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0

By the CLT, we have the convergence in distribution $$ \frac{S_n}{\sqrt{n}}\Longrightarrow N(0,1). $$ Thus since $\sqrt{n}\nearrow \infty$, it follows that $\limsup_n S_n=\infty$ a.s. More details: Since $\lim_{n\to\infty}\mathbb P(S_n>\sqrt{n})=\int_1^{\infty}e^{-x^2/2}>0$, it follows that $\mathbb P(\limsup_n S_n=\infty)>0$. Thus by the ...


0

It is just binomial with $r$ trials and probability $p$ of being white. Every time you toss a coin, immediately set aside a white or black ball. When you choose $r$ from $k$ balls, you are choose $r$ from $k$ independent coin tosses, which is just $r$ independent coin tosses.


3

Let $Z=E[X\ | \ Y]$. Your equation states: $E[X \ | \ Z]=Z$. This follows from the following fact. Tower Property of Conditional Expectation: $$E[E[X\ | \ \mathcal{F}]\ | \ \mathcal{G}]=E[X\ | \ \mathcal{G}],\text{ whenever }\mathcal{G}\subset \mathcal{F}.$$ Proof of your equation: We apply the tower property with $\mathcal{G}=\sigma(Z)$ and ...


-2

Essentially the law of iterated expectation, perhaps more commonly written like $$\operatorname{E_X} [X] = \operatorname{E}_Y [ \operatorname{E}_{X \mid Y} [ X \mid Y]].$$ For a discrete case, the essence of the proof is $$\operatorname{E}_Y [ \operatorname{E}_{X \mid Y} [ X \mid Y]] = \sum_y \sum_x x \cdot \operatorname{P}(X=x \mid Y=y) \cdot ...


1

The lindeberg CLT can be applied for this sequence of random variables. Hint: figure out a convinient lower bound for the cumulative variance $s_n^2=\sum_{i=1}^n \sigma_k^2$


0

You've done part a correctly so far.   You just need to integrate the transformed pdf over the appropriate range. Given, density function of $f_R(r) = 25 \cdot \mathbf 1_{0.03< r< 0.07}$, and change of variable $R= g(X)$ where $g(X) = \ln \frac{X}{50000}$ $$\begin{align} g'(x) & = \frac 1 x \\[2ex] f_X(x) & = f_R(g(x)) \cdot |g'(x)| ...


1

Unfortunately, the difference is not negligible (though of course it depends on how much one is actually willing to neglect). Take for example the sets $X,Y=\{0,100\}$. Then over a uniform distribution, $$E[\min(x,y)]=25$$ while $$\min(E[X],E[Y])=50$$ One is double the other! To see $E[\min(x,y)]=25$, we note that there are $4$ possibilities which ...


-1

It's a reference to a fundamental flaw with HitchHikers Guide to the Galaxy 'Infinite Improbability Drive': http://hitchhikers.wikia.com/wiki/Infinite_Improbability_Drive Although in theory the principle works, it leaves behind a residual 1/probability, which (un)fortunately has to be accounted for.


1

Hints: $\mathbb E(X-Y)=\mathbb EX-\mathbb EY$ if $X$ and $Y$ are independent then $\operatorname{Var}(X-Y)=\operatorname{Var}X+\operatorname{Var}Y$


0

We can use the fact that, on any normed vector space, the norm is a continuous function with respect to the norm topology. Namely, if $V$ is a normed vector space with norm $\|\cdot\|,$ then we have for any $x,y\in V$ the "inverse" triangle inequalitiy $$ \bigl| \|x\| - \|y\| \bigr| \leq \|x-y\|. $$ If you haven't seen this before, it follows "easily" from ...


2

For $p\geqslant 1$, we have $\lVert X_n\rVert_p^p=2/n$, hence $X_n\to 0$ in $\mathbb L^p$. Since (in general) convergence in $\mathbb L^1$ implies convergences in probability, which implies in turn convergence in distribution, we can answer questions 2), 3) and 4). Note that for these question, we did not use independence. Question 1) can be solved using ...


-1

The first answer is correct because the distribution function of your random variable converges to a function describing a non (Kolmogorovian) probability measure: uniform over the real line. The second answer is wrong. The average converges to 1 with probability 1. That is the distribution functions converges to the distribution function of a constant ...


0

For your question in the title (which is different from the one inside) a way to view is,a grocery store: events correspond to packed items for sale (they may not sell a single apple, for example). Probability space corresponds to these packed items ALONG with their price.


