Tag Info

New answers tagged

1

A martingale is characterized by its characteristic function. apply the property with $f(x) = x$ proves that $W$ is a martingale fix $u$ and apply with $f(x) = \exp(iux)$; define $g(u, t) = \mathrm E(\exp(iuW_t)); M(t) = f(W_t) - \frac 12\int_0^t f''(W_s) ds$ and you get $$ 1 = M(0) = \mathrm EM(t) = g(u,t) + \frac 12 u^2\int_o^t g(u, s) ds $$ whose ...


0

Back for a second try. Running a couple of simulations suggests that the normalized sum $S_n/s_n$ indeed converges, but strangely the histogram of $S_n/s_n$ seems more rectangular than bell-shaped. To understand what the limiting distribution is, calculate the characteristic function of the partial sum $S_n=X_1+\cdots+X_n$: $$ \begin{align} E\exp(itS_n) ...


0

If $P(A)>0$ then $P(B|A)=\frac{P(A \cap B)}{P(A)}$, as you saw in elementary probability. So if $X=\chi_C$ then $$\int_\Omega X(\omega) dP(\omega|A) = \frac{P(C \cap A)}{P(A)} = \frac{1}{P(A)} \int_A X dP.$$ Now extend to simple functions and finally random variables (as usual).


1

If you want to show the CLT holds, you'll need a criterion other than the Lindeberg condition. Note that $s_n^2< 4^{n+1}$. Use this inequality to bound the Lindeberg expression away from zero: $$\sum\limits_{j=1}^{n}E[X_{j}^2\chi_{\{|X_{j}|>\epsilon s_{n}\}}]= \sum_{j=1}^n 4^jP(|X_j|>\epsilon s_n)\ge 4^nP(|X_n|>\epsilon s_n) \ge ...


0

Let $$F(x):=x^{13}\cdot(\frac{(-x)^0}{3!}+\frac{(-x)^1}{2!}+\frac{(-x)^2}{1!}+\frac{(-x)^3}{0!})^{13}$$ Then $$4!^{13}\cdot\sum_{k=13}^{52} \frac{1}{k!}[x^k]F(x)=\frac{50972203946555791528902451677555189167087762981}{92024242230271040357108320801872044844750000000000} =0.000553899741\cdots$$ is the required probability.


1

As I said in my comment, the prime ($'$) is being used to denote the complement of the set—in this case, $(A \cup B)'$ means not $(A \cup B)$. To provide some intuition behind thanasissdr's answer: Consider that $P(A) = P(B) = 1/2$. In some sense, both $A$ and $B$ cover half of the probability space. But we are told $P((A \cup B)') \equiv P(\text{not }(A ...


1

Some basic identities $P(A\cup B)= P(A) + P(B) - P(A\cap B)$ $P(A-B) = P(A\cap B') = P(A) - P(A\cap B)$ $P(A')= 1-P(A)$ $P(A\cup B)= 1-P\big((A\cup B)'\big) = 1- 0.2 = 0.8$ Then, $P(A\cap B) = P(A) + P(B) - P(A\cup B) = 0.2 $ Thus, $P(A\cap B')= P(A) - P(A\cap B) = 0.5-0.2 =0.3$


2

Let $A$ be the event that $\int_0^\infty e^{B_t} \,dt < +\infty$. By the Kolomorgov 0-1 law, $P(A) = 0$ or $1$. Now let $B$ be the event that $\int_0^\infty e^{-B_t}\,dt < +\infty$. By symmetry, $P(A) = P(B)$. Moreover, on $A \cap B$ we have $$\int_0^\infty (e^{B_t} + e^{-B_t}) \,dt < +\infty$$ which is absurd since $e^x + e^{-x} \ge 1$ for all ...


1

If $X_i \sim \operatorname{Exponential}(\lambda)$ are iid such that $$f_{X_i}(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ then $T = \sum_{i=1}^n X_i \sim \operatorname{Gamma}(n,\lambda)$ with $$f_T(x) = \frac{\lambda^n x^{n-1} e^{-\lambda x}}{\Gamma(n)}, \quad x > 0.$$ Then it is easy to see that $$\operatorname{E}[n/T] = \int_{x=0}^\infty \frac{n}{x} ...


