Tag Info

4

Here is a solution that uses SLLN and the additive property of the Poisson distribution. Let $(N_t)_{t \geq 0}$ be a Poisson process of unit rate. Then by the SLLN, together with the inequality $$\frac{N_{[t]}}{[t]+1} \leq \frac{N_t}{t} \leq \frac{N_{[t]+1}}{[t]},$$ it is easy to check that $N_t / t \to 1$ as $t \to \infty$ a.s. Now let T_k = \lambda_1 ... 3 First, rewrite the exponenent as \begin{align*} \exp \left( n\left( a\frac{S_n}{n} - b \right) \right) \end{align*} By the Law of Large Numbers, we havea S_n/n- b \to a \mathbb{E}(\xi_1) - b$a.s.. So$a S_n -bn \to - \infty$a.s. if and ony if$b>a \mathbb{E}(\xi_1) =0, and thus \begin{align*} \exp( a S_n - b n ) \to 0 \text{ iff } b > 0. \end{... 3 Depends on what valid claim numbers are. If the digits are randomly generated then the odds of them putting that one down are one in100$million. If the claim number needs to start in$34-$then it's one in a million. But, I think this is just a couple of nails in a big coffin for your case. Having your company's name on the document is much more ... 2 We can show that for any integer$k$, the sequence$\left(\mathbb E\left[\left(\frac 1{\sqrt n}\sum_{i=1}^nX_i\right)^{2k}\right]\right)_{n\geqslant 1}$is bounded, by expanding the sum. Therefore, we have the uniform integrability of the sequence$\left(f\left(n^{-1/2}\sum_{i=1}^n X_i\right)\right)_{n\geqslant 1}$where$f(x)=\left|x\right|^p$for any$p\...

2

We assume that \begin{align*} d\left(\! \begin{array}{c} S^1(t)\\ S^2(t) \end{array} \!\right) =\textrm{diag}\left(S^1(t), S^2(t)\right)\bigg[\left(\! \begin{array}{c} r\\ r \end{array} \!\right)dt + \left(\! \begin{array}{cc} \sigma_{1,1} &\sigma_{1,2}\\ \sigma_{2,1} &\sigma_{2,2} \end{array} \!\right)d \left(\! \begin{array}{c} W_t^1\\ W_t^2 \end{...

