Hot answers tagged

20

Consider all of the $6\times 5$ ways to pick two pieces of fruit.   That's $30$: $$\boxed{\begin{array}{|l|ccc:ccc|}\hline ~ & A_1 & A_2 & A_3 & O_1 & O_2 & O_3 \\ \hline A_1 & \times & \color{green}{A_1,A_2} & \color{green}{A_1,A_3} & \color{blue}{A_1,O_1} & \color{blue}{A_1,O_2} & \color{blue}{A_1,O_3} ...


18

The only reason you are multiplying by 2 in the second case is because you are using a shortcut due to the fact that the two scenarios that you are adding have a probability found with the same formula. You just need to add up the probabilities you are seeking. Case 1) $\frac{3}{6}*\frac{2}{5}$ Case 2) $\frac{3}{6}*\frac{3}{5}+\frac{3}{6}*\frac{3}{5}$ ...


17

the answer is indeed $\frac 12$ . As an alternative way to see that: let's pause just before Bob tosses his final (extra) toss. At this point, there are three possible states: either Bob is ahead, Alice is ahead, or they are tied. Let $p$ be the probability that Bob is ahead. By symmetry, $p$ is also the probability that Alice is ahead (so the ...


14

When all you have is the raw set structure, the only limit concept that really makes sense is: $S$ is the limit of the sequence $S_1, S_2, S_3,\ldots$ iff $$ \forall x\; \exists N\in\mathbb N\; \forall n>N : x\in S\Leftrightarrow x \in S_n $$ In other words every possible element is either in all but finitely many $S_n$ (in which case it is in the ...


14

In terms of the sample space of events $Ω$, an event $E$ happens almost surely if $P(E) = 1$, whereas an event happens surely if $E=Ω $. An example: suppose we are (independently) flipping a (fair) coin infinitely many times. The event $$\{ \text{I will get heads infinitely often}\}$$ is an almost sure event (because it is possible get only a finite ...


14

Let $p_n$ be the $n$th prime. The events $p_n\mathbb N$ are independent*, because $$P(p_n\mathbb N \cap p_m\mathbb N)=P(p_np_m\mathbb N)=\frac 1{p_np_m}=P(p_n\mathbb N)P(p_m\mathbb N)$$ The sum of the reciprocals of the primes $$\sum_n \frac 1{p_n}$$ famously diverges. So, by the second Borel-Cantelli lemma, the event that infinitely many of the events ...


12

You want to prove the statement: $$\lim_{n\to\infty}\sum_{i=1}^{n}c=c \implies c=0$$ Instead, you can prove the equivalent statement: $$c\neq0 \implies \lim_{n\to\infty}\sum_{i=1}^{n}c \neq c$$ And this is rather simple, as you can use the exact trick that you were trying to avoid: $$c\neq0 \implies ...


12

If the covariance matrix is not positive definite, we have some $a \in \mathbf R^n \setminus \{0\}$ with $\def\C{\mathop{\rm Cov}}\C(X)a = 0$. Hence \begin{align*} 0 &= a^t \C(X)a\\ &= \sum_{ij} a_j \C(X_i, X_j) a_i\\ &= \mathop{\rm Var}\left(\sum_i a_i X_i\right) \end{align*} So there is some linear combination of the $X_i$ which has ...


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.


10

It would generally not be true even if they were independent. For example if $X,Y,Z$ were identically and independently continuously distributed then they can come in any order with equal probability so $ P(X \leq Y \leq Z) = \frac16$ but similarly $P(X \leq Y)P(Y \leq Z) = \frac12 \times \frac12 = \frac14$.


10

In order to give more strength to the induction hypothese let us prove more generally: $\exists\alpha\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\alpha_{i}\right)}\right\} _{i\in I}\text{ is independent}\implies\forall\beta\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\beta_{i}\right)}\right\} _{i\in I}\text{ is independent}$ Assume that the ...


10

If we divide every roll by $n$, rolling the die and dividing by $n$ approximates the uniform distribution on $[0,1]$ for arbitrarily large $n$. We then are looking for the expected number of samples from a uniform distribution required to get a sum above $1$. For any integer $k \in \mathbb{N}$, Let $X_1, \ldots , X_k, \ldots$ be the random variables in ...


9

Define $X_n$ to be such that $X_n$ is $0$ with probability $1-\frac{1}{n}$ and $n^2$ with probability $\frac{1}{n}$. It is the case that $E[X_n]=n \to \infty$. But for any positive $k$ we have $\mathbb{P}(X_n > k) = \frac{1}{n} \to 0$. Thus showing a counter example.


9

Here is a proof of the first question using Zorn's lemma, and a counterexample to the second question using an ultrafilter. (So both cases used some form of the axiom of choice!) Theorem (Sierpinski): For a non-atomic probability space $(\Omega, \mathcal{F}, \mu)$, $\mu$ is surjective onto $[0,1]$. Proof: Let $x \in [0,1]$, and let $$ \mathcal{G} = ...


