# Tag Info

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I'm sorry I can't show it to you because I don't see it either. But you seem to study the same script like me. If I read the definitions correctly, the following should hold: $$p'_t(x)=p'_t(0,x)=\mathbb P[S_t=x|S_0=0]=p(x)t.$$ This implies $\frac{\mathrm d}{\mathrm d t}p'_t(x)=p(x)$ and ...

1

Fix $f \in C(E)$ and $\eta \in E$ such that $$f(\eta) = \min_{\zeta \in E} f(\zeta).$$ Then $$g(\zeta) := f(\zeta)-f(\eta) \in C(E)$$ defines a non-negative function and therefore, since $T$ is a positive operator, $$Tg \geq 0. \tag{1}$$ On the other hand, it follows from the linearity of $T$ and $T1 = 1$ that $$Tg = Tf - f(\eta). \tag{2}$$ Combining ...

1

Since $2^{-m} m^{-3/2}$ is a decreasing function of $m$, $m(n)$ is approximately the value that makes the inequality into an equality, that is $$2^{-m(n)} m(n)^{-3 / 2} \approx \frac{1}{n} ,$$ or equivalently $$m(n) + \tfrac{3}{2} \log_2 m(n) \approx \log_2 n .$$ For large $m$ we have that $$m(n) \gg \tfrac{3}{2} \log_2 m(n) ,$$ so the contribution of the ...

1

Geometric approach Note that $\frac{t^n}{n!}$ is the volume of the set $S_t=\{(t_1,t_2,\dots,t_n)\in\mathbb R^{n}\mid t_i\geq 0\text{ and } t_1+t_2+\cdots+t_n\leq t\}$. So we can perform a change of variables in the integral, replacing $t=t_1+\dots + t_{n+1}$, as: \begin{align}\int_{0}^\infty \frac{t^n}{n!}e^{-t}\,dt &= ... 2 This is - as so often - done by the good sets principle. Let \def\A{\mathscr A} \A' := \left\{B\in \sigma(\A) : \exists \A_B \subseteq \A, \A_B \text{ countable }, B \in \sigma(\A_B) \right\} $$Now we show that \A' is a \sigma-algebra: \emptyset\in \A', as \emptyset \in \sigma(\emptyset) and \emptyset is countable. If B\in \A', then there ... 2 The way you worded that is a little strange. I have an alternative approach. Let I be the event that a person is infected, D be the event that a person develops the disease, and \bar I, \bar D, not those events. Then$$P(I\bar D)= P(\bar D|I)P(I) = (1-P(D|I))P(I) = (1-.30)(.5) = 0.35 = 35\%$$where in the second step I used the product rule. 0 As far as I am understanding this, when it says "I have a deck of cards marked from 1 to 4" they actually mean they only have one of each card in the deck (i.e. a deck of four cards), as such, if a card has been drawn in the first draw, it can not be drawn in the second. 1 If the probability of drawing the first card marked 1 is 1/4, then the probability of drawing the second card marked 1 is 0. I did not understand this point. How can it be certain that first card drawn is 1 to make the probability of drawing the second card marked 1 is 0. It can't be certain, but sometimes it does happen, and ... 1 I think what they mean is if the first card is a 1 then the probability of the second card is 0 since that card has already been drawn. This is a conditional probability. But it doesn't make sense to say if "the probability" the first is 1 "is 1/4" then the probability the second is a 1 is 0. You have to remove the parts in quotes for it to ... -1 Yes, that is the definition of independence of two discrete random variables. And since X and Y are independent then if A is any event for X and B is any event for Y then A and B are independent. In particular the two events you specify. 0 Let us denote events: A_{n,m} = \{|X_n - X_m| > \epsilon\} = \{|X_n - X + X - X_m| > \epsilon\} A_{n} = \{|X_n - X| > \epsilon/2\} A_{m} = \{|X_m - X| > \epsilon/2\} Then we need to show that P(A_{n,m}) \le P(A_n) + P(A_m). Proof: Now that's that out of the way. P(A_{n,m}) = P(|X_n - X_m| > \epsilon) = P(|X_n - X + X - X_m| ... 1 You're on the right track. Now argue that if |X_n-X|+|X_m-X|>\epsilon, then |X_n-X|>\epsilon/2, or |X_m-X|>\epsilon/2. (Hint: Suppose not.) The desired result then follows from P(A\cup B)\le P(A)+ P(B). 1 Yeah, applying Etemadi's inequality is a good idea. The following identity, which holds for any non-negative random variable X, will also be useful:$$\mathbb{E}(X) = \int_0^{\infty} \mathbb{P}(X > r) \, dr.$$Because of the monotonicity of r \mapsto \mathbb{P}(X > r) this implies$$b \sum_{k=0}^{\infty} \mathbb{P}(X > kb) \leq \mathbb{E}(X) ...