2

Suppose that $p_i$ and $p_j$ are prime. Then $$A_i\cap A_j=\{X=p_i^2p_j^2k\mid k\in\mathbb{N}\}.$$ So, \begin{align*} P(A_i\cap A_j)&=\sum_{k=1}^{\infty}P(X=p_i^2p_j^2k)\\ &=\frac{1}{\zeta(s)p_i^{2s}p_j^{2s}}\sum_{k=1}^{\infty}\frac{1}{k^s}\\ &=\frac{1}{p_i^{2s}p_j^{2s}}. \end{align*} On the other hand, \begin{align*} P(A_i)&=P\{X=... 2 We give a very weak result: sometimes the only functions g are the trivial ones. Let X take on values 0 and 1, and let \Pr(X=1)=p, where p is transcendental. Let g(0,0)=a, g(0,1)=b, g(1,0)=c, g(1,1)=d. Here a,b,c,d only take on values 0 and 1. The corresponding probabilities are (1-p)^2, p(1-p), p(1-p),p^2. There are two ways the ... 2 Yes, we can choose such a K, or just notice that\lim_{K\to +\infty}K^{1-p}A=0$$and that the bound of \mathbb E\left(\left|X\right|;\left|X\right|\gt K\right) is uniform with respect to X\in\mathcal C. The follows from the inclusion$$\left\{\left|X\right|\gt K+k^{-1}\right\}\subset \left\{\left|X-X_n\right|\gt k^{-1}\right\}.$$To see this, it ... 2 In addition to the identical distribution independence is used to obtain that (Z_1,Z_1+Z_2) and (Z_2,Z_1+Z_2) are identically distributed which is needed to conclude \mathbb E(Z_1 \mid Z_1+Z_2) = \mathbb E(Z_2 \mid Z_1+Z_2). Edit. The distribution of (Z_1,Z_1+Z_2) is the image (push forward) of \mathbb P^{(Z_1,Z_2)}= \mathbb P^{Z_1} \otimes \... 2 \newcommand\E{\Bbb E}Say -1\le X\le 1. If p is an odd polynomial then \E[p(X)]=0. Suppose f:[-1,1]\to\Bbb R is continuous and odd. There exist polynomials q_n which converge to f uniformly. Hence p_n\to f uniformly on [-1,1], if p_n(t)=(p(t)-p(-t))/2. But p_n is an odd polynomial, hence \E[f(X)]=0. Now suppose -1\le a< b\le 1 ... 2 \mathcal{C} is UI \Rightarrow (i) Let \epsilon>0. Then, since \mathcal{C} UI there exists K>0 s.t. $$E[|X|] = E[|X|1_{|X|>K}+|X|1_{|X|\le K}]\le \epsilon + K$$ for all X\in\mathcal{C}. Thus (i). \mathcal{C} is UI \Rightarrow (ii) You want to find for a given \epsilon>0 a \delta>0 such ... 2 Suppose W = X. Then E[X\mid W] = X whereas E[Y\mid W] = E[Y]. There is too much freedom in the choice of W. You have to control for some property to arrive at a useful conclusion. Otherwise, any conclusion is possible as the example above shows. 2 To be clear, you want to find P_n(t) for a birth-death process with constant birth rate q(i,i+1)=\lambda and constant death rate q(i,i-1)=\mu? This particular birth and death process is exactly the M/M/1 queue. As you can see, under "Transient solution", there is a solution for the probability mass function dependent on time for a particular state. 2 \emptyset \in \cal A because \chi_\emptyset \equiv 0. If A \in \cal A then X \setminus A \in \cal A because \chi_{X \setminus A} = 1 - \chi_A = 1 + (-1) \cdot \chi_A. If A,B \in \cal A then A \cap B \in \cal A because \chi_{A \cap B} = \chi_A \chi_B. DeMorgan's rules and that fact that \cal A is closed under complementation imply that ... 1 By the L^1-convergence, we have$$L^1-\lim_{s \to \infty} \mathbb{E}(X(t+s) \mid \mathcal{F}_t) = \mathbb{E} \left( X(\infty) \mid \mathcal{F}_t \right). \tag{1}$$On the other hand, as (X(t))_{t \geq 0} is a martingale, we have$$\mathbb{E}(X(t+s) \mid \mathcal{F}_t) = X(t)$$for all s \geq 0. Hence, trivially,$$L^1-\lim_{s \to \infty} \mathbb{E}...

1

If $U \sim Unif(0,1),$ notice that $Y = U^2 \sim Beta(.5, 1),$ which concentrates probability near $0$. Because the density function of $Beta(.5, 1)$ is $f_Y(y) = 1/(2\sqrt{y})$ for $0 <y <1,$ the point at which the densities of $U$ and $Y$ cross is $1/4$. Also, $P(U > 1/4) = 3/4$ and $P(Y < 1/4) = 1/2.$ More generally, $P(U > a) = 1 - a$ ...

