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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. ...


7

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


6

From the stochastic integral analogue of integration by part, which is a corollary of Ito's Lemma, $$d(W_t\ln t) = \frac{W_t}{t}dt+(\ln t)dW_t,$$ so $$I:=\int_0^t \frac{W_s}{s}ds=W_t\ln t-\int_0^t (\ln s)dW_s=\int_0^t \ln\frac{t}{s}dW_s$$ as $W_t\ln t\to 0$ in probability, as $t\to 0^+$. Since the random variable $I$ is a linear combination of independent ...


6

$\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\operatorname{E}}$Markov's inequality says if $Y$ is a non-negative random variable and $y>0$, then $$ \Pr(Y\ge y) \le \frac{\E(Y)} y. $$ (See the bottom of this posting for a proof of Markov's inequality.) Applying Markov's inequality to $(X-\mu)^2$, where $\mu=\E X$, we get $\Pr((X-\mu)^2\ge d^2)\le ...


6

There is no need to apply Bayes Rule. The numerator of your expression is incorrect. If the owner has $2$ Rs $100$ notes, then there is only one way for the maid to take two $100$ Rs notes. So the numerator is not $\binom{4}{2}$ but $1$. Extra: I'd like to point out that in this problem, the maid taking two notes and the owner taking two notes are ...


6

Hints: Since $\mathbb{P}(X_1 \neq 0)>0$, there exists $\epsilon>0$ such that $\mathbb{P}(|X_1|>\epsilon)>0$. Conclude from $$\sum_{n \geq 1} \mathbb{P}(|X_n|>\epsilon)= \sum_{n \geq 1} \mathbb{P}(|X_1|>\epsilon) =\infty$$ and the Borel-Cantelli lemma that $|X_n(\omega)|>\epsilon$ happens infinitely often for almost all $\omega \in ...


5

$\{X_k\}$ cannot be a nonnegative MTG with $X_0=0$. So, I assume it is just a MTG. Using the original Doob's weak inequality, for some $z >-r$ (and noticing that $\mathbb{E}X_n=0$): $$P\{\max_{1\le k\le n}{X_k}\ge r\}\le P\{\max_{1\le k\le n}{(X_k+z)^2}\ge (r+z)^2\}\le\frac{1}{(r+z)^2}\mathbb{E}(X_n+z)^2=\frac{1}{(r+z)^2}(\mathbb{E}X_n^2+z^2)$$ ...


5

I agree with you and Nate Eldredge that the limit doesn't exist. However, if we replace $0$ with $+\infty$, then the limit exists and equals $e^{-a}$. Let $F$ denote the cdf of a standard normal, and let $\;f=F'$ be the pdf. Then $$\lim_{x \to \infty} \frac{P(X>x+\frac{a}{x})}{P(X>x)} = \lim_{x \to \infty} \frac{1-F(x+\frac{a}{x})}{1-F(x)}$$Now as $x ...


4

There isn't a "closed form" solution in terms of standard functions. If you write $S(u) = F(z)/2$ with $z = e^{iu}$, $F(z) =\displaystyle \sum_{n=0}^\infty \dfrac{z^{2^n}}{2^n}$, then $ z F'(z) = \displaystyle \sum_{n=0}^\infty z^{2^n}$ is a commonly-used example of a lacunary function.


4

As you have guessed, the answer is NO in general. And a single counter example is enough: take $\{X_n\}$ i.i.d. uniform $(0,1)$ then $|X_n|< n$ for all $n\geq 1$ so that $$ Y_n=X_n\implies\text{Var}{Y_n}=\text{Var}{X_n}=\frac{1}{12}\implies\sum_{n=1}^\infty\frac{\text{Var}{Y_n}}{n}=\frac{1}{12}\sum_{n=1}^\infty\frac{1}{n}=\infty. $$


4

Here's a hands-on intuitive approach to the problem. Suppose we have an alphabet $\mathcal{A}=\{a_1,\ldots,a_k\}$. Given nonnegative integers $n_1,\ldots,n_k$ and $m_1,\ldots, m_k$, let $$\left[\begin{array}{c}n_1,\ldots,n_k\\m_1,\ldots,m_k\end{array}\right]$$ denote the number of words that can be formed from $n_i$ copies of $a_i$, such that the word ...


4

This is a weakened form of Bernstein's Theorem (weakened by unnecessarily assuming identical distributions having finite variances), so the proof is shorter. Here's a sketch, adapted from "On three characterizations of the normal distribution" by M. P. Quine: Define $\quad U = X+Y, \quad V=(X-Y)^2$ and characteristic functions $$\phi(t) = E e^{itX}\\ ...


4

Let $A$ be any Borel set in $S$. By Lusin's theorem, we can choose a sequence of continuous functions $f_m$ with $f_m \to 1_A$ $\mu$-almost everywhere. Moreover, if we set $g_m = \max(\min(f_m, 1), 0)$ then $g_m$ is also continuous, $0 \le g_m \le 1$ and $g_m \to 1_A$ $\mu$-almost everywhere as well. Since $\int g_m\,d\mu = \lim_{n \to \infty} \int ...


3

Well, the probability of having at least $1$ person with the birthday of $q$ is $n/365$. Unfortunately, this isn't quite true. If you could guarantee that the $n$ people didn't share a birthday, then it would be true, but if you could guarantee that, there wouldn't be much of a birthday question! The actual probability that at least one person out ...


3

If we coloured the dice one red and the other blue then the probability that the red has 1 and the blue has 2 is $\frac{1}{36}$. Similarly if we wanted the blue to have 1 and the red 2 the probability is $\frac{1}{36}$. As these events are mutually exclusive the probability of getting 1 or 2 regardless of which die they are on is $\frac{2}{36}$


3

You seem to understand the concept pretty well. Just like you, I would have said that the time of ruin is $τ = \min\{n : X_n \leq 0\}$ instead of $τ = \min\{n : X_n = 0\}$. But in this precise example, the time of ruin is the first time that you have exactly 0. The concept of stopping time is closely related to that of filtration of a stochastic process. In ...


3

$τ = \min\{n : X_n = 0\}$ is the first $n$ such that $X_n =0.$ i.e. the first time that the process hits zero, as you said. A non-stopping time would the first time $n$ such that $X_n = \max X_j$. The reason being that at a given time you don't know where the process will go next.


3

I like Brian’s answer, but if you’re uncomfortable with the idea of counting the number of ways the maid can do something in the past, here’s another way to look at it. Call the bills the maid took $b_1$ and $b_2$, and the ones the owner took $b_3$ and $b_4$. The question is then this: Given an arrangement $b_1, b_2,\dots,b_{16}$ of the bills where ...


3

The approaches suggested earlier appear more efficient, but I think you are right in the sense that you can also solve it with the Bayes theorem, e.g. as suggested below: (I try to follow the terminology used at this source: http://www.math.uah.edu/stat/dist/Conditional.html) Define the set of notes stolen by the maid as discrete random variable $A_i$ ...


3

To put it another way: The owner took two $100$ bills. That means there were $14$ other bills left for the maid to have taken previously, and thus $_{14}C_2 = 91$ different ways for her to have chosen bills from the wallet. Only one of those ways also yields two $100$ bills, so the desired probability is $1/91$.


3

Edit: This answer shows the identity $$\mathbb{E} \left( \left| \sum_{i=0}^{n-1} \int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt \right|^2 \right) = \mathbb{E} \left( \sum_{i=0}^{n-1} \left[\int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt\right]^2 \right). \tag{1}$$ Fix $i<j$. Then, by the tower property, $$\begin{align*} ...


3

Yes, this follows directly from Markov's inequality: $$\mathbb{P}(|X_n| \geq R ) =\mathbb{P}(X_n \geq R) \leq \frac{\mu}{R}\qquad \text{for any $R>0$}.$$ Consequently, $$\sup_{n \in \mathbb{N}} \mathbb{P}(|X_n| \geq R) \stackrel{R \to \infty}{\to} 0.$$


3

This is not an easy question. Let $Z_0(n)$ be a random variable denoting the longest sequence of heads in a sequence of $n$ flips. In 1980, Guibas and Odlyzko showed that $$\mathbb{E}(Z_0(n)) = \log_2(n)+\frac{\gamma}{\log 2} -\frac{3}{2} +\rho_0(n)+o(1)$$ where $\gamma$ is the Euler-Mascheroni constant, and $\rho_0(n)$ is an osscilatory function of $\log ...


3

If you are guessing randomly, then yes, the probability of getting the sequence correct is just $6^{-4}$. You can think of guessing each peg one at a time. The probability of getting any peg correct is $6^{-1}$, and as an individual peg does not give you information on any other peg, the probability of getting all $4$ correct is just $6^{-1}\times ...


3

First off the size of the total sample space is $10^7$. Not if the first three digits are fixed as 452 and only the remaining four can vary. With that out of the way, i am i terpreting the question as saying that 452 have to be at the beginning of my phone number. In which case there would be (3)(2)(1) ways of selecting those digits, No. There is ...


3

No. The expression you displayed will have to be integrated over $[0,1]$: $$\mathbb P(X=k)=\int_0^1{{n\choose k}p^k(1-p)^{n-k}}dp.$$ This is so because $${n\choose k}P^k(1-P)^{n-k}=\mathbb P(X=k\mid P)$$ and $$\mathbb P(X=k)= \mathbb E[\mathbb P(X=k\mid P)].$$


3

A first observation is that $|x+y|\geqslant|x-y|$ if and only if $xy\geqslant 0$, so defining $$f(x,y):=|x+y|-|x-y|,$$ the positivity of $f$ is linked to that of $xy$. We would like to find a more tractable expression for $f$. Assume that $x\gt 0$ and $ y\gt 0$. Then $f(x,y)=x+y-|x-y|=2\min\{x,y\} =\min\{|x|,|y|\}$. Since $f(-x,-y)=f(x,y)$ we get ...


3

Yes, it's true by Cauchy-Schwarz inequality: $$\left(\int_{\Omega}X d\mu\right)^2=\left(\int_{\Omega}X \chi_{[0,\infty)} d\mu\right)^2\leq\left(\int_{\Omega}X^2 d\mu\right)\left(\int_{\Omega}\chi_{[0,\infty)}^2 d\mu \right)=\\=\left(\int_{\Omega}X^2 d\mu\right)\left(\int_{\Omega}\chi_{[0,\infty)} d\mu \right)=\\=\left(\int_{\Omega}X^2 d\mu\right)P(X>0)$$ ...


3

The thing which breaks this (and other things like martingale convergence) is the fact that you can have things be very large on small sets. If you have $L^p$ bounds for $p > 1$, this just won't happen. But for $p = 1$, it can! Neal provided a good answer, but allow me to add one that won't require much thinking for you! $X_n = 2^{n} 1_{[0,2^{-n}]}$ ...


3

Picture a (regular) polygon with so many sides it is virtually indistinguishable from a circle. The set of points that are closer to the centroid than to the any of the edges is, approximately, a circle of half the radius, hence roughly a quarter the area. So your assumption appears to be incorrect.



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