# Tag Info

## New answers tagged probability-theory

0

Some observations on i.i.d. vectors $\{X_i\}_{i=1}^{\infty}$ through a function $f:\mathbb{R}^k\rightarrow\mathbb{R}$. 1) As in my above comments: Lipschitz-like property: If there is a real-valued constant $L>0$ such that: $$||f(X_i+\delta_i)-f(X_i)|| \leq L||\delta_i||\quad \forall i \in \{1, 2, 3, ...\}$$ then the desired probability 1 ...

1

Notice that if $X < +\infty$ a.s., then $Y < \infty$ a.s. follows easily. Indeed, let $A = \{Y = +\infty\}$, we need to show that $\mathbb{P}(A) = 0$. It follows from the condition on $Y$ that $$\mathbb{E}(Y \mathbb{I}_{ \{X \leq p\}} \mathbb{I}_A) \leq \mathbb{E}(Y \mathbb{I}_{\{X \leq p\}}) <\infty,$$ hence Y \mathbb{I}_{X \leq p} \mathbb{I}... 2 \begin{align*} \mathbb E(\min(X,Y)&=\int_{400}^{800}\int_{500}^{600}\min(x,y)\dfrac{d x}{100}\dfrac{d y}{400}\\ &=\dfrac{1}{40000}\left(\int_{400}^{500}\int_{500}^{600}\min(x,y) dxdy + \int_{500}^{600}\int_{500}^{600}\min(x,y) dxdy + \int_{600}^{800}\int_{500}^{600}\min(x,y) dxdy\right) \\ &=\dfrac{1}{40000}\left(\int_{400}^{500} 100ydy +\int_{... 0 You may getf$from$g$. $$f_\alpha(x) = \frac{\alpha}{(x+1)^{\alpha+1}}$$ so $$f_1(x)dx = \frac{1}{(x+1)^2}dx = \frac{\lambda d\tilde{x}}{(\lambda \tilde{x}+1)^2} = g(\tilde{x})d\tilde{x}$$ where we used the substitution$x = \lambda\tilde{x}$. 0 After a bit of research told me that nothing of the above is true; see for example the book of Dudley "Real analysis and Probability" (section 11.4). The empirical measure on a measure space$(\Omega,\mathcal F,\mu)$is defined via a sequence of iid random variables$X_i$defined on$\Omega^{\mathbb N}$, by the mapping $$A\mapsto \frac{1}{n}\sum_{i=1}^n\... 1 The total variation distance is a metric on the space of all complex Borel measures and its restriction to the class of probability measures is a nice complete metric. [It is not separable in general]. 1 The second one is correct if the random variables are positive. The first one seems to be wrong. 4 Here is another intuitive argument: One of the 110 initial swimmers went faster than the other 109. Let's look at who was fastest and how long they took By symmetry, each of them was equally likely to be fastest, each with probability \frac{109!}{110!}=\frac1{110}, since each of the 110! orders of these swimmers are assumed equally likely By a ... 1 Hints: (a) Show that the chain is irreducible (this is easy) and that the chain is positive recurrent by finding a stationary distribution. Show that the period is 1 (by irreducibility you have to check this for one state). (b) Already done when you solved (a) using the suggested approach. (c) The chain is positive recurrent, thus E_3[T_3] is equal to ... 2 I'll try to get you started with (a) and (b): (a) A chain is ergodic if some power of its transition matrix has all positive elements. For your chain, the second power has all positive elements. A = matrix(c(0.5, 0.5, 0, 0.5, 0, 0.5, 0, 0.5, 0.5), nrow = 3, byrow=T) A [,1] [,2] [,3] [1,] 0.5 0.5 0.0 [2,] 0.5 0.0 ... 2 Just integrate the pdf with t>0:$$F(t) = \int\limits_{-\infty}^t f(x)dx = \int\limits_{-\infty}^0 0dx + \int\limits_{0}^t e^{-x}dx = 0 + -e^{-x}|_0^t = -e^{-t} + 1 = 1-e^{-t}$$and if t\leq 0:$$F(t) = \int\limits_{-\infty}^t f(x)dx = \int\limits_{-\infty}^t 0dx = 0$$1 In other words, you want to compute in a closed form$$\frac{1}{2 \sqrt{2 \pi }}\int_a^\infty e^{-\frac{x^2}{2}} \,\,\text{erfc}\left(-\frac{\alpha x+\beta }{\sqrt{2}}\right)\,dx$$Having made, in the past, an extensive search on the Internet of "integrals involving the error function" I did not find anything for a\neq 0. In my humble opinion, beside ... 0 So, you want to show that$$\mathbb{E}_\mathbb{Q} \left|{1 \over \mathbb E_{\mathbb P}[{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]}\mathbb E_{\mathbb P}\left[X{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G\right]\right| < \infty.$$By the conditional triangle inequality, it is enough to show$$\mathbb{E}_\mathbb{Q} \left[{1 \over \mathbb E_{\mathbb P}[{{... 0 In general, if$(S, \mathcal{S}, \nu)$is a measure space,$\mu$is another measure on$(S, \mathcal{S})$, and$f : S \to [0,\infty]$is an$\mathcal{S}$-measurable function, we say that$f$is the density of$\mu$with respect to$\nu$if, for every set$A \in \mathcal{S}$, we have $$\mu(A) = \int_A f\,d\nu.$$ (If such$f$exists then it is unique up to$\...