1

It may help to use the fact that $$\mathbb{E}\left[X\right] = \int_0^{\infty} \mathbb{P}(X \geq x) dx$$ for positive continuous variables so that $$\mathbb{E}\left[\frac{n}{X_1+\ldots+X_n}\right] = \int_0^{\infty} \mathbb{P}\left(\frac{n}{X_1+\ldots+X_n} \geq x\right) dx = \int_0^{\infty} \mathbb{P}\left(X_1+\ldots+X_n \leq \frac{n}{x}\right) dx$$ Now use ...


1

(Not a proof, but here's some intuition at least) Standard Brownian motion has mean 0 and variance t for $0 \le t \lt \infty$ Thus, on average the integral becomes... $$\int_0^{\infty} e^0 \ ds=\int_0^{\infty} 1 \ ds$$ Which clearly diverges. What you should aim for proving is that Brownian motion crosses between negative and positive values in such a ...


0

The number of partitions of $17$ into up to three parts is $33$ (see for example A001399), so the number of partitions of $20$ into exactly three positive parts is $33$, so the number of partitions of $23$ into exactly three distinct positive parts is $33$, so the number of partitions of $20$ into exactly three distinct non-negative positive parts is ...


2

You have three distinct boxes, and want to count ways to put 20 indistinct balls into them. The total count of solutions is, as you have calculated: $$\frac{22!}{2!20!} = 231$$ Now, to generate forbidden solutions you can choose 2 boxes, then put the same number of balls $n$ into each of them, and the remainder into the other. Since the number $n$ can vary ...


0

No, the joint probability distribution does not have to be stationary w.r.t. time. Just imagine a precessing top with no friction. The probability distribution for $l^2$ is just a delta peaked at some particular value. But the joint probability distribution of $(l_1,l_2,l_3)$ is a delta peak that moves around in the angular momentum space.


1

In calculating $F_Y$, you did wrong. Note that $X^2$ and $X$ are not independent. But $X^2 \le X$ iff $X \le 1$. We have for $t \in [0,1]$: \begin{align*} \def\P{\mathbf P}\P(Y \le t) &= \P(X \le t)\\ &= \frac 12 t\\ \end{align*} and for $t \in [1,4]$: \begin{align*} \P(Y \le t) &= \P(Y \le 1) + \P(1 < Y \le t)\\ ...


1

Take $k=2$, $t_1=1$ and $t_2=2$. If $(Y_n(1),Y_n(2))$ converged in distribution to $(X,X)$, then in particular $X_n-X_{2n}$ would converge to $0$ in distribution. But it is not always the case. For example, if $X$ is a symmetric non-degenerated random variable and $X_n=X$ if $n$ is even, $-X$ if $n$ is odd, then taking $n$ odd we would have $2X=0$ in ...


2

Finite additivity and countable sub-additivity is equivalent to countable additivity. The proof is below. Let $\mu$ be a measure. It is clear that if $\mu$ is countably additive then it is finitely additive and countably sub-additive. Assume that $\mu$ is countably sub-additive and finitely additive. Consider a collection $\{A_n \}_{n=1}^\infty$ of ...


1

No, not quite that.   Conditional Mutual Information is: $$\begin{align} I(X;Y\mid Z) & =D_{KL} (p(X,Y,Z)\parallel p(X\mid Z)\;p(Y\mid Z)\;p(Z)) \\[2ex] & = \sum_{z\in Z} p_{_Z}(z)\; D_{KL}(p(X,Y\mid Z=z)\parallel p(X\mid Z=z)\;p(Y\mid Z=z)) \\[3ex] & = \sum_{x\in X}\sum_{y\in Y}\sum_{z\in Z} p_{_Z}(z)p_{_{X,Y\mid Z}}(x,y\mid ...


1

Mutual information is often written $I(X;Y)=D_{KL}(p(X,Y)||p(X)p(Y))$. Your instructor has provided a slightly generalised version which depends on some other variable, $Z$. You are free to take expectations over this. Note that if $Z$ is constant, then that leads to the usual definition.


