HW6.tex

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\fancyhead[R]{Swupnil Sahai}
\fancyhead[L]{Stat G6108 HW 6}
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% CUSTOM SHORTCUTS

\def\ci{\perp\!\!\!\perp}
\def\ex{\mathbb{E}}
\def\prob{\mathbb{P}}
\def\ind{\mathbb{I}}
\def\grad{\triangledown}
\def\bigo{\mathcal{O}}

% PROBLEM 1
\noindent 
(1) Firstly, we know from Inference I that the join distribution of the order statistics is $n!$. Additionally, we know that for every $t>0$, if the number of arrivals in the time interval $[0,t]$ follows a Poisson$(\lambda t)$ distribution, then we have that the sequence of inter-arrival times are independent and identically distributed Exponential$(\lambda)$ random variables. Hence, the conditional distribution we are attempting to find is equivalent to finding the conditional distribution of the (unit normalized) arrival times of the first $n$ arrivals, given that we know the arrival time of the $(n+1)th$ arrival.\\

\noindent
Hence, if we take $0 < t_1 < \cdots < t_n < 1$, it follows that the conditional probability that there is one arrival in each interval of the form $(t_i,t_i+dt_i)$ and no arrivals outside of the union of these intervals $(1-\sum_{i=1}^n dt_i)$, given that there are definitely $n$ arrivals in the (0,1) interval is approximately:

$$f \biggl(\frac{S_1}{S_{n+1}},\cdots,\frac{S_n}{S_{n+1}} \biggr| S_{n+1} \biggr) 
\approx \frac{e^{-(1-\sum_{i=1}^n dt_i)} \prod_{i=1}^n \frac{e^{-dt_i} dt_i}{1!}} {\frac{e^{-1}1^n}{n!}} 
= \frac{n! e^{-(1-\sum_{i=1}^n dt_i)}e^{-\sum_{i=1}^n dt_i} dt_1\cdots dt_n}{e^{-1}}$$
$$= n! dt_1\cdots dt_n. 
\hspace{5 pt} \square$$ 

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% PROBLEM 2
%PART A
\noindent
(2) (a) Since we just showed that $X_{(k)}$ is equivalent in distribution to $\frac{S_k}{S_{n+1}}$, it follows that if $\frac{S_k}{S_{n+1}} \overset{p}\to \alpha$ then $X_{(k)}\overset{p}\to \alpha$ as well. $\surd$\\

\noindent
Now, using the Weak Law of Large Numbers and the definition of $\alpha$, we find:
$$\frac{S_k}{S_{n+1}} 
= \frac{\sum_{i=1}^k Y_i}{\sum_{i=1}^n Y_i}
= \frac{\frac{1}{n}\sum_{i=1}^k Y_i}{\frac{1}{n}\sum_{i=1}^n Y_i}
= \frac{\frac{\alpha}{k}\sum_{i=1}^k Y_i}{\frac{1}{n}\sum_{i=1}^n Y_i}
= \alpha \frac{\frac{1}{k}\sum_{i=1}^k Y_i}{\frac{1}{n}\sum_{i=1}^n Y_i}
\overset{p}\to \alpha \frac{\ex Y_1}{\ex Y_1}
= \alpha. 
\hspace{5 pt} \square$$

%PART B
\noindent
(b) 
$$\prob(\sqrt{n}(X_{(k)}-\alpha) \leq t)
= \prob(X_{(k)} \leq \frac{t}{\sqrt {n}}+\alpha)
= \prob(\sum_{i=1}^n \ind(X_i \leq \frac{t}{\sqrt {n}}+\alpha) \geq k)$$
$$= \prob \biggl(\frac{1}{n} \sum_{i=1}^n \ind (X_i \leq \frac{t}{\sqrt {n}}+\alpha) \geq \alpha \biggr)$$
$$= \prob \biggl\{ \sqrt{n} \biggl(\hat{F}_n \biggl(\frac{t}{\sqrt {n}}+\alpha \biggr)-F \biggl(\frac{t}{\sqrt {n}}+\alpha \biggr) \biggr)
\geq \sqrt{n} \biggl(\alpha-F\biggl(\frac{t}{\sqrt {n}}+\alpha \biggr) \biggr) \biggr\}$$
$$\approx 1-\Phi \biggl(\frac{\sqrt{n}(\alpha-F(\frac{t}{\sqrt {n}}+\alpha))}{\sqrt{F(\frac{t}{\sqrt {n}}+\alpha)(1-F(\frac{t}{\sqrt {n}}+\alpha))}}\biggr)$$
Now we know that $F(\frac{t}{\sqrt {n}}+\alpha) 
\to F(\alