New answers tagged

0

The problem is not to construct an infinite product of sigma-algebras. This always exixsts and is defined as you have stated, as the smallest sigma-algebra that makes all the projections measurable. The problem is to find measures on this space. Of course a measure on the infinite product always pushes foreward to a measure on the finite products contained ...


1

This is because both $X_n$ and $Y$ only take values from a set of two points $\{3, 8\}$, which means given any positive $\epsilon \in (0, 5]$, the set \begin{align} \{\omega: |X_n(\omega) - Y(\omega)| \geq \epsilon\} & = \{\omega: X_n(\omega) = 8, Y(\omega) = 3\} \cup \{\omega: X_n(\omega) = 3, Y(\omega) = 8\} \\ & = \{\omega: X_n(\omega) \neq ...


2

Method 1: Using the Law of Total Probability. Let $S$ be the event of selecting a child with a sister. Let $F_{\rm GGG}, F_{\rm GGB}, F_{\rm GBB}, F_{\rm BBB}$ be the event of selecting a child from the relevant family structure; given that we know we are selecting from families with three children. The count of girls in a family has a Binomial ...


0

The facts about the correlation matrix tell us that it has ones on the diagonal and the specified correlation on the non-diagonals. You can generate 3 i.i.d standard normal variables ($X,Y,Z$). Define 3 new R.Vs ($U,V,W$) as $U= aX + bY + bZ$ $V= bX + aY + bZ$ $W=bX+bY+aZ$ where $(a,b)$ solves $a^2 +2b^2 = 1$ and $2ab + b^2 = p$


1

Obviously we need $\rho \le 1$. Take a random vector $\mathbf{X}\sim \mathcal{N}(0,I_{n\times n})$ and a real RV $Z\sim \mathcal{N}(0,1), Z \perp \mathbf{X}.$ Then it is easy to check that $\sqrt{\rho}\cdot Z \cdot \mathbf{1}_{n\times 1}+ \sqrt{1-\rho}\cdot \mathbf{X}$ has the distribution you want.


2

Both methods are fine. You didn't really specify the problem, but I gather you meant to imply that a child is selected from the family uniformly randomly, and that the genders of the three children are independent. In that case, both your arguments are correct.


0

The methodology for such a process is as follows: 1) Generate i.i.d standard normals (X,Y,Z). You can think of this as a vector $z$ 2) Do a Cholesky Decomposition on your specified correlation matrix $R$. 3) Multiple your vector $z$ by your newly decomposed vector. This will generate 3 correlated standard normals. 4) These correlated standard normals can ...


2

Note that the graph is distributed as $G(n, p')$ for some $p'$. Why? Clearly the appearance of an edge in the final graph is independent of the choices for the other edges. Fix an edge $e$ and so $p' = P(e \text{ appears in final graph})$. Now, note that $$ \{e \text{ appears in final graph}\} = \{e \in \text{intermediate graph}\} \cup \{e \notin ...


1

With the well deserved criticism of the problem in my Comments, let's see how one would solve it if we ignore some of the unnecessary clauses in the problem and we make it work by changing one of the inputs. Let's say only 60 customers buy cake at Cafo (to make the numbers work with the other conditions), and the problem doesn't say stupid things like "each ...


0

Your $\sigma / \sqrt n$ should be $\sigma \sqrt n$ for the standard deviation of the sum $S_n$ (not the standard deviation of the mean) If you want the probability that $S_n$ is positive to exceed $0.99$ then you need the probability it is non-positive to be less than $0.01$ so using the Central Limit Theorem to give an approximation, you want to ...


0

As @carmichael561 remarks, independence between random variable $X$ and sigma-algebra $\cal G$ amounts to the assertion $$ P(X\in B\mid A)=P(X\in B)\tag1$$ for every Borel $B$ and every $A\in\cal G$. In particular it implies $$ E(X\mid A)=E(X).\tag2 $$ for every $A\in\cal G$. Regarding your question whether (2) can serve as the definition, the answer is that ...


0

Apologies for writing $H(s)$ instead of $H(\beta)$; it's $s$ in my notes, if I changed it to $\beta$ I'd miss one. First Question Yes, $s^\rho H(s)\to0$ implies $t^{1-\rho}h(t)\to0$. Suppose not; there exist $t_n\to\infty$ with wlog $$t_n^{1-\rho}h(t_n)\ge1.$$Let $s_n=1/t_n$. Now, $$h(t)\ge t_n^{\rho-1} \quad(t>t_n),$$so $$H(s_n)\ge ...


