# Tag Info

0

Your proof is correct, but the usage of the characteristic functions seems rather artifical to me. You have shown that, by the Skorohod theorem, there exist random variables $X_n' \sim X_n$ and $X' \sim X$ such that $X_n' \to X'$ a.s. and, therefore, $a_n X_n' +b_n \to a X'+b$ a.s. Now Skorohod's theorem automatically yields $a_n X_n + b_n \stackrel{d}{\to} ... 2 Since$\text{Var}[X] \geq 0$(this is a well-known fact),$E[X^2] - (E[X])^2 \geq 0$and the inequality desired follows. 1 The case when f and g are increasing functions, for which the inequality holds, is well explained in Appendix 9.9, titled "Verification of Antithetic Variable Approach When Estimating the Expected Value of Monotone Functions", of the book Simulation by Sheldon M. Ross, 5th Ed. Academic Press, 2012. 1 For (b), one we got the existence of$k$such that$I\subseteq \bigcup_{j=1}^k I_j$, we use (a) to obtain$\mathbb P(I)\leqslant\sum_{j=1}^k\mathbb P(I_j)\leqslant \sum_{j=1}^{+\infty}\mathbb P(I_j)$. (c) We have to show that if$I$is an interval and$(I_j)_{j\geqslant 1}$is a collection of intervals such that$I\subset\bigcup_{j=1}^\infty I_j$, then ... 1 If$\phi_n$is the characteristic function for$X_n$and$\phi$the characteristic function for$X$then you know$\phi_n(t) \to \phi(t)$for all$t$. Let$\epsilon > 0$. Choose$N_1$large enough so that$n \geq N_1$implies $$|\phi_n(at) - \phi(at)| < \frac{\epsilon}{2}.$$ Since the$\phi_n$are uniformly continuous, for each$n$there is a ... 0 No. For each combination of {male, female}, {married, single}, {grad, not grad}, you can calculate the number who gave that combination of responses (lumping in "no answer" with "opposite answer" for simplicity). If you do so, you end up with a negative number of single female non-college-grads. There's a simple shortcut to calculate that by adding and ... 1 Each visitor is of type 1 with probability$p=80\%$and of type 2 with probability$1-p$hence the number of visitors of type 1, from a given time instant, before a type 2 visitor arrives, is the number$N$of heads before the first tail in a heads-or-tails game of probabilities$p$and$1-p$. This is the geometric distribution such that$P(N=n)=p^n(1-p)$, ... 0 You should apply a transform to the time derivative, remember that$F = \exp(r(T-t))\,S_t$So $$\frac{\partial V}{\partial t}_s = \frac{\partial V}{\partial t}_t - \frac{\partial V}{\partial F}_t \frac{\partial F}{\partial t}_t$$ Which should remove the spare term. 0 Thank you Bryon Schmuland, the answer of "The strong Markov property with an uncountable index set" does answer my question too. I was first put off why$T$should be finite, but since then$\{T<\infty \}=\Omega, X_T$is well defined for every$\omega$without having to look at the Limits. 0 Here is one way of doing this. To make things simpler, let's call the label the colors$1$through$n$, and let's say you have$c_{i}$stones of color$i$(so in your example,$n = 9$). Let$C = \sum_{i} c_{i}$be the total number of stones, and let's say you choose$m$stones from this set. For question 1, without loss of generality let's say the desirable ... 1 From the multiset,$S = \{\mathrm{Blue}^3, \mathrm{Red}^3, \mathrm{Green}^3, \mathrm{White}^3, \mathrm{Yellow}^3, \mathrm{Orange}^3, \mathrm{blacK}^1, \mathrm{grEy}^{18}\}$Select 6 stones, find$P(B \geq 1, R \geq 1)$.$P(B \geq 1, R \geq 1) = P(B = 1, R \geq 1) + P(B = 2, R \geq 1) + P(B = 3, R \geq 1) \\ \quad = P(B=1)P(R\geq1 \mid B=1) +P(B=2)P(R\geq1 ...

2

This might be one of the rare cases I know where the CDF is the most convenient approach... For every $n\geqslant0$, let $M_n=\max_{0\leqslant k\leqslant n}X_k$, then, for every $x$, $$[X_N\leqslant x]=\bigcup_{n\geqslant1}[x\geqslant X_n\gt X_0\geqslant M_{n-1}]=\bigcup_{n\geqslant1}[x\geqslant M_n,A_n],$$ where $$A_n=[X_n\gt X_0\geqslant M_{n-1}].$$ ...