1

We have that \begin{align} \operatorname E\mathrm{sgn}(X-Y) & =-1\cdot\Pr(X<Y)+0\cdot\Pr(X<Y)+1\cdot\Pr(X>Y) \\ & =\Pr(X>Y)-\Pr(X<Y) \end{align} and \begin{align} \operatorname E\mathrm{sgn}^2(X-Y) & =(-1)^2\cdot\Pr(X<Y)+0^2\cdot\Pr(X<Y)+1^2\cdot\Pr(X>Y) \\ & =\Pr(X>Y)+\Pr(X<Y) \end{align} using the law ...

2

Let $W_n=n^{-1/2}Y_n$. Then $$n^{-1/2}E[Y_n\mid Y_n>0]=E[W_n\mid W_n>0]\le \frac{\sqrt{E|W_n|^2}}{P\{W_n>0\}}=\frac{\sigma}{P\{W_n>0\}},$$ and $P\{W_n>0\}\to 1/2$ (by the CLT), i.e. for any $\epsilon>0$, $P\{W_n>0\}\ge \frac{1}{2}-\epsilon$ for $n$ large enough.

1

$$|f|1\{|f|>n\}\le |f|\in L^1.$$ Thus, $$\lim_{n\to\infty}\int|f|1\{|f|>n\}d\mu=\int \lim_{n\to\infty}|f|1\{|f|>n\}d\mu.$$