0

A random variable $X$ is independent of a $\sigma$-algebra $\mathcal{G}$ if the $\sigma$-algebras $\sigma(X)$ and $\mathcal{G}$ are independent. Recall that $\sigma(X)$ consists of all sets of the form $\{X\in B\}$, where $B$ is a Borel subset of $\mathbb{R}$. If $X$ has finite expectation and is independent of $\mathcal{G}$, then in fact $$ ...


0

First question: You are basically right, but let's lay some notation. I assume your graph is finite (finite edges and finite vertices). Let a path be a set $\{e_i\}$ for $i=1,2,...,k$ where each $e_i$ is an edge and the $e_i$ lands where $e_{i+1}$ starts, for all $i=1,2,...,k-1$. Now, given this path, call the probability of following it ...


0

I and II give precise asymptotics for $H$ and $h$. However, the statement $I^{\prime}$ below is imprecise. With some additional assumptions, you may be able to make some statement with inequalities, but I doubt it will be as nice as the original equivalence. I$^{\prime}$: $\beta^{\rho}H(\beta)\rightarrow 0$ as $\beta\rightarrow 0$ Note that $H(\beta)\sim ...


3

Your axiom list states that Additivity: $P(E_1 \cup E_2)=P(E_1)+P(E_2)$, where $E_1$ and $E_2$ are mutually exclusive. Note that $A \wedge \neg A$ is mutually exlusive with itself. Hence $$P(A \wedge\neg A)=P\left((A \wedge \neg A)\cup(A \wedge \neg A)\right) = P(A \wedge\neg A) + P(A \wedge\neg A) $$


3

As booleans, boolean-valued functions, boolean-valued random variables, or any other similar sort of thing, $A \wedge \neg A$ is the same thing as "false", and $A \vee \neg A$ is the same thing as "true". Applying this, your question reduces to Why is $P(\text{true}) = 1$ and $P(\text{false}) = 0$? Answering this depends on the exact details you want ...


1

What about $X_n = n$ a.s., and $S\colon \mathbb{R}\to\mathbb{R}_+$ defined by $$S(x) = \frac{x^2}{x^4+1}$$? You will get $S(X_n) \to 0$ a.s., but clearly not what you want for $(X_n)_n$. Yet $S\geq 0$ is continuous and cancels only at $0$. It looks like a sufficient condition for what you want is that $S$ be continuous and strictly monotone with ...


0

We have been given: $$f_{X,Y}(x,y) = y^{-1}e^{-y} ~[0\leq x\leq y]$$ So then we know: $$\begin{align}f_Y(y) ~ = ~& \int_0^y y^{-1}e^{-y}~\operatorname d x~~[0\leq y]\\[1ex] =~& e^{-y}~[ 0\leq y]\end{align}$$ Then, by change of variables (chain rule, Jaccobian, etc.): $$\begin{align}f_{XY\mid Y}(z\mid y) ~=~ & \lvert \dfrac{\mathrm d ...


0

A logical contradiction can never be true.   Something that surely will never happen (an impossibility) has probability measure of zero. A tautology will always be true.   Something that surely will happen (a certainty) has a probability measure of one. That is all.


47

We see that in probability, we represent the event $A$ as a set of elements in our sample space, and $\neg A$ as the complement of $A$ in our sample space. Thus, in probabilistic terms, $$P(A \wedge \neg A) = P\left(A \cap \overline{A}\right) = P(\emptyset)$$ by the definition of the complement. And by the Kolmogorov Axioms, we see that $$P(\emptyset) = ...


2

Check out George Lowther's blog: https://almostsure.wordpress.com/2010/01/03/the-stochastic-integral/ Especially Lemma 3, answers your question "why cadlag?"


2

The domain of $f(X)$, or actually $f\circ X$ is $\Omega$. It must be looked at it as the function $\Omega\to\mathbb R$ prescribed by $$\omega\mapsto f(X(\omega))$$ Defined like that also $f(X)$ is a random variable if $f:\mathbb R\to\mathbb R$ is a Borel-measurable function.


1

You were mostly okay, except for the cases where you have three ones or three fives.   (Also note the multinomial coefficient in the other cases .) $$\begin{equation*} \begin{split} &\sum_{i\in\{2,3,4,6\}} \tfrac{5!}{2! \cdot 2! \cdot 1!}~ (\tfrac{2}{10})^2~(\tfrac{2}{15})^2 ~\mathsf P(X{=}i)^1 ~+~\tfrac{5!}{3!\cdot ...


0

I can't tell you what you did wrong, because I'm not sure what you did right. In any case, split it into disjoint events, and then add up their probabilities. The probability of exactly $2$ ones and $2$ fives: ...


3

Your intuition is pointing you in the right direction. Let $\mu_Y$ be the distribution of $Y$. Then $$ \infty>E|X+Y|=\int_{\Bbb R}E|X+y|\,\mu_Y(dy), $$ which means that $E|X+y|<\infty$ for $\mu_Y$-a.e. $y\in\Bbb R$. You should be able to finish it from there.