2

You need to be more careful: The support of $X$ is $(0,1)$; consequently, $-\infty < \log X < 0$, and $\infty > -2\theta \log X > 0$. So the transformation is order-reversing (larger values of $X$ give smaller values of $Y$), and you need to take this into account: $$\begin{align*} \Pr[Y \le y] &= \Pr[0 < Y \le y] = \Pr[0 < -2\theta ...


1

Here is a partial answer that just looks at corellations. We have: \begin{align} x_t &= e_t^2 + 2e_tC_{t-1} \\ z_t &= e_{t-1}^2 + 2e_{t-1}C_{t-2} \end{align} First note that $E[x_t]=E[z_t]=1$ for all $t \geq 2$. Now we show that $\{x_2, x_3, x_4, \ldots\}$ are pairwise uncorrelated. Fix $t>n\geq 2$. Then: $$ E[x_tx_n] = E[(e_t^2 + ...


0

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1

Let's write out the transition matrix of $X$ when $a = 1$: $$\mathcal P = \frac{1}{10} \begin{bmatrix} 1 & 3 & 0 & 1 & 5 & 0 \\ 2 & 2 & 1 & 0 & 1 & 4 \\ 6 & 0 & 1 & 1 & 2 & 0 \\ 6 & 0 & 1 & 1 & 0 & 2 \\ 3 & 0 & 4 & 1 & 0 & 2 \\ 3 & 0 & 1 & 4 & ...


1

It is not true. Say you toss two coins. Let $X_i$ be the number of "heads" you get from the $i$th coin. Then $$ (X_1,X_2)=\begin{cases} (1,1) \\ (1,0) \\ (0,1) \\ (0,0) \end{cases} $$ each with probability $1/4$. In three cases, the minimum is $0$ and in one case it is $1$. Notice that ...


1

Let $C_a$ and $C_b$ be independent random variables that take value $0$ and $1$ each with probability $\frac{1}{2}$. Then $\min(C_a,C_b)=0$ with probability $\frac{3}{4}$ and $\min(C_a,C_b)=1$ with probability $\frac{1}{4}$, giving expectation $\frac{1}{4}$. But $\min(E(C_a,C_b)=\frac{1}{2}$. There are many other counterexamples.


5

You are on a good start. We already know that $\mathbb{P} (A) \in \{0, 1\}$, but how do we decide which one? First remark: a sequence $(X_n)$ is in $A^c$ if and only if its sum is bounded from above, i.e., if and only if there exists $N \in \mathbb{N}$ such that $S_n \leq N$ for all $n$. Now, let's use the symmetry of the random walk. The map $(X_n)_{n \in ...


0

Now I see this is quite obvious... but did not saw it 3 hours ago. So here is a proof. As each $\mathcal A_n \subseteq \mathcal A$ we have $\sigma\left( \bigcup_n \mathcal A_n \right) \subseteq \mathcal A$. This works for each filtration. For the other inclusion, note that $\mathcal A^{\mathbb N}$ is generated by the countable product of sets from $\mathcal ...


1

I can varify that $F$ is correctly calculated based on given definitions in the question. It can however be furher simplified as: $$\dfrac{1}{2}\int_{-c}^{c} f_X(x)\, dx+ \int_{-2}^{-c} f_X(x)\,dx\,+\int_{c}^{2} f_X(x)\,dx=1-\dfrac{1}{2}\int_{-c}^{c} f_X(x)\, dx$$


0

Perhaps $P(B|A,C)=\dfrac{P(A,B,C)}{P(A,B,C)+P(A,\text{not }B,C)}=\dfrac{P(B|A)P(C|B)}{P(B|A)P(C|B)+P(\text{not }B|A)P(C|\text{not }B)}$


0

The following conditions guarantee the exsitence of a unique median $1.$ CDF of the random variable $X$ is continuous. $2.$ CDF of the random variable $X$ is strictly increasing. Let $$a(m)=\int_{-\infty}^{m}f_X(x)\mathrm{d}x,\quad \mathcal{A}=\{m:a(m)\geq 1/2\}$$ and $$b(m)=\int_{m}^{\infty}f_X(x)\mathrm{d}x,\quad \mathcal{B}=\{m:b(m)\geq 1/2\}$$ If ...


0

Hint: Assuming Riemann integration, how do you justify the existence of integrals like $$ \int\limits_{0}^{x/n}f(t)\,\mathrm{d}t\,? $$


2

The point of contention is $x{}={}2$ and what happens there. Being careful about how we approach this point of discontinuity for $F(x)$, using a limiting sequence in terms of the $F_n(x)$ and the right continuity of $F(x)$, note that for all $k>0$, $$ F_n\left(2+\dfrac{1}{k}\right){}={}1\,,\ \ \forall\ n>k\,. $$ So, for all $k>0$, $$ ...