0

It seems to me the result is not true. I think you may have tried simulations in which $|\mu|$ is small, $\sigma$ larger. You are dealing with relative errors, which can be large for estimated quantities near $0.$ If I follow correctly what you say, then here is a simulation of it in R based on $B = 100,000$ samples with $n = 10,\; \mu = 1000,\; \sigma = ... 1 Multivariate standard normal density function is in the form$f(x)=\exp(-\|x\|^2)$where$x$is the vector. Observe that$f(x)$doesn't care about$\frac{x}{\|x\|}$and only cares$\|x\|$, hence conditioned on any slice$\|x\|=a$,$f(x)$is a constant which means all directions are equally-likely (i.e. uniform distribution). Now, the conditional ... 2 Let$B_{n}=\cup_{i=1}^{n} E_{i}$. Notice, that$\lim\limits_{n \to \infty} B_{n} =E$(The details are left to you!, but as a hint,$B_{n}$is ncreasing!). $$\mu (E)=\mu(\lim B_{n})= \lim \mu(B_{n})=\lim \mu(\cup_{i=1}^{n} E_{i})= \lim \sum\limits_{i=1}^{n} \mu (E_{i}) =\sum\limits_{i=1 }^{\infty}\mu(E_{i}).$$ The seccond lime is because$B_{n}$is increasing ... 2 Recall that the arrival times in a Poisson process with rate$\lambda$follow an Exponential distribution with mean$1/\lambda$. Since inter-arrival times are independent, and you have two Poisson process, then it sounds like you are asking for $$P(T<S)$$ where$T = X_1+X_2$is the sum of two waiting times for one process, and$S = Y_1+Y_2$is the sum ... 1 Fix$\epsilon>0$. Since$(X_t)_{t \geq 0}$satisfies the SLLN, there exists for almost all$\omega \in \Omega$a constant$S=S(\omega)>0$such that $$\left| \frac{X_s(\omega)}{s} - \mathbb{E}(X_1) \right| \leq \epsilon \quad \text{for all s \geq S}.$$ As$U_t \uparrow \infty$, we have$U_t(\omega) \geq S$for$t \geq T=T(\omega)$sufficiently ... 0 $$P(C|D)=\frac{P(C\cap D)}{P(D)}=\frac13$$ $$P(C|D^C)=\frac{P(C\cap D^C)}{P(D^C)}=\frac13$$ So $$P(D)=3P(C\cap D)$$ and $$P(D^C)=3P(C\cap D^C)$$ Now $$P(D)+P(D^C)=1$$ Thus, $$P(C\cap D)+P(C\cap D^C)=\frac13$$ $$P(C)=\frac13$$ 1 Hint for alternative route: If random variable$X$only takes values in finite set$S$and random variable$Y$only in finite set$T$then checking independence comes to checking$P\left(X=s\wedge Y=t\right)=P\left(X=s\right)P\left(Y=t\right)$for$s\in S$and$t\in T$. In your case the sets$S$and$T$can be taken quite small, so the verification will ... 1 Toss a fair coin. For a single drawing the expected proportion of heads is distributed as follows (for$0$and$1$): $$\frac12, \frac12$$ For ten drawings in a row, by simple probability computation (from$0$to$10$): $$\frac{1}{1024}, \frac{10}{1024}, \frac{45}{1024}, \frac{120}{1024}, \frac{210}{1024}, \frac{252}{1024}, \frac{210}{1024}, ... 2 Which one is true, a high probability or 0.5? If indeed every new and discrete event is independent of the results of the previous independent events, then the probability is indeed 0.5. So the only question remaining is - are those events really "physically" independent of each other? I don't think that science has a decisive answer for this ... 1 It's surely 0.5. By no chance is the probability of tail going to increase. We tend to look at it that way only because, we have a feeling that successes and failures don't come continuously, but come pair-wise. That is a notion of probability where all the partition events are exhausted before any event starts repeating. However, actually, it doesn't happen ... 1 the l_2 norm of rv Y is equivalent to E[Y^2]^{\frac{1}{2}} thus you must prove that$$E[Y_1^2]\leq E[(Y_1-Y_2)^2]=E[Y_1^2]-2E[Y_1Y_2]+E[Y_2^2]$$letting E[Y^2]=a and E[Y]=b you would be proving that$$a \leq 2(a-b)$$with Jensens you have a\geq b but I dont know how you could prove statement above is generally true. 0 U^tAU=D where D is diagonal and U^tU=I. Use U to change variables in the multi-dimensional transform integral. Starting with$$ \int_{\mathbb{R}^n}e^{-(Ax,x)}e^{ix\cdot\xi}dx, $$Let x=Uy. Because U is orthogonal, then \|x\|=\|Uy\|=\|y\| gives a constant Jacobian equal to 1. Therefore, the above is transformed to$$ ... 0 What the book it hinting at is the knowledge that if $$X_t=X_0+\int_0^t X_s\,ds +B_t,$$ then$X$is a continuous semimartingale and $$\int_0^t e^{-s}\,dX_s=\int_0^t e^{-s}X_s\,ds+\int_0^t e^{-s}\,dB_s.$$ The differential form of an SDE is more than just a way to save ink (or electrons); it can (as here) serve as a guide to deducing one integral ... 0 As proposed by @T.Bongers, it follows by Fatou' lemma. Indeed,$X_n,~n\geq 1$are non negative and converge ae to$X$, so by Fatou's lemma, we get$Ε(X)\leq \liminf E(X_n)$. Since$E(X_n)\leq c$for all$n,$we get$Ε(X)\leq \liminf E(X_n)\leq c,$as desired. 3 (From my comment above.) This follows from Fatou's lemma. 0 The statement is not clear, because you don't specify how$Z$is (jointly) related with the other variables. Assuming you mean that the conditional are the same ($P(X|Y,Z)=P(X|Y',Z')$) : $$H(X|Y,Z)=-\sum_{y,z} p(y,z) \sum_{x} p(x|y,z)\log p(x|y,z)$$ This implies that for both scenarios the inner sum (the entropies conditioned on particular values: ... 1 $$X_{n+1} = X_n + T_{n+1}$$ where$T_{n+1}$is independent of$X_1,\ldots,X_n$and $$\Pr(T_{n+1}\in A) = \int_A \frac{dt}{(2+t^2)^{3/2}}$$ for every measurable set$A$. (Thus$T_{n+1}$has a t-distribution with$2$degrees of freedom.) 