8

In the context of probability theory, you can write $$ \Pr(\varnothing) = \Pr(\varnothing \cup \varnothing) = \Pr(\varnothing)+\Pr(\varnothing). $$ Then if you subtract $\Pr(\varnothing)$ from both sides, you get $$ 0 = \Pr(\varnothing). $$ Thus you don't need infinite series. That $\infty\cdot0=0$ does not make sense in certain broad contexts, since for ...


8

You have some event, which you typically don't know when occurs, but that can/will occur some time in the future. The time that this event occurs is random, and it is a stopping time if, at any point in time, you know whether the event has occurred or not. A few quick examples. 1) Your own (a stopping time): Let $\tau$ denote the time that I'm ruined (i.e. ...


8

Consider two normal distributions with the same variance and different means.


8

There is a difference between "almost surely" and "surely." Consider choosing a real number uniformly at random from the interval $[0,1]$. The event "$1/2$ will not be chosen" has probability $1$, but is not impossible. I recommend reading the relevant Wikipedia article, which I found very clarifying when I was learning probability.


8

Definitely incorrect. Let F = X+Y. Suppose X and Y are IID normal. Then E[X|F] and E[Y|F] are both linear in F, and hence perfectly correlated.


8

Use the strong law of large numbers. Choose $B\in M$ so that $\lambda(B)\neq \mu(B)$, and consider the disjoint sets $$\left\{x\in X^\infty: {1\over n}\sum_{j=1}^n 1_{[x_j\in B]}\to \lambda(B)\right\}\mbox{ and }\left\{x\in X^\infty: {1\over n}\sum_{j=1}^n 1_{[x_j\in B]}\to \mu(B)\right\}.$$


8

By definition: $E[X]$ is only defined (EDIT: as a real number) for $X$ such that $E[|X|] < \infty$.


8

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 ...


8

Hint: It is $E(X(X-1))=E(X^2)-E(X)$. And $Var(X)=E(X^2)- [E(X)]^2 \Rightarrow E(X^2)=Var(X)+[E(X)]^2$


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 $$1+x \leq e^x.$$


7

If I toss a coin an infinite amount of times, can I be sure to get an infinite amount of heads? According to the Borel-Cantelli lemma, since each coin toss is an event of probability $\frac12$ and a sum of $\frac12$ diverges, the probability of $\limsup_{n\to\infty}\{\text{heads at $n$-th flip}\}$ is 1. But the $\limsup$ is precisely the event of ...


7

The vector $\vec X = [X_1,\dots,X_N]^T$ has a rotationally invariant distribution. That is, if $A$ is any orthogonal matrix, then the distribution of $\vec X$ and $A \vec X$ are the same. Hence by letting $A$ be an orthogonal matrix that takes $[1,\dots,1]^T$ to $[\sqrt N,0,\dots,0]^T$, your problem is the same as computing $$ \Pr(\sqrt N X_1 \mid \sum ...


7

Notice that if $A,B,C$ are pairwise disjoint, you have \begin{align} \Pr(A\cup B\cup C) & = \Pr(A\cup B)+\Pr(C) & & (\text{since }A\cup B\text{ is disjoint from }C) \\[6pt] & = \Pr(A) + \Pr(B) + \Pr(C) & & (\text{since }A\text{ is disjoint from }B) \end{align} and the same can be done with any finite sequence of pairwise disjoint ...


7

The $\sigma$-algebra generated by the events $\{\omega \in \Omega: \omega_n = W \}$ is the so-called Borel $\sigma$-algebra on $\Omega = \{H,T\}^\mathbb{N}$. One can show, by transfinite induction (so you need some set-theory background) that there are at most $|\mathbb{R}| = 2^{\aleph_0}$ many Borel sets, while the power set of $\Omega$ has $2^{|\Omega|} = ...


7

Simply follow the hint... First note that, since $E(X\mid Y)=Y$ almost surely, for every $c$, $$E(X-Y;Y\leqslant c)=E(E(X\mid Y)-Y;Y\leqslant c)=0,$$ and that, decomposing the event $[Y\leqslant c]$ into the disjoint union of the events $[X>c,Y\leqslant c]$ and $[X\leqslant c,Y\leqslant c]$, one has $$E(X-Y;Y\leqslant c)=U_c+E(X-Y;X\leqslant c,Y\leqslant ...


7

I give 2 proofs, that show that there is indeed something more than the pure analytical formula, beyond its intrinsic beauty. First proof: Let us consider the formula under the form: $$\frac{\sin \pi x}{\pi x} = \prod_{k=1}^{\infty}\cos{\frac{\pi x}{2^k}}= \lim_{n \rightarrow \infty}{\prod_{k=1}^{n}\cos{\frac{\pi x}{2^k}}}$$ Its Fourier Transform (using ...



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