Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

New answers tagged

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 get $f$ 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 $X=Y$ ...


Top 50 recent answers are included