2

Consider $([0,1],\mathscr F ($borel sets on [0,1]$), m($lebesgue measure$))$ Let $X_n(x)=n1_{[0,\frac{1}{n}]}(x)$. This R.V. converges to 0 a.s. (it only does not converge to 0 for x=0) but $E(X_n(x)) = 1$ for every $n$ so $E(X_n) \not \rightarrow E(X) =0 $ Let $X_1=1_{[0,\frac{1}{2}]}$, $X_2=1_{[\frac{1}{2},1]}$, $X_3=1_{[0,\frac{1}{3}]}$, ...


0

For the case $|cx|<2$, the result follows by showing that the integral is non-negative, as per stocasticboy's comment. To see this, by sketching the graph, it should be obvious that for $|t|<2$, the function $\tfrac 1 t\sin t$ is non-negative (see sketch on Wolfram Alpha).


1

Note that this might be what @Did suggest in the comments. However, since I haven't dealt with conditional probability/expectation yet I will rewrite it in my own words. Note that for all $a>0$ we have $a \mathbb{1}_{|X| \geq a} \leq |X|$. Applying this to the problem at hand we easily see that \begin{align}\mathbb{E} (|S_n|) \geq \mathbb{E}( \sqrt{n} ...


4

The second player has 8 choices for his first move, 6, for the second, etc., so in any game he has a total of $8 \times 6 \times 4 \times 2 = 384$ possible sequences of moves. If only one choice is correct at each point, which is an underestimation, then in 5 million games he would be expected to draw at least $\lambda = 5 \times 10^6 / 384 \approx 1.3 ...


2

Consider $X = \frac{1}{2}$ and $X_{\max} > \frac{1}{2}$. Then $a \in (0,1]$ and so \begin{align*} aX \le X =\frac{1}{2} \implies 1 - aX \ge \frac{1}{2} >0. \end{align*} Thus the random variable $\frac{X}{1-aX}$ is strictly positive, and so is it expectation. For a more general answer, if $X >0$, then we have \begin{align*} \begin{cases}1 -aX \ge 1 ...


1

Since $W\leq Z$, if $W>\varepsilon$ then $Z>\varepsilon$. In other words $$ \{W>\varepsilon\}\subset \{Z>\varepsilon\}$$ hence $\mathbb{P}(W>\varepsilon)\leq \mathbb{P}(Z>\varepsilon)$.


2

Let $X_n=B(n,p)$ be a binomially distributed random variable. Also notice that $X_n=Y_1+Y_2+\cdots+ Y_n$ where $Y_i$ are i.i.d. Bernoulli with parameter $p$. Now observe that \begin{align} \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}&= \operatorname{E}(X_n)\\ &= \operatorname{E}( Y_1+Y_2+\cdots Y_n)\\ &=\operatorname{E}( ...


5

In discrete time at least, the definitions I'm familiar with are fairly straightforward. Given an increasing filtration $\{\mathcal{F}_n\}_{n=0}^{\infty}$, a process $\{X_n\}_{n=0}^{\infty}$ is adapted if each $X_n$ is $\mathcal{F}_n$-measurable. For predictable processes, the random variables are measurable with respect to slightly smaller ...


12

We have this sum: $$ \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} \tag 1 $$ First notice that when $k=0$, the term $k\dfrac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$ is $0$, and next notice that when $k\ne0$ then $$ \frac k {k!} = \frac 1 {(k-1)!} $$ so that $$ k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} = \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}. $$ The two expressions ...


7

The trick is the using the identity $k { n \choose k} = n {n-1 \choose k-1}$. $$\begin{align*} &\sum_{k=1}^n k { n \choose k } p^k (1-p)^{n-k}\\ &= \sum_{k=1}^n n { n-1 \choose k-1} p^k (1-p)^{n-k}\\ &=np \sum_{k=1}^n {n-1 \choose k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\ &=np (p+(1-p))^n\\ &=np \end{align*}$$


2

Regarding the indicator functions, just use their definition, i.e. they are equal to 1 on the event they indicate, and $0$ on its complement. Then $$ 1_{\{S\ge K_1\}}1_{\{S\ge K_2\}}=1_{\{S\ge K_1\}\cap{\{S\ge K_2\}}}=1_{\{S\ge \max\{K_1,K_2\}\}}. $$


0

Here’s another solution, basically the same as already given, but perhaps easier to follow, and at least more colorful. The ranges over which $\color{blue}X$, $\color{brown}Y$, and $\color{green}Z$ are uniformly distributed are shown above. Partition what can happen into the following three disjoint cases. Case 1 ($p=\frac{1}{2}$): ...