0

I do not know whether it is elegant or not, but what I'd try is as follows: first, $$\mathbb{P}\{N=1\} = \mathbb{P}\{X_1 > X_0\} = \int_{\mathbb{R}} \mathbb{P}\{X_1 > x,\ X_0 \in dx\} = \int_{\mathbb{R}} dx f(x)\left(1-F(x)\right)$$ the last step by independence. Then, \begin{align*} \mathbb{P}\{N=2\} &= \mathbb{P}\{X_2 > X_0,\ X_1 \leq ...

4

Given two numbers $x,y$ it is known that $$\min\{x,y\}=\frac{x+y-|x-y|}{2}$$ If you replace $p$ and $q$ in the formula your formula follows (after an integration with respect to $\nu$ and assuming that $P$ and $Q$ are probability measures).

2

I assume you're referring to distributions with unbounded tails. That rules out uniform, beta and so forth. The exponential distributions are a good watermark in the distributions manifold, splitting it into 3 subspaces: exponential, subexponential and superexponential. Examples of each are: Exponential: Normal Superexponential: Subexponential: Lévy, ...

1

Because we call it a "game" does not it make a game in the sense that is studied by Game Theory. Roulette, slot machines, and other games of chance are not strategic games. Game Theory studies games where what is the best for you (usually) depends on what the other players are doing are doing. For Game Theory, you will need basic probability: Bayes rule, ...

0

Because you asked for some solution using Jensen's inequality (is this homework?), here is my attempt: $$E(\log^2p(X))=\sum_{i=1}^rp_i\log^2\frac{1}{p_i}$$ Let's use natural logarithms. Now, $\log^2(y)$ is concave in the range $y \ge e$. Let's divide the full support in two sets of "high and low probabilities", $R_L$ and $R_H$, such that $i\in R_L$ iff ...

0

I agree with the other respondent. The focus on machine learning is more geared towards ready-to-implement algorithms. However there are some Simple results like the Bias-Variance Tradeoff and other sort of discussions of "limiting" performance. From what I've seen, your question might be most practically addressed by the area of model selection (i.e. how to ...

1

Some years ago I had this kind of situation. I took a class with the name "Probability" I thought that was ok, I am fine with general probability, nothing more than that. Turned out this course was going to be more with game theory "a la Las Vegas" style then just conventional probability. And so I passed with a B. The reason for missing my A was because I ...

0

I'll address your questions in order: The $x$ disappears from the second equation because it is being provided by the Gibbs Sampler, hence it is not a explicit variable of integration. The second equation is indeed confusing, because the LHS does not actually mathematically follow from the RHS! They are relying on the behavior of the Gibbs Sampler to make ...

2

Even for uniform distributions on disjoint convex subsets of the plane it is not necessarily true. For a simple example, let $f_1$ be a uniform distribution on the segment from (0, 0) to (2, 0) and let $f_2$ be the Dirac measure at (1, 1/10). Now $$E_{X,Y}\|X-Y\|_1 = 2/3 > 3/5 = E_{X,Z}\|X -Z\|_1$$ in the $\ell^1$ norm. In the Euclidean norm the ...

1

Actually, your second statement is false. This would only be true if $X,Y$ are jointly normal, which you do not assume. And if you're assuming joint normality then uncorrelated=independent.

1

to the second question: The first two tosses must not be considered because the results already confirmed. greetings, calculus

1

If the sequence $(X_n)$ is independent and $X_n\to X$ almost surely then $X$ is an almost sure constant. Hence the sequence $(X_n-X)$ is independent.

0

Yes, you're just reparametrizing. Let $\gamma=\mu^2$ and let $L^*$ be the likelihood using $\gamma$ and $L$ the likelihood using $\mu$. Then $L^*(\gamma)=L(\sqrt{\gamma})$. Now just recall that the Cramer--Rao lower bound is the inverse of minus the expectation of the Hessian of the loglikelihood divided by $n$, which equals $\sigma^2$ if you're ...

1

No, for (a) you are looking for the probability that $A < B$ and $A < C$. You don't care who comes second. Hint: These exponential random variables could be the first arrival times for independent Poisson processes with rates $r_A$, $r_B$, $r_C$; the rates are the reciprocals of the mean arrival times. This could be realized using a single ...