3

It can be shown that nonnegative random variables $X$ and $Y$ have the same distribution so long as $\mathbb{E}[X^\alpha]=\mathbb{E}[Y^\alpha]$ is finite for all $\alpha\in(a,b]$, any $0\le a <b$. Setting $U=\log X$ and $V=\log Y$, define the functions $$ f(\alpha)=\mathbb{E}[1_{\{X > 0\}}e^{\alpha U}],\ g(\alpha)=\mathbb{E}[1_{\{Y > 0\}}e^{\alpha ...


1

Let the RV $X$ take value $n^2$ with probability $\frac{c}{n^2}$ where $c$ is a norming constant. Then $$E \left( \frac{X}{n^2+X} \right) \geq \sum_{k=n}^{\infty} \frac{c}{k^2} \frac{k^2}{n^2+k^2} \geq \sum_{k=n}^{\infty} \frac{c}{k^2} \frac{n^2}{n^2+n^2}= \frac{1}{2}\sum_{k=n}^{\infty} \frac{c}{k^2} \sim \frac{c}{2n} $$ which is not summable. Note that ...


2

Three-dimensional case (n=3) The Wikipedia article on area and volume element states that the area element of the unit sphere is $\sin\phi\,\mathrm d\phi\,\mathrm d\theta$ where I use $\phi$ to denote inclination and $\theta$ to denote azimuth. If you consider your fixed vector as the zenith direction, then my $ \phi$ will be the same as yours. To turn ...


1

Hints: Let $\arg\min_{{b_0,b_{-0}}}E\left[(X_{n+1}-b_0-b_{-0}'X)^2\right]=(\beta_0,\beta_{-0}')'=\beta\in\mathbb{R}^{n+1}, \quad X=(X_1,\dots,X_n)'.$ $\beta_0=E[X_{n+1}]-\beta_{-0}'E[X], \quad \beta_{-0}=Var(X)^{-1}Cov(X,X_{n+1}).$ $Cov(X, X_{n+1}-\beta_0-\beta_{-0}'X)=0.$ Use normality of $(X_1,\dots,X_{n+1})'.$ Show that ...


2

The claim holds only true if $X$ is assumed to be real-valued, i.e. $\mathbb{P}(|X|<\infty)=1$. If this assumption holds true, then the continuity of the measure $\mathbb{P}$ implies $$\mathbb{P}(|X| \geq r) \leq \delta$$ for $r>0$ sufficiently large where $\delta$ is chosen as in the assumption for $\epsilon := 1$. Hence, by assumption, ...


1

It is false. Suppose $X$ is uniformly distributed in the interval $[0,1]$ and $Y=1-X$. Or that $X\sim N(0,1)$ and $Y=-X$. Or that $X\sim\mathrm{Bernoulli}(1/2)$ and $Y=1-X$. Or that $X,Y\sim\text{i.i.d. }\mathrm{Exponential}(1)$. In each of these cases, $X$ has the same distribution, and hence the same expected value as $Y$, and in each of these cases, ...


0

It is not bounded sorry for the inconvenience. Here is the proof a friend gave me (thanks a lot to him): (It can also be found here) Let consider the set $W_1 = \{w = (s,w_1,\dots)| w_1 \leq n\}$. Then $\mathcal{E}^n_n$ is bigger than $\displaystyle\sum_{w\in W_1}O^n_n(w)*P(w)$. \begin{align} \sum_{w\in W_1}O^n_n(w)*P(w) &= \sum_{k=0}^n \sum_{i=0}^k ...


1

Let $P\{Z>0\}=1$ and realize that $\{Z>0\}=\bigcup_{n=1}^{\infty}\{Z\geq\frac1{n}\}$. That implies that $P\{Z\geq\frac1{n}\}$ converges to $P\{Z>0\}=1$ so that $P\{Z\geq\frac1{n}\}>\frac12$ for some $n$. Consequently $\mathbb EZ\geq\frac1{n}\frac12>0$. Apply this on $Z=X-Y$. Edit: Alternatively $1=P\left\{ Z>0\right\} ...