2

For consistency note that by taking logs we get $$a\times\frac{X_1+\dotsb+X_n}{n}\times\frac{\log(1-\frac{a}{n})}{a/n}\stackrel{\text{a.s.}}{\to}a\times\lambda\times-1=-a\lambda$$ as $n\to \infty$ by the strong law of large numbers and the fact that $$\lim_{x\to0}\frac{\log(1-x)}{x}=-1.$$ Hence Y_n=(1-\frac{a}{n})^{X_1+X_2+...+X_n} \stackrel{\text{a.... 0 \sigma-finite is a particular kind of measure that is the countable union of measurable sets with finite measure. The other part of the density is about the Radon-Nikodym Theorem. One definition of a probability density function is as the Radon-Nikodym derivative of the induced measure with respect to a base measure, which is what is talked about in your ... 1 I believe that using the The Central Limit Theorem and conducting some Hypothesis Tests can help you out. Recall that the CLT states that if x_{1},...,x_{n} is an independent and identically distributed sample coming from some distribution where E[x]=\mu and Var[x]=\sigma^2<\infty then we can say that \frac{\sqrt{n}(\bar{x}-\mu)}{\sigma} ... 1 Both are wrong . For the first one X_1=1,X_2=2 and \alpha =2 gives a counterexample. For the second one you need continuity of F_{X_2} at the point \frac {\alpha} 2. In general you should take the left hand limit of F_{X_2} at the point \frac {\alpha} 2 on RHS. 1 According to "A first course in Stochastic Processes" by Samuel Karlin and Howard M. Taylor, this process is called "Multi-Type Branching Process". The extinction criterion for it is: \lim\limits_{n \to \infty} P(v_n = 0) = 1 iff the absolute value of all eigenvalues of EA does not exceed 1. 0 I assume your "first probability" is the one with k. One way to to look at it: If the first probability < 1/n^2 (what you assumed), and second probability \le first probability (since k \le n), then second prob \le first prob < 1/n^2. What's the problem? Another way to look at it: If the first prob < 1/n^2, for all k \le n, then ... 1 Note that B_T = B_0e^{rT}, not e^{-rT}! What I think your professor has done is the following (I will set X = \mathbb{I}_{S_T\geq K} for simplicity): \begin{align} e^{-rT}\text{E}_Q\left[S_T X\right] &= S_0 e^{-rT}\text{E}_Q\left[\frac{1}{S_0}S_T X\right]\\ &= S_0\text{E}_Q\left[\frac{S_T}{S_0} e^{-rT} X\right]\\ \end{align} Now, using the ... 0 This is the proof of professor. I tried to write: E^{\mathbb{Q}}[S_T \mathbb{I}_{S_T \geq K}]e^{-rT}=E^{\mathbb{Q}^S}[\frac{S_T}{B_TS_0}\mathbb{I}_{S_T \geq K} \frac{d \mathbb{Q}^S}{d \mathbb{Q}}]e^{-rT}=E^{\mathbb{Q}^S}[\frac{S_T \cdot 1}{B_TS_0}\mathbb{I}_{S_T \geq K} \frac{d \mathbb{Q}^S}{d \mathbb{Q}}]e^{-rT}=E^{\mathbb{Q}^S}[\frac{S_T \cdot B_0}{... 0 The book is called "Foundations of Modern Probability". By definition (check page 10) \mu f=\int f\,d\mu=\int f(\omega)\,\mu(d\omega). $$So on the left hand side of (4), you have$$ \mu(g\circ f)=\int (g\circ f)(\omega)\,\mu(d\omega). $$On the other hand, \mu\circ f^{-1} is a measure (check the bottom of page 9):$$ (\mu\circ f^{-1})B=\mu(f^{-1}(B)...

1

Let $(a,b)$ be a pair of cards in your deck. The probability that $(a,b)$ is a black-black pair is $25/102$: as @lulu pointed it out $$\mathbb P((a,b) \ \textrm{black})=\mathbb P(a \ \textrm{black}) \times \mathbb P(b \ \textrm{black} | a \ \textrm{black})=(1/2)\times (25/51)=25/102.$$ Let $X_{(a,b)}$ be the random variable defined as $1$ if $(a,b)$ is a ...