2

You are given $\displaystyle \lim_{n \to \infty} P(\{|X_n - 1| \ge \epsilon\}) = 0$ for every $\epsilon > 0$. If $0 < \epsilon < 1$ then $$\left| \frac{1}{X_n} - 1 \right| \ge \epsilon \iff -\epsilon \le \frac{1}{X_n} - 1 \le \epsilon \iff \frac{1}{1 + \epsilon} \le X_n \le \frac{1}{1 - \epsilon}$$ and $$ \frac{1}{1 + \epsilon} \le X_n \le ...


2

We have to show that $$\mathbb{P} \left( \left| \frac{1}{X_n} - 1 \right|> \varepsilon \right) \to 0 \qquad \text{as} \, \, n \to \infty$$ for any $\varepsilon>0$. To this end, we note that $$\left| \frac{1}{X_n} - 1 \right|> \varepsilon \iff |X_n-1|> \varepsilon \cdot |X_n|.$$ This implies $$\begin{align*} \left\{ \left| \frac{1}{X_n} - 1 ...


1

Yes, the statement holds true. Two ways to go about this: Proof 1: By Skorohod's representation theorem, there exists a probability space and random variables $X_n',Y_n'$, $X',Y'$ on this probability space with $$(X_n,Y_n) \sim (X_n',Y_n'), \qquad (X,Y) \sim (X',Y')$$ such that $$(X_n',Y_n') \to (X',Y') \qquad \text{almost surely.}$$ Since $a_n \to 1$ and ...


1

If you have a lottery with $10^6$ tickets, the probability you win with a ticket is $10^{-6}$. If $10^3$ people buy one ticket for each, then the probability that someone wins is $10^{-3}$, that is $1000$ times your probability of winning. But clearly if you buy $10^3$ tickets... When you hear, frequently, that someone has won the lottery, the event ...


0

However you hear every day in the news that someone has won $100,000,000 or similar somewhere in the world. Those people are extremely lucky. That's exactly why you hear about them in the news in the first place.


0

It could also work for you. But the probability is very low. One criteria for playing lottery is the expected value. If the expected value is negative, then it is maybe not useful to play lottery. For example: You can win 1,000,000 dollars with a probability of 1:200,000 and you have to pay 5 dollars. Then the expeceted value is $ ...


1

The thing is that if you take one specific person, say me, then the chance of me winning the lottery is very small. But the chance of someone winning the lottery is much larger, because so many people play the lottery. This is because the chance of someone winning the lottery is equal to the chance of me winning OR you winning OR my neighbor winning OR ... ...


0

When they bought the ticket, they were also not having 100% chances of winning, because probability is same for everyone, which neglects randomness(*) involved in the process and takes everyone equally likely. Probability just gives a likelihood of an event happening and becomes truer as the number of experiments approaches infinity, for one or two events, ...


4

They don't mean you literally can't win the lottery, they just mean that the expected value of buying a ticket is negative, or that winning is so unlikely that buying a ticket is a bad investment. However, you should read How Not to be Wrong by Jordan Ellenberg, which (among other things) tells the story of some students from MIT who noticed that some ...


1

I would say that the difference is merely that the first one denotes a probability mass function and the latter denotes just a probability, that is a value. For instance $$ P(A|B)=\begin{cases} 1/3 \mbox{ if A is true}\\ 2/3 \mbox{ if A is false}\end{cases} $$ and hence $$P(A=t | B)=1/3.$$ It does not appear to have anything to do with Bayesian theory.


3

Recall the following statement from probability theory: Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of pairwise independent random variables. Then $Y:=\limsup_{n \to \infty} Y_n$ is constant almost surely. Fix $t \geq 0$ and a sequence $(t_n)_{n \in \mathbb{N}}$, $t_n \neq t$, such that $t_n \to t$. By the continuity of the process $(X_s)_{s \geq 0}$, ...


0

To mark the question as answered, I do a copypaste from the deleted answer by voldemort: You actually need to show that the union is at most countable, as if the complement of a set in your space is finite, then the set must be uncountable. – voldemort So, what I understand is that; If $A_n$ are finite then they are countable, then the union is ...


1

Something that you might find helpful are Blaschke products. If a sequence of numbers $|z_n| < 1$ satisfy the condition $\sum_{n=0}(1-|z_n|) < \infty$ then there is a function analytic in the disc for which $f(z_n)=0$. In particular, the Blaschke product is such a function $$B(z)=\prod_{n=0}^\infty \frac{|z_n|}{z_n} \frac{z_n - z}{1-\bar{z_n}z}.$$ In ...