1

Yes. The proof for the general case is a bit messier than the other two, because as you know we usually only retain nonstrict inequalities under a limit process. The idea is that the set $X>Y$ can be broken up into a union of sets $X>Y+1/n$. If any of these has positive probability, and $X<Y$ has zero probability, then $E[X]>E[Y]$ (why?). The way ...


0

I would say the most interesting thing is that your metric is equal to the $L_1$ Kantorovich/Wasserstein/transport distance. And there is a lot of research related to this thing (although most of the research is for $L_2$ versions). Wiki1, Wiki2 Using right version of the distance you can prove path-connectedness of the space of prob. measures with weak ...


1

\begin{align} E[B_t^3|\mathcal{F}_s]&=E[(B_t-B_s+B_s)^3|\mathcal{F}_s]\\ &=E[(B_t-B_s)^3+3(B_t-B_s)^2B_s+3(B_t-B_s)B_s^2+B_s^3|\mathcal{F}_s]\\ &=E[3(B_t-B_s)^2B_s|\mathcal{F}_s]+B_s^3\neq B_s^3\\ E[3tB_t|\mathcal{F}_s]&=E[3t(B_t-B_s+B_s)|\mathcal{F}_s]\\ &=E[3t(B_t-B_s)|\mathcal{F}_s]+E[3tB_s|\mathcal{F}_s]\\ &=3tB_s \neq 3sB_s ...


0

Let $\Omega_X$ be the set of full measure on which $X_n \rightarrow X$. Define $\Omega_Y$ likewise. Note that $\Omega_0:=\Omega_X \cap \Omega_Y$ has full measure as well. On $\Omega_0$, $X_n \rightarrow X$ AND $X_n \rightarrow Y$. Hence $X=Y$ a.s due to uniqueness of limit.


1

Suppose that $X_n\to X$ almost surely as $n\to\infty$ and $X_n\to Y$ almost surely as $n\to\infty$. Then there exists $\Omega'\subset\Omega$ such that $\Pr(\Omega')=1$ and for each $\omega\in\Omega'$ $$ |X_n(\omega)-X(\omega)|\to0 $$ as $n\to\infty$. Similarly, there exists $\Omega''\subset\Omega$ such that $\Pr(\Omega'')=1$ and for each $\omega\in\Omega''$ ...


1

first you can get $b$ simply be computing the time zero expectation: $$\mathbb{E}(e^{5B_t})= e^{0.5 \times 25 \times t}.$$ So $b = 12.5.$ With this value, your final equation $$\exp\{\frac{25(t-s)}{2}+5B_s-bt\} = \exp\{5B_s-bs\}$$ holds and we are done.


2

I believe that $\{\tau\leq 1\}$ is a shorthand for \begin{align*} \{\omega\in\Omega\,|\,\tau(\omega)\leq 1\}. \end{align*} To see that this is consistent with the statement that $\{\tau\leq 1\}=\{1\}$, note that \begin{align*} \tau(1)=&\,\inf\{t\geq 0\,|\,\max\{t-1,0\}>0\}=\inf\{(1,\infty)\}=1,\\ \tau(2)=&\,\inf\{t\geq ...


3

Hint: Condition on the result of the coin-flip.


0

The formula for a 95% confidence interval with a lower bound will be $p - z^*\sqrt{p(1-p)/n},$ where $p = .45,$ $n = 1060,$ and $z^* = 1.645$ cuts 95% from the upper tail of a standard normal distribution. This gives $0.5148181 \approx 51.5\%.$ The interpretation is we have 95% confidence that more than 51.5% of individuals in the population (from which ...


0

If $E(X)$ is finite, the inequality $|e^{ihx}-1|\le |hx|$ gets you uniform continuity right away: $$|\varphi(t+h)-\varphi(t)|\le\int|hx|dF_X(x)=|h|E(|X|)\;.$$ If $X$ is not integrable, you've already found an upper bound that is free of $t$, so it suffices to show that $$\lim_{h\to0}\int|e^{ihx}-1|dF_X(x)=0\;,\tag1$$ since (1) then implies that for every ...


