## A latex test for homework 1

1. Find a counter example shows that ${\forall}$ n, ${P(|Y_n|\leq K^*)\geq 1-\epsilon}$ does not imply ${P(|Y_n|\leq K^*, \forall n)\geq1-\epsilon}$

Let ${{X_n}}$ denote a sequence of i.i.d. exp(1) random variables. For ${\forall \epsilon}$ and any n, ${\exists K^*}$ such that ${P(|Y_n|\leq K^*)\geq 1-\epsilon}$, where ${K^*=z_{\epsilon/2}}$. But the event ${\{|Y_n|\leq K^*, \forall n)\}}$ is equivalent to ${\{\sup\limits_n |Y_n|\leq K^*\}}$. From exponential we see that ${|Y_n|=Y_n}$ so ${\sup\limits_n|Y_n|}$ is just the n-th order statistic of the random variables ${Y_1, Y_2, \ldots, Y_n}$. The CDF of this order statistic is ${P_{Y_{(n)}}(x)=F_{Y_1}^n(x)}$, so ${P(|Y_n|\leq K^*, \forall n)=F^n_{Y_1}(K^*)}$. We cannot find any finite ${K^*}$ such that ${\lim\limits_n F^n_{Y_1}(K^*)\geq 1-\epsilon}$ for ${\epsilon\in [0,1]}$. So a counter example is found.

2. Proof that if ${r_n(Y_n-\theta)}$ converge to random variable ${Y}$, ${\lim\limits r_n=\infty}$ then ${Y_n-\theta}$ converge to 0 in probability.

By the statement ${(vi)}$ in Portmanteau Lemma in the textbook, for any fixed M satisfies ${P(|Y|\geq M)\leq \epsilon}$, ${P(|r_n(Y_n-\theta)|\geq M)}$ exceeds ${P(|Y|\geq M)}$ arbitrarily little for large n. Then ${\forall \epsilon}$, ${\exists N}$ such that when ${n\geq N}$ we have ${P(|Y|\geq M)\leq P(|r_n(Y_n-\theta)|\geq M)\leq P(|Y|\geq M)+\epsilon}$. By the triangular inequality, we have ${P(|r_n(Y_n-\theta)|\geq M)N}$. ${P(|Y_n-\theta|\geq M/r_n)<2\epsilon}$. As n goes to infinity, we have ${M/r_n}$ goes to 0. So ${Y_n-\theta}$ converge to zero in probability.

3. Problem 3.2

Define a function from a subset of ${\mathbb{R}^5}$ to ${\mathbb{R}}$.

$\displaystyle \phi(x_1,x_2,x_3,x_4,x_5)=\frac{x_1-x_2x_3}{(x_4-x_2^2)(x_5-x_3^2)}$
Then ${\phi(\cdot)}$ is a continuous function. Let ${(\mathbf{X}, \mathbf{Y})}$ be the bivariate sample of sample size n. Without loss of generality, suppose ${\mu_X=\mu_Y=0}$, ${\sigma_X=\sigma_Y=1}$ and with ${\rho}$ being the correlation coefficient. Let ${\overline{XY}=\frac1n\sum X_iY_i}$, ${\overline{X}=\frac1n\sum X_i}$,${\overline{Y}=\frac1n\sum Y_i}$, ${\overline{XX}=\frac1n\sum{X_i^2}}$, ${\overline{YY}=\frac1n\sum{Y_i^2}}$. Then

$\displaystyle \rho=\frac{\rho\sigma_X\sigma_Y-\mu_X\mu_Y}{\sigma_X\sigma_Y}=\frac{\mathsf{E}XY-\mathsf{E}X\mathsf{E}Y}{(\mathsf{E}X^2-n(\mathsf{E}X)^2)(\mathsf{E}Y^2-n(\mathsf{E}Y)^2)}=\phi(\mathsf{E}XY,\mathsf{E}X,\mathsf{E}Y,\mathsf{E}X^2,\mathsf{E}Y^2)$

