# Tag Info

3

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 $T_k = \lambda_1 ... 2 In addition to the identical distribution independence is used to obtain that$(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

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

$\newcommand\E{\Bbb E}$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 ...

1

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.

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

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

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\...

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