5

Step $I$: Pick $6$ balls and weigh them $3$ on each side. Step $II$: $1$. If the balls balance each other, then weigh the remaining two balls and the heavier one will tilt the balance on its side. $2$. If the balls do not balance each other, then the heavier ball should be one of the three balls on the side the balance tilts. From these three balls, ...


0

heropup -> Don't you think we have $\hat{\theta}\sim InvGamma(n,n\theta^{-1})$ instead of $\hat{\theta}\sim InvGamma(n,n\theta)$ ?


1

The assertion is not true. A counterexample: Pick a random variable $Y$ and define $Y_n:=Y$ and $X_n:=Y+1/n$. Then $X_n$ converges in distribution to $Y$, and obviously $Y_n$ converges in distribution to $Y$. But $X_n$ and $Y_n$ are equal nowhere.


2

Denote by $$p(t,x) := \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{x^2}{2t} \right), \qquad x \in \mathbb{R},$$ the density of the normal distribution with mean $0$ and variance $t$.As you already noted, this function solves the heat kernel equation, i.e. $$\frac{\partial}{\partial t} p(t,x) = \frac{1}{2} \frac{\partial^2}{\partial x^2} p(t,x).$$ For $f ...


1

For concreteness, let $\pi$ be the random permutation such that $X_t = R_{\pi(t)}$. Then we have $$ S = \sum_{t=1}^N a_t X_t = \sum_{t=1}^N a_t R_{\pi(t)} = \sum_{t=1}^N a_{\pi^{-1}(t)} R_t. $$ It remains to show that for each $\pi$, the vector $(a_{\pi^{-1}(1)},\ldots,a_{\pi^{-1}(n)})$ has the same distribution as $(a_1,\ldots,a_n)$. One way to show this is ...


2

I claim that $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;E\big[XE[Y|\mathcal{G}]\big]=E\big[E[X|\mathcal{G}]\cdot E[Y|\mathcal{G}] \big]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(*)$$ If we can prove this then we will be finished, because the right-hand side of equation $(*)$ is symmetric in $X$ and $Y$, and so the ...


0

The probability of exactly $6$ dead people at the end of the month is: $$\binom{10}{6}\times(0.6)^6\times(1-0.6)^{10-6}\approx0.25$$ More generally, the probability of exactly $n$ dead people at the end of the month is: $$\binom{10}{n}\times(0.6)^n\times(1-0.6)^{10-n}$$


1

It follows from the Cauchy-Schwarz inequality that $X \cdot Y \in L^1$. Therefore, (iii) is a direct consequence of the dominated convergence theorem.


1

It is not certain that 6 people will be dead and 4 alive. That is the most likely outcome, but, you could imagine a very unlucky month where each person died. The likelihood for each number of deaths is plotted below.


0

Hint: if you had ten coins each of which came up heads with probability 60% and you flipped them would you always see exactly 6 heads and 4 tails? If the coins were fair (50-50) would you always see just 5 heads and 5 tails?


0

Assume $X$ and $Y$ are independent. Let's redefine $Y$ and $X$ to $\tilde Y$ and $\tilde X$ so that $$ E( \tilde Y^n )=E(Y ^n g(Y))/E(g(Y))\\ E( \tilde X^n )=E(X ^n f(X))/E(f(X))\\ $$ Since $g(\ )/E(g(Y))$ and $f(\ )/E(f(X))$ are densities with respect to the distributions of $Y$ and $X$ respectively, this is possible. Then $$ E((\tilde Y-\tilde ...


0

I've found an answer for my question: Theorem 3.4 of Pertti Mattila's Geometry of sets and measures in Euclidean spaces: Fractals and rectifiability.


0

If $A$ has an inverse and if $Y=A X$ then $X=A^{-1}Y$. So for a (measurable) set $B$ in $R^n$ $$Q_Y(B)=P(Y\in B)=P(X\in A^{-1}B)=Q_X(A^{-1}B).$$ So, the relationship between the distributions is simple. If the pdf's exist then the special formula can be derived based on the dollowing integral: $$Q_Y(B)=\iint\cdots ...



Top 50 recent answers are included