4

First remark : the events "$S_n^X$ hits $A$ before $B$" are negligible. Let's be rigorous. Let $\tau_A := \inf \{n \geq 0: S_n \geq A\}$, and let $\tau_B := \inf \{n \geq 0: S_n \leq B\}$, both with values in $\mathbb{N} \cup \{+\infty\}$. Then you wan to prove that: $$\mathbb{P}_{(S_n^X)} (\tau_A < \tau_B) > \mathbb{P}_{(S_n^Y)} (\tau_A < ... 1 You went almost as far as one can go with this approach... Assume that all the coefficients \mu_k are different (otherwise things become more complicated to write down). Starting from your recursion relations, one can guess that, for every 0\leqslant j\leqslant i there exists some coefficients (\alpha_{ij}^k)_{j\leqslant k\leqslant i} such that, for ... 0 It is true for absolutely continuous distributions (but not for discrete distributions). The empirical distribution function is nothing else but the sample mean of Bernoulli random variables that has expected value F(x) and variance (1/n)F(x)\cdot[1-F(x)]. Therefore the Central Limit Theorem applies and we have$$\sqrt n\Big(\hat F_n(x) - F(x)\Big) ...

3

The function $u:p\mapsto p\log^2p$ is nonnegative and bounded by $u^*=4/\mathrm e^2$ on $[0,1]$ hence a first easy upper bound is that, for every random variable $X$ taking at most $r$ values, $$E(\log^2p(X))=\sum_{i=1}^rp_i\log^2p_i\leqslant ru^*=4r/\mathrm e^2.$$ To refine this, one can follow the proof of the classical entropy bound, since one tries to ...

2

Let $f_n$ be defined on $[0,1]$ with graph consisting of the polygon with vertices $(0,0)$, $(1/n,1)$, $(2/n,0)$, $(1,0)$, and let $f(x) = 0$. For every Borel probability measure $\mu$ on $[0,1]$, $\int f_n(x) \; d\mu(x) \le \mu((0,2/n)) \to 0$ as $n \to \infty$. But if $\nu_n$ is the point mass at $1/n$, $\nu_n$ converges weakly to the point mass at ...

0

$P[h_{bad} \text{makes no error on } S_n]$ is the probability that a particular bad classifier would work on the training data. To get a bound on the probability that any bad classifier would work on the training data, we need to multiply by the number of potentially bad classifiers, $K$.

0

Whenever you have a function $f \,:\, A \to B$ and some $\sigma$-Field $\mathcal{B} \subset \mathcal{P}(B)$ on $B$, then the set of all preimages of sets in $\mathcal{B}$, i.e. $$\mathcal{A_f} := \left\{ f^{-1}(Y) \,:\; Y \in \mathcal{B}\right\}$$ is a $\sigma$-field on $A$. Note the use of $f^{-1}$ here doesn't imply that $f$ is invertible - $f^{-1}(Y)$ ...

1

If the variables are jointly normal, then their sum, denote it $S_n$ will be a normal random variable, with mean zero and a variance, denote it simply by $v_s$. Then the random variable $Z = S_n^2/v_s$ will follow a chi-square distribution with one degree of freedom. Then, the random variable $W = v_sZ =S^2_n$ follows a Gamma distribution with shape ...

2

In fact $\lim_N Y_N$ exists almost surely is equivalent to $E(|X_1|) < +\infty$, which is not the case for Cauchy variable. One direction of above assertion is the famous law of large number. To prove the other direction, note that if $\lim_N Y_N$ exsits alomst surely, then $\lim_N\frac{X_N}{N} = 0$ alomst surely, which we will see is impossible if ...

1

We can use the chain rule for conditional probabilities on $P(X,Y|Z)$ to condition twice, as follows:- $$P(X,Y|Z)=P(X|Y,Z)P(Y|Z)$$ This follows from the following - where you know that $P(X,Y)=P(X|Y)P(Y)$, so that $P(X,Y,Z)=P(X|Y,Z)P(Y,Z)$:- $$P(X,Y|Z)=\frac{P(X,Y,Z)}{P(Z)}=\frac{P(X|Y,Z)P(Y,Z)}{P(Z)}=P(X|Y,Z)P(Y|Z)$$

2

There are many possibilities. Here is one: Suppose $X$ can take values between $1$ and $10$ with a density function $f(x)= \dfrac{\log_{10} e}{x}$. Here is an extension: Suppose $X$ can take values between $10^a$ and $10^b$ where $b-a$ is a positive integer, with a density function $f(x)= \dfrac{\log_{10} e}{(b-a)x}.$

2

I've not worked out the details but I am sure it will go through. Calculate the moment generating function $E[exp(s \sum X_i)]$. Differentiate this enough times and you get your answer. Taking the covariance matrix to have compound symmetry then I get the generating function for $\sum X_i$ to be $exp(n(1+(n-1)\rho)\sigma^2 s^2 /2)$. This is at least right ...