0

For absolutely continuous radom variables you always have a unique median. Indeed, let $F$ denote the distribution function of a continuous random variable $X$, and $f$ its density function. We have $$F(x) = \int_{-\infty}^x f(y)dy.$$ The median is given by the value $x^\ast$ such that $$F(x^\ast)=1/2.$$ Definte the function $G(x) := F(x) - 1/2$. We have ...


0

The density of $S$ is given by the convolution of the densities of $X$ and $Y$: $$f_S(s) = \int_{\mathbb R}f_X(s-y)f_Y(y)\mathsf dy. $$ Now $$ f_X(s-y) = \begin{cases} \frac12,& 0\leqslant s-y\leqslant2\\ 0,&\text{otherwise} \end{cases} $$ and $$ f_Y(y) = \begin{cases} \frac13,& 0\leqslant y\leqslant1\\ 0,&\text{otherwise.} \end{cases} $$ So ...


0

If $F$ denotes the CDF of $S$ then it is clear that $F(s)=1$ if $s\geq 5$ and $F(s)=0$ if $s\leq0$. Now let $s\in(0,5)$. Prescribe function $g_s$ on $\mathbb R$ by $\langle x,y\rangle\mapsto1$ if $x+y\leq s$ and $\langle x,y\rangle\mapsto0$ otherwise. $$F(s)=P(X+Y\leq s)=\int\int f_X(x)f_Y(y)g_s(x,y)dxdy=\frac16\int_0^2\int_0^3g_s(x,y)dydx$$ Here $f_X$ and ...


0

Correction required in the answer given by Jhon D. Cook. The sum of two normal random variables is normal provided R.Vs are independent


0

An "engineering" explanation: say you have a sampling grid in 2D using $2$ samples per dimension, you will have a $2\times 2$ grid with $4$ samples. $4^{1/2}=2$ Now you do the "equivalent" for 3D, you will have $2\times 2\times 2$ grid with $8$ samples with the same sampling density $8^{1/3}=2$. A 3D grid of $3\times 3\times3=27$ samples would have a ...


1

Suppose $F(\cdot)$ is the distribution function of the random variable $X$. Then $H(a) = F(a) - F(a^-) = p >0$ means $P(X = a) =p >0$. If $H(a_0) = p_0 >0$, then since $H(\cdot)$ is continuous, there exists $\delta >0$, such that for all $a \in [a_0 - \delta, a_0 + \delta]$, we have $H(a )> \frac{p}{2}$. This is to say that for uncountably ...


0

The problem you run into with the $X\ge0$ part of your proof is when you convert to $EX_n$ to $\int x\mu_n(dx)$, you can't use MCT on the right side (here, $\mu_n$ is the measure induced by $X_n$, for simple $X_n$ approaching $X$). I think it is easier to prove the more general fact that $$ Ef(X)=\int f(x)\,\mu(dx) $$ for any measurable $f$. The proof ...


0

The definition of $\mu$ is that $\mu(B)=P(X^{-1}(B))$, for $B\subset\mathbb{R}$ and measurable. You've got a sum of indicators, and you've proven the result for single indicators. So when looking at $\int_\mathbb{R}xd\mu(x)$, just split $\mathbb{R}$ into sets $B_1,B_2,\cdots$, where $B_i=X(A_i)$. You can always rewrite any simple function into one that's a ...


2

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $X:\Omega\to\mathbb R$ a random variable. Let $S_n=\{\omega\in\Omega : |X(\omega)|<n\}$. Then $S_n\subset S_{n+1}$ and $$\bigcup_{n=1}^\infty S_n = \Omega, $$ so $$\mathbb P\left(\bigcup_{n=1}^\infty S_n\right)=\lim_{n\to\infty} \mathbb P(S_n)=\mathbb P(\Omega) = 1. $$ $\mathbb P(S_n)$ is an ...


2

If $X:\Omega\rightarrow\mathbb R$, i.e. if $X$ takes values in $\mathbb R$, then the statement is always true. We have $\{X\in\mathbb R\}=\bigcup_{n\in\mathbb N}\{|X|<n\}$ where $\{|X|<1\}\subseteq\{|X|<2\}\subseteq\cdots$. Based on that it can be shown that $P(\{|X|<n\})$ converges to $P(\{X\in\mathbb R\})=1$. Edit: For a proof of that ...


2

If a probability space is given then the conditional expectation does not depend on the values taken by the random variable in the condition; it depends only on the sets on which this random variable takes these values as constants. You can experience this if you calculate a very simple example when the random variables involved are discrete. Since only ...



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