1

Replacing $X_n$ by $X'_n:=\sqrt nX_n$, the question can be rephrased as follows: if $\left(X_n\right)_{n\geqslant 1}$ is a sequence of random variables such that $\{X_n(1+\varepsilon_n),n\geqslant 1\}$ is uniformly integrable for a sequence $(\varepsilon_n)_n$ converging to $0$ in probability, is $\{X_n,n\geqslant 1\}$ uniformly integrable? The answer is: ...

0

Probability distribution can be used to define two objects : a probability density function (for a continuous r.v.) or a probability mass function (for a discrete r.v.), a cumulative distribution. The first one is defined over the range of the random variable. For example, if you consider a real random variable, then it is defined over $\mathbb{R}$. The ...

2

Given $a<b$ it is easy to construct a sequence of continuous functions $(f_n)$ such that $0\leq f_n \leq 1$ and $f_n(x) \to 1$ for $x \in [a,b]$, $f_n(x) \to 0$ for $x \notin [a,b]$. Similarly take $c<d$ and choose continuous functions $(g_n)$ such that $0\leq g_n \leq 1$ and $g_n(x) \to 1$ for $x \in [c,d]$, $g_n(x) \to 0$ for $x \notin [c,d]$. Then ...

0

The joint CDF of $X_{(i)},X_{(j)}$ is given by $F_{X_{(i)},X_{(j)}}(u,v) = \sum\limits_{k=i}^{j-1}\sum\limits_{m=j-k}^{n-k}\frac{n!}{k!m!(n-k-m)!}[F_{X}(u)]^k[F_{x}(v)-F_{x}(u)]^m[1-F_{X}(v)]^{n-k-m} + P(U \geq j)$. The argument for this is explained in Statistical Inference by Casella and Berger problem 5.26 pretty clearly. I can elaborate more on it if I ...

1

If $\{X_i\}$ is an i.i.d sequence of random variables, then so is $\{|X_i|\}$. Now you can apply the weak law of large numbers to the sequence $\{|X_i|\}$.

0

The second half seems trivial. $$\mathbb{E}( (|X_1| + |X_2| + \dots) /n ) = \frac{1}{n}(\mathbb{E}(|X_1|)+\mathbb{E}(|X_2|)+\dots) = \mathbb{E}(|X_1|)$$ as they are identically distributed. (We don't even need independence). I guess this is convergence but this is just a constant sequence.

0

Yes, once such case is when all the random variables are uniform in $[0,1]$. The sum of uniform random variables has the Irwin-Hall distribution and it's tail bounds match the ones given by Hoeffding's inequality. See Corollary 5 here. Also see Bernstein's inequality which is a generalization of Hoeffding.