1

First observation: The random variables $X$, $Y$ and $Z$ are independent and uniform on $[2.9,3.1]$, $[2.7,3.1]$ and $[2.9,3.3]$ respectively hence $$X=2.9+0.2U,\qquad Y=2.9+0.2R,\qquad Z=2.9+0.2S,$$ where $U$, $R$ and $S$ are independent and uniform on $(0,1)$, $(-1,1)$ and $(0,2)$ respectively. Consequence: $E(\max(X,Y,Z))=2.9+0.2\cdot E(\max(U,R,S))$. ...


0

Here is what I got for a compact metric space: We must show that any sequence $\mu_n$ in $\mathscr{M}(X)$ has a $\omega^*$-convergent subsequence (this shows that $\mathscr{M}(X)$ is actually sequentially compact). Let $\{f_i\}_{i=1}^{\infty}$ be a countable dense subset of functions in $C(X)$. For any sequence $\mu_n$ in $\mathscr{M}(X)$ we have that ...


1

As Wikipedia points out, there are several results called Riesz (representation) theorem. A common feature of some of them is that on some spaces, every linear functional is integration against something. The result that the author uses here is the representation of linear functionals on $L^q$: they are integrals against $L^p$ functions. More precisely, ...


1

I'll assume that you intended to imply that each coupon independently has probability $\frac1p$ to be in any given black box. Then we have $n$ independent Bernoulli trials with probability $\frac1p$ and are looking for the expected time for all of them to have succeeded at least once. The probability for a given coupon not to have been collected after $k$ ...


1

I think there are some issues with the independence assumptions. Independence of $A$ from $\cal{H}$ From "$\{X=i\}$ independent from $\cal{H}$" and "$A\subset \{X=i^*\}$" you cannot conclude that $A$ is independent from $\cal{H}$. Counterexample: Let $\Omega=\{1,2,3,4\}$, $\cal{F}=2^\Omega$ and $P(\{i\})=1/4$. Let $X$ be the random variable with ...


0

$| B_{\tau \wedge t} | \leq \max(\alpha, \beta)$ Stopping times may be seen as hitting times of your process $(B_t)_{t \geq 0}$ into a set $A=\{\alpha,-\beta\}$. That is, our process stops whenever $B_t=\alpha$ or $B_t=-\beta$. Think about a particle moving up or down, on a plot where $B_t$ is on the y-axis and $t$ is on the x-axis. Also, wlg suppose ...


1

First, $\phi(x)=x$ is not a bounded function in general (for example $X=\mathbb{R}$). But even if it would be a bounded function, then the statement would not imply $$\int_X x \, d \mu(dx) = \int_X x \, d \nu(dx) \Rightarrow \mu = \nu .$$ The statement says that if $$ \int_X \phi(x) \, d\mu(x) = \int_X \phi(x) \, d\nu (x) $$ holds for all bounded, ...


3

Hint: You can write $X_{n}=Y_{1}+\cdots+Y_{n}$ where the $Y_{i}$ are iid with $P\left\{ Y_{1}=1\right\} =p$ and $P\left\{ Y_{1}=-1\right\} =1-p$. This makes it easier to find $\mu$ (linearity of expectation) and allows application of the strong law of large numbers.


0

Let's talk about the expected fraction of boys in the population, and let's just consider one family. The fraction of boys is $F = \frac{B}{B + G}$. If we follow your rule, this family will stop having children precisely when they get a boy, so $B = 1$, implying $F = 1/(1 + G)$. Now, let's note that $$P(G = 0) = \frac{1}{2},\quad P(G=1) = ...


1

Let the expected number of girls in any given family is $g$. If a boy is born first (probability =$\frac{1}{2}$) then $g=0$, but if a girl is born (probability =$\frac{1}{2}$) then $g=1 + g$ ($1$ for the girl already born to the couple and $g$ for the fact that state is reset to original state where couple have to keep breeding until a boy is born) ...


2

Here is the current edition of the question: If $X$ is a symmetric $n$-dimensional random vector with mean $0$ then is it true that: \begin{align*} & X \text{ follows a multivariate normal law} \\ & \text{iff} \\ & \|X\| \text{is a chi random variable with $n$ degrees of freedom?} \end{align*} Let $X=(X_1,\ldots,X_n)$. As phrased above, ...


6

$$\sum_{k,\ell}\xi_k\bar{\xi_{\ell}}\varphi_X(t_k-t_{\ell})=E\left(\sum_{k,\ell}\xi_k\bar{\xi_{\ell}}e^{i(t_k-t_{\ell})X}\right)=E\left(\left|\sum_k\xi_ke^{it_kX}\right|^2\right)\geqslant0$$



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