$\displaystyle r_n=\phi(\overline{XY},\overline{X},\overline{Y},\overline{XX},\overline{YY})$
Here we have ${\mathsf{E}\overline{X}=\mathsf{E}\overline{Y}=0}$, ${\mathsf{E}\overline{XX}=\mathsf{E}\overline{YY}=1}$, ${\mathsf{E}\overline{XY}=\rho}$ and ${\mbox{var}\overline{X}=\mbox{var}\overline{Y}=\frac1n}$. Let ${\mathbf{X^*}=(X_1^2, X_2^2,\ldots, X_n^2)^T}$ and ${\mathbf{Y^*}=(Y_1^2, Y_2^2,\ldots, Y_n^2)^T}$. Then

$\displaystyle \begin{array}{rcl} \mbox{var}(\overline{XX})&=&\frac1{n^2}\mbox{var}\sum(X_i^2-\overline{X^2})^2\\&=&\frac1{n^2}\mbox{var}({X^*}^TAX^*)=\frac1n[\frac1n\mbox{var}({X^*}^TAX^*)] \end{array}$

$\displaystyle \begin{array}{rcl} \mbox{var}(\overline{YY})&=&\frac1{n^2}\mbox{var}\sum(Y_i^2-\overline{Y^2})^2\\&=&\frac1{n^2}\mbox{var}({Y^*}^TAY^*)=\frac1n[\frac1n\mbox{var}({Y^*}^TAY^*)]\\ \mbox{var}(\overline{XY})&=&\frac1n\mbox{var}(XY) \end{array}$
Let ${a_1=\frac1n\mbox{var}({X^*}^TAX^*)}$, ${a_2=\frac1n\mbox{var}({Y^*}^TAY^*)}$, ${a_3=\mbox{var}(XY)\leq\frac1n\mbox{var}(X)\mbox{var}(Y)<\infty}$. A is a certain constant matrix. It is easy to see that ${a_1}$ and ${a_2}$ are linear functions of ${\mathsf{E}X^4}$, ${\mathsf{E}X^3}$, ${\mathsf{E}X^2}$ and ${\mathsf{E}X}$, which are all finite. Thus

$\displaystyle \sqrt{n}\:\overline{X}\sim N(0,1)$

$\displaystyle \sqrt{n}\:\overline{Y}\sim N(0,1)$

$\displaystyle \sqrt{n}\:\overline{XX}\sim N(1,a_1)$

$\displaystyle \sqrt{n}\:\overline{YY}\sim N(1,a_2)$

$\displaystyle \sqrt{n}\:\overline{XY}\sim N(\rho,a_3)$
What's more, we will take partial derivative with respect to each variable in the function ${\phi}$ and evaluated at ${(x_1=\rho, x_2=0, x_3=0, x_4=1, x_5=1)}$, which are

$\displaystyle \frac{\partial \phi}{\partial x_1}=\frac{1}{(x_4-x_2^2)(x_5-x_3^2)}=1$

$\displaystyle \frac{\partial \phi}{\partial x_2}=\frac{-x_3(x_4-x_2^2)(x_5-x_3^2)+2x_2(x_1-x_2x_3)(x_5-x_3^2)}{(x_4-x_2^2)^2(x_5-x_3^2)^2}=0$

$\displaystyle \frac{\partial \phi}{\partial x_3}=\frac{-x_2(x_4-x_2^2)(x_5-x_3^2)+2x_3(x_1-x_2x_3)(x_4-x_2^2)}{(x_4-x_2^2)^2(x_5-x_3^2)^2}=0$

$\displaystyle \frac{\partial \phi}{\partial x_4}=\frac{-(x_1-x_2x_3)(x_5-x_3^2)}{(x_4-x_2^2)^2(x_5-x_3^2)^2}=-\rho$

$\displaystyle \frac{\partial \phi}{\partial x_5}=\frac{-(x_1-x_2x_3)(x_4-x_2^2)}{(x_4-x_2^2)^2(x_5-x_3^2)^2}=-\rho$
So according to Delta method, ${\sqrt{n}(r_n-\rho)\rightarrow N(0, a_3+\rho^2a_1+\rho^2a_2)}$