5 Neither of them holds, in general, for stochastic integrals. The trouble already starts if you consider measures which need not to be non-negative, i.e. signed measures. For a signed measure$\mu: (\Omega,\mathcal{A}) \to \mathbb{R}$we cannot expect that the triangle inequality $$\left| \int f(x) \, \mu(dx) \right| \leq \int |f(x)| \, \mu(dx) \tag{1}$$ ... 1 So the thing is that what you are looking for is exactly the content of lemma 3. I report the claim here : Lemma 3 : Let$X$be a càdlàg square integrable martingale and${\xi}$be a bounded predictable process. Then,$\int\xi\,dX$is a square integrable martingale. In your case$W$is a Brownian motion, so you fit the conditions for the lemma as a ... 0 Hint: Since$A$is an SPD matrix, it has a unique SPD square root. Apply the change of variables $$y = A^{1/2}x$$ Noting that $$\langle Ax,x \rangle = \langle A^{1/2}x,A^{1/2}x \rangle$$ 0 It is not possible because your uniform distribution should have a constant density function over$A$. But the Lebesgue integral of such a function over$A$is$0$. This follows from the definition of Lebesgue integral for simple functions. In that case the integral would be the product of a constatnt times$\mathcal{L}(A)$which is$0$. 2 This probability is$0$. You can find it by conditioning on$g_1,\dots,g_{N-1}$.$span(g_1,\dots,g_{N-1})$is an$N-1$dimensional subspace. Let$S$be the subspace orthogonal to this and pick a line$L$in$S$. Observe that$L$always exists since$N\leq M$. Due to rotational invariance of normal distribution, the projection of$g_N$onto$L$is a standard ... 2 I believe you are overthinking this assignment, and perhaps you feel the need to do some calculations before giving any kind of answer. You should not approach this task in that manner, as I hope my answers to your questions will illustrate. (i) is correct as long as you have "seniors" in your answer. They are the people contacted for responses to the ... 3 Call three heads and three tails when tossing six coins a success. Then the probability of success is$\frac{\binom{6}{3}}{2^6}$, which simplifies to$\frac{5}{16}$. Let$X$be the number of trials until the first success. Then$X$has geometric distribution with parameter$p=\frac{5}{16}$. It is a standard result that if$X$has geometric distribution ... 1 For each$\omega$choose a symbol, either$T$or$H$, that appears infinitely often in the sequence$\omega_1,\omega_2,\dots$. Now let$(\alpha(i))_{i\geq 1}$consist of the indices of that symbol in$\omega$. 1 No. Covariance,$\mathsf{Cov}(X,Y)$, cannot be expressed exclusively in terms of$\mathsf E(X), \mathsf E(Y), \mathsf {Var}(X), \mathsf {Var}(Y)$. Covariance involves the term of$\mathsf E(XY)$, which cannot be expressed in those other terms (unless the rv are uncorrelated, but in which case covariance is simply zero). $$\mathsf {Cov}(X,Y) = \mathsf ... 1 Sure, you can use the calculus formula of the covariance. Cov(X,Y)= E(XY)-E(X)E(Y) Or, you can also have Cov(X,Y) =\frac{Var(X)+Var(Y)-Var(X+Y)}{2}= \frac{E(X^2)-E^2(X)+E(Y^2)-E^2(Y)-Var(X+Y)}{2} 1 First, it is not the same as the probability of having k/0.65 customers walk into the shop. For example, taking k = 5, it is really unlikely that exactly 2/0.65 \approx 7.692 customers will walk into the shop. (Well, maybe less unlikely if the shop is located in King's Landing...) What you have is a rate of customers walking in and buying coffee of ... 0 This is how I would view the symmetry associated with the stationary distribution: Put each ball in either urn with equal probability. Pick that ball up a random number of times unrelated to which urn it is in, and move it to the other urn. The probability that that ball started in urn A was \frac12 and in urn B also \frac12. Similarly with its ... 1 Consider the sequence \{H,T,H,T,H,T,\ldots\} that infinitely alternates between heads and tails. Clearly, the proportion of heads for this sequence converges to 1/2. However, the subsequence made up of only the even terms, \{T,T,T,\ldots\}, does not have a convergent proportion of heads converging to 1/2, despite there being other sequences with ... 2 For the first part, (i):$$ \mathbb{P}\{X \geq Y\}\} = \mathbb{E}[\mathbb{1}_{\{X \geq Y\}}] = \mathbb{E}[\mathbb{E}[\mathbb{1}_{\{X \geq Y\}}\mid Y] ]. $$Dealing with the inner conditional expectation first,$$ \mathbb{E}[\mathbb{1}_{\{X \geq Y\}}\mid Y] = e^{-\lambda} \sum_{k=0}^\infty \mathbb{1}_{\{k \geq Y\}} \frac{\lambda^k}{k!} = e^{-\lambda} ... 2 For the first one, for any$t>0$, $$P(X\geq Y)=P(tX\geq tY)=P(t(X-Y)\geq0)=P(e^{t(X-Y)}\geq1)\leq E(e^{tX})E(e^{-tY})$$ by Markov inequality and independence. Noticing that$E(e^{tX})$is the moment generating function of$X$(and noticing the same about$Y$), and substituting the forms of the moment general function, we get a bound that is equal to ... 0 For the first problem, i would try this way (but i'm not sure it will work though ;) :$\mathbb{P}(X \geq Y) = \sum_{k \geq 0} \mathbb{P}(X \geq k, Y=k) = e^{-3\lambda}\sum_{k \geq 0} \sum_{i=k}^{\infty} \frac{\lambda^{i+k}2^k}{i! k!}$. 1 Choose a random variable$Y$such that$X=Y\mathbf P$-a.s. and$X_n \to Y$pointwise. As all$X_n$are$\sigma(\{X_n : n \in \mathbf N\})$-measurable,$Y$is also. By assumption$F$and$\sigma(X_n: n \in \mathbf N)$are independent, hence so are$F$and$\sigma(Y) \subseteq \sigma(X_n: n \in \mathbf N)$. Now let$f \in F$and$A \subseteq \mathbf R\$ ...

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