $P(X=12)$. For the highest ball to be $12$, you have to pick one ball from the set $\{12\}$ and two balls from the set $\{1,2,3,\dots,11\}$. So there should be a total of $1\cdot {11\choose 2}$ ways to do this. You can generalize this idea to continue the problem.


3

$E[X|Y=y]$ is just a numerical function of $y$, it's not random. By contrast, $E[X|Y]$ is a random variable (though it can potentially be constant, even then it should be thought of as a random variable). The interesting but intuitive fact is that when you define $E[X|Y]$ as $E[X|\sigma(Y)]$, you have $E[X|Y]=E[X|Y=y]$ when $Y=y$. In other words, $E[X|Y]$ ...


3

Two approaches: In the base-$d$ expansion, the decimals are i.i.d. random variables uniformly distributed in $\{0,\dots, d-1\}$ (this does require proving). From there, the probability is $\frac{1}{d^{10}} = \frac{1}{10^{10}}$. This is computing the probability that the number falls in the interval $[a,b]$ with $a = 0.1111111111$ and ...


3

I think everything you're doing is correct, but perhaps it would benefit from exploiting some cancellations so things work out more smoothly. In your second line, with six terms, I claim if you take the first, fourth, fifth, and sixth terms and sum them, you get $$\boxed{ \lambda^2 s^2 -\lambda s -2\lambda t X_s}.$$ I used this grouping because lots of ...


3

Assume without loss of generality that $r \leq s \leq t$. By conditioning with respect to $\mathcal{F}_s$, we find using the martingale property $$\mathbb{E}(M_r M_s M_t) = \mathbb{E}(M_r M_s \mathbb{E}(M_t \mid \mathcal{F}_s)) = \mathbb{E}(M_r M_s^2).$$ If we denote by $\langle M \rangle_t$ the quadratic variation (i.e. the unique increasing process ...


3

For fixed $t_0 \geq 0$, we have $$\{\omega; t_0 \, \text{is not a local maximum for $t \mapsto B_t(\omega)$}\} \supseteq \{\omega; \forall \delta>0 \exists t \in [t_0,t_0+\delta]: B_{t}(\omega)>B_{t_0}(\omega)\}.$$ If we set $W_s := B_{s+t_0}-B_{t_0}$, $s \geq 0$, then we can rewrite this as $$\begin{align*} \{\omega; t_0 \, \text{is not a local ...


3

It's always $1$. Standard argument: Let $E_n$ be the answer, for a collection of $n$ letters. For $i\in \{1,\cdots,n\}$ let $X_i$ be $1$ if $i$ is a fixed point of a randomly chosen permutation. $0$ otherwise. Of course the probability that $X_i$ is fixed is $\frac 1n$ so $E[X_i]=\frac 1n$. As expectations are linear (regardless of independence) we have ...


3

If $U\sim\mathsf U(0,1)$, then $2U\sim\mathsf(0,2)$. Assuming $X$ and $Y:=2U$ are independent, we compute \begin{align} \mathbb P(X>Y) &= \int\limits_{\{(x,y)\ :\ x>y\}} f_{X,Y}(x,y)\ \mathsf d(x\times y)\\ &= \int_0^2\int_y^\infty f_X(x)\ \mathsf dx\ f_Y(y)\ \mathsf dy\\ &= \int_0^2\int_y^\infty \lambda e^{-\lambda x}\ \mathsf dx\ \frac12\ ...


3

First, why is it true that $$ \tag{1} \left\{\sup_{n\in\mathbb{N}} X_n < \infty\right\} = \left\{\sup_{n\in\mathbb{N}} X_{n+m} < \infty\right\} $$ for all $m \geq 0$? Fix $m \geq 0$. If $\sup_{n\in\mathbb{N}} X_n < \infty$, then there exists some $C$ (a real-valued random variable) such that $X_n \leq C$ for all $n \in \mathbb{N}$. In particular, ...



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