1

Let it be that $p$ denotes the probability that some fixed day will appear to be a sunny day. Then the probability that the day after this day is sunny equals: $$p\times0.7+\left(1-p\right)\times0.2$$ so that:$$p=p\times0.7+\left(1-p\right)\times0.2$$ and consequently $p=0.4$. The expectation of the number of sunny days among $360$ is then $360\times0.4=... 1 If you can approximate$P(z)$via samples, then the sum approximates$\int P(X|z)P(z)dz= \int P(X,z)dz = P(X) $1 It looks like a question to be quickly answered within a multiple-choice questionnaire. So, an "optical" solution without any further calculation or reasoning might be helpful, as well: 1 Given that the length of the stick is$1$. We break the stick into two pieces at some point$x$. Then we will be left with one large piece and one small piece. Let the length of one piece of stick be$x$, then the other piece will be$1-x$. When we add up, we get length$1$. So far so good. Note that one piece will always be less than$\dfrac12$and the ... 1 Let's let$x$denote the length of the smaller piece. Then$x$is uniform on$[0, 1/2]$, as the split-point$s$is uniform on$[0, 1]$, but for split points past$1/2$, the "smaller piece"$x$becomes$1-s$instead of$s$. The formula you've got is great when the probability space is discrete; in this case, it's continuous, and we use $$E(x) = \int x \... 1 Since H\in \mathcal H\subset \mathcal G,$$\boldsymbol 1_H\mathbb E[X\mid \mathcal G]=\mathbb E[\boldsymbol 1_HX\mid \mathcal G],\quad \quad \text{and}\quad \boldsymbol 1_H\mathbb E[X\mid \mathcal H]=\mathbb E[\boldsymbol 1_HX\mid \mathcal H].$$Therefore$$\mathbb E\big[\mathbb E[\boldsymbol 1_HX\mid \mathcal H]\big]=\mathbb E[\boldsymbol 1_HX]=\mathbb E\... 1 By definition of conditional expectation (https://en.wikipedia.org/wiki/Conditional_expectation):$\int_{H}\mathbb E[X|\mathcal{H}]dP=\int_{H}\mathbb X dP\int_{H}\mathbb E[X|\mathcal{G}]dP=\int_{H}\mathbb X dP$(note that H is both$\mathcal{G}$and$\mathcal{H}$measurable) so the two expressions are both equal to the integral of the original function.... 1 The comprehensive statement is: Let$G$be a finite group and$\nu \in M_p(G)$with support$\Sigma$. The convolution powers$\nu^{\star k}$converge to the uniform distribution if and only if$\Sigma \not\subset K$for any proper subgroup$K$of$G$, and$\Sigma\not\subset Hg$for any coset of any proper normal subgroup$H \rhd G$. ... 1 The result is correct. Let$S = \{1, 2, 3, ...\}$be countably infinite. 1) First prove that $$\sum_{j \in S} \pi_j \leq 1$$ 2) Next prove that$\pi \geq \pi P$(via the hint in my comment above), equivalently: $$\pi_j \geq \sum_{k \in S} \pi_k P_{kj} \quad \forall j \in S \quad (Eq. 1)$$ 3) Suppose (Eq. 1) holds with strict inequality for at least ... 0 Let$X_n$be any Markov chain. Let the reward for stopping after$n$steps be$1 - 1/n$. 3 You don't even need the condition$\int |\xi| dP <\infty$. For any$n$the interval$[-n,n]$can have at most$n$points$a_1,a_2,..,a_n$with$P(\xi=a_i) >\frac 1n$for each$i$. [ Because if we had$n+1$such points$a_1,a_2,..,a_{n+1}$the$ P(\xi \in \{a_1,a_2,..,a_n\}) \geq \frac {n+1} n>1$which is a contradiction. Now take the union of all ... 0 Consider two independent rolls of a fair six-sided die, and the following events:$A$= {First roll is 1, 2 or 3},$B$= {First roll is 3, 4, or 5} ,$C$= {The sum of the two rolls is 9}. Then,$P(A)=\frac{1}{2}$,$P(B)=\frac{1}{2}$and$P(C)=\frac{4}{36}$. It can easily be seen that$P(A \cap B \cap C)=\frac{1}{36} = P(A)P(B)P(C)$However,$P(A \cap ...

1

$X=n$ iff $n \leq \frac {\ln\, U} {\ln(1-p)} <n+1$ iff $(n+1) \ln(1-p) \leq \ln\, U <n\ln(1-p)$ iff $(1-p)^{n+1} \leq U < (1-p)^{n}$ so $P(X=n)=(1-p)^{n} -(1-p)^{n+1} =(1-p)^{n}(1-(1-p))=p(1-p)^{n}$

1

For some positive integer $r$, we have that $P(X>r)$ $=P(\lceil\frac{lnU}{ln(1-p)}\rceil>r)$ $=P(\frac{lnU}{ln(1-p)}>r)$ $=P(log_{1-p}^{}U>r)$ $=P(U<(1-p)^{r})=(1-p)^{r}$ so $X$ is the geometric distribution.

4

Yes, they are different. Suppose $p$ is irrational. Then for every fixed $n$, the even $A_n=p$ would never happen (because $A_n(\omega)$ must be one of $0,\frac1n,\frac2n,\dots,\frac{n-1}n,1$ for each $\omega$). Similarly if $p\in\mathbb{Q}\cap(0,1)$ with least denominator $m$, then $A_n=p$ would never happen for $n$ which are not a multiple of $m$. So $\... 0 Comment: @JMoravitz has given you excellent guidance. Here is a simulation in R statistical software that gives some answers (accurate to 2 or 3 places), along with some formulas you may have seen in your course (or may see soon). set.seed(611) x = rbinom(10^6, 1, .5); y = rbinom(10^6, 1, .5) z = pmax(x,y) mean(x); mean(z); mean(x*z) [1] 0.500266 # aprx ... 1 Note that$E[X|A]$is a constant and hence $$E[E[X|A]1_{A}|B]=E[X|A]\cdot E[1_{A}|B]=E[X|A]P(A|B),$$ where the first equality follows from the linearity of conditional expectation with respect to an event. You do not need to use independence here. 0 You are not quite correct, but not far away. Apart from your typographical confusion between$1$and$l$, you have some inequalities the wrong way round and some major issues in the density function for$Z$: it cannot be less than$l/2$since it is the longer of the two pieces, and in fact it has a uniform distribution between$l/2$and$l$. I think you ... 0 Hint: Establish the identity $$(X-Y)I(X> Y)=\int I(Y<x\le X)\,dx,\tag1$$ where$I(A)$is the indicator of event$A$: it takes value$1$when$A$is true,$0$otherwise. That is, you have to prove that the random variable on the LHS of (1) equals the random variable on the RHS. Next, establish a similar identity for the case$X<Y$. (The case$X=Y\$ ...

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