2

Note that $\mathbf x$ and $\hat{\mathbf x}$ are vectors indexed by the vertex set $V$. The assumption in (1) is that every entry of $\mathbf x$ and $\hat{\mathbf x}$ corresponding to some vertex in the clique $C$ coïncide. Example: if $V=\{1,2,3,4,5\}$ and $C=\{1,2\}$, condition (1) asks that some function $g$ on $\Lambda^5$ is such that ...

0

There is no $X_C$ here and, by definition, $$x_C=(x_v)_{v\in C}.$$ (This notation is used everyuwhere on the page linked to, and well before the paragraph reproduced in the question.)

0

We are given that $X,Y\sim \mathrm{Exp}(1)$ so that the probability density function (pdf) of $X$ is $f(x)=e^{-x}$ and the pdf of $Y$ is $f(y)=e^{-y}$ For all $z_1,z_2>0$ we need to evaluate the joint density of $X$ and $Y$ with the appropriate limits as follows:- ...

4

Cauchy-Schwarz inequality applied to the product $Y\cdot\mathbf 1_A$ with $A=[Y\ne0]$ yields $$E[Y\cdot\mathbf 1_A]^2\leqslant E[Y^2]\cdot E[\mathbf 1_A^2].$$ The LHS is $E[Y]^2$ because $Y\cdot\mathbf 1_A=Y$, and, in the RHS, $E[\mathbf 1_A^2]=P[A]$, hence $$E[Y]^2\leqslant E[Y^2]\cdot P[A],$$ or, equivalently, $$... 0 Using theorem for transformation variable, it will hold relation$$ g_Y(y)=f_X(x)\cdot|J|, $$where J is Jacobian. Since it only one variable, then the Jacobian is$$ J=\frac{dx}{dy}. $$We have X\sim\mathcal{U}(a,b), then the pdf of X is$$ f_X(x)=\frac1{b-a}. $$The transformation is Y=2-4X\;\Rightarrow\;X=\dfrac{2-Y}{4}, then the Jacobian is$$ ...

0

$$g(x,t) = ∫_0^t \theta(u) \exp(-\alpha(t-u) ) dB(u) \\ = e^{-\alpha t} ∫_0^t \theta(u) \exp(\alpha u ) dB(u)$$ Now use the fact that $t \to e^{-\alpha t}$ is of finite variation and the Ito formula for a product: If $X,Y$ are two Ito processes then so is $XY$: $$dX_t = a_t dt + \sigma_t dB_t \\ dX_t = b_t dt + \sigma'_t dB_t \\ d(XY)_ t = X_t dY_t ... 0 Yes, it is. For a formal proof you could apply integration by substitution, e.g., in case Y=4X$$P[x_0\le Y\le x_1]=P[\frac{x_0}{4}\le X\le \frac{x_1}{4}]=\int_{x_0/4}^{x_1/4}\frac{1}{b-a}\,dx=\int_{x_0}^{x_1} \frac{1}{b-a}\frac{1}{4}\,dx Then you see that $4X$ is uniformly distributed on $[4a,4b]$. I think you can solve the more general case on your ...

0

Questions 1 and 3 are basic Poisson Process results. Let $(N(t):t>0)$ be a Poisson-Process with rate $\lambda$. Then the probability distribution of the waiting time is exponential$(\lambda)$, so using that you can solve question one. Also from any given moment the total of occurences in a timespan of $t$ time is poisson$(\lambda t)$ distributed, which ...

1

This solution adapts the proof in Heinz Bauer's Measure and Integration Theory. For the first item, fix $\mu \in M(X)$ and without loss of generality take $\mu$ to be unsigned (taking components later resolves this). Fix an open ball $V$ of $\mu$ in the vague topology, say, $V$ is the intersection of balls under the seminorms $\mu(f_1), \dots, \mu(f_n)$ ...

1

We do not. Let $X$ be any nontrivial integrable random variable on $(\Omega, \mathcal{F}, P)$ with $EX = 0$. Set $\mathcal{F_k} = \{\Omega, \emptyset\}$ for $k \le 42$ and $\mathcal{F}_k = \mathcal{F}$ for $k > 42$. Then $E[X \mid \mathcal{F}_{42}] = EX = 0$ a.s, but $E[X \mid \mathcal{F}] = X$ which is not zero.

1

Define $X':=X-\mathbb E[X\mid F_{k_0}]$ for $X$ an integrable $\mathcal F$-measurable random variable: we do not have in general that $\mathbb E[X'\mid\mathcal F]=0$ a.s. , otherwise $X$ would be almost everywhere equal to a $\mathcal F_{k_0}$-measurable random variable, which is not necessarily the case.

0

Hint: By symmetry, this is $E(Y|X+Y)$. Given that $X+Y=w$, what is $E(X+Y)$?

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