1

Assuming that $X$ and $Y$ are independent, their joint pdf is $f(x,y)=\frac{1}{150}$ if $0\leq x\leq 15$ and $0\leq y\leq 10$, and $f(x,y)=0$ otherwise. Therefore $$\mathbb{P}(Y<X)=\frac{1}{150}\int_0^{10}\int_0^x\;dydx+\frac{1}{150}\int_{10}^{15}\int_0^{10}\;dydx$$ $$=\frac{1}{150}\int_0^{10}x\;dx+\frac{1}{3}=\frac{x^2}{300}\Big|_0^{10}+\frac{1}{3}=\... 1 This is a comment, not a proof. Feller, vol. II, page 260, section VIII.4, central limit theorem, has this example: Example. (a) Central limit theorem with infinite variances. It is of methodological interest to note that the proof of theorem 1 applies without change to certain distributions without variance, provided appropriate norming constants are ... 1 Yes, \mathcal{A} is a \sigma-algebra. It is clear that \chi_{X} =1 \in \mathcal{F} since it is a constant function, so X\in \mathcal{A} Note that \chi_{A\cap B} = \chi_{A} \chi_{B} so if A,B\in \mathcal{A} it follows that A\cap B\in \mathcal{A}. Then \chi_{A\cup B} = \chi_{A} +\chi_{B} - \chi_{A\cap B} \in \mathcal{F}. Also \chi_{A\... 1 \vec z=\vec x\Vert\vec y~ is the vector formed by concatenating vectors \vec x and \vec y. That is that:$$\vec x=\begin{pmatrix}a_1 \\ a_2\\ \vdots \\ a_n\end{pmatrix}, \vec y=\begin{pmatrix}b_1 \\ b_2\\ \vdots \\ b_m\end{pmatrix}\\ \vec z= \vec x\Vert \vec y = \begin{pmatrix}a_1 \\ a_2\\ \vdots \\ a_n\\b_1 \\\vdots \\ b_m\end{pmatrix}$$Then ... 1 We see that$$P(Y|X) \cdot P(X) = \frac{P(Y,X)}{P(X)}\cdot P(X) = P(Y,X) = P(Z),Where Z is defined as the variable over the cartesian product of the supports of Y and X (as in, a realization z of Z is in S_Y \times S_X). Hence, 1. is the more appropriate answer. 1 These imply E[X]=c and E[Y]=d. However, a constant conditional expectation does not imply independence. Consider any independent random variable (X,Z), and by defining Y=ZX create (X,Y) for which E[ZX|X]=XE[Z|X]=XE[Z]=d.X and Y=ZX are not necessarily independent. 1 Thanks for @Did's help. Now I have got the answer. Because X_t-X_{s+1/n} is independent of {\mathcal{ F_{s+1/n} }} and \mathcal{ F_{s+} }\subset\mathcal{F_{s+1/n}}\  , X_t-X_{s+1/n}\  is independent of {\mathcal{ F_{s+} }} . Now the question is to prove "Assume that \left\{Y_n\right\} is independent of \mathcal{G} and that Y_n\overset{a.s.}{\... 1 But with you attempt, you do not reach the conclusion. Let K>0, then we have $$E[|X|1_{F}] = E[|X|1_F1_{|X|>K}+|X|1_F1_{|X|\le K}]\le E[|X|1_{|X|>K}]+E[K1_F]= E[|X|1_{|X|>K}]+KP(F).$$ 1 We need only show that, for any Borel set A \in \mathbb{R}, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP. \end{align*} We denote by F the common cumulative distribution function of Z_1 and Z_2. Then, from the independence assumption, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP &= E\left(Z_1 \pmb{1}_{Z_1+Z_2 \in A} \... 1 Let (Z_1,Z_2) be uniformly distributed on \{(-1,0),(0,1),(1,-1)\}. Then Z_1 and Z_2 are identically distributed but dependent. The value of Z_1+Z_2 fixes the (different) values of Z_1 and Z_2, so \mathbb E[Z_1\mid Z_1+Z_2]\ne\mathbb E[Z_2\mid Z_1+Z_2]. 1 Both results are actually equivalent. You can prove one from the other. Regarding the first result: Let \mathcal{C} be a class of subsets of \Omega under finite intersections and containing \Omega. Let \mathcal{B} be the smallest class containing \mathcal{C} which is closed under increasing limits and by difference. Then \mathcal{B} = \... 1 The problem is actually already solved by Jyrki's comment, since the only way to satisfy the given conditions is that the 3 voters have the cyclical preferences given in his example, or the opposite preferences. Thus you just need to count the number of times each of the candidates wins under the proposed "solution", given such a preference set. As you ... 1 By the stationarity of the increments, this is equivalent to\min_{s \in [0,t]} \mathbb{P}(|X_s| \leq \epsilon)>0$$for all t>0. Without any additional assumptions this does, in general, not hold true. Consider for example X_t := t (i.e. a deterministic drift process), then$$\mathbb{P}(|X_t| \leq \epsilon) = 0 for all $t>\epsilon$. ...

1

I think 'probability theory' is an extremely vague topic to list as your field of interest on a PhD application, but maybe it is just right in your case. It seems clear from what you say that you have not really settled on any specific part of probability as a possible area for your PhD thesis. So you may not be ready to be more specific, and may look ...

Only top voted, non community-wiki answers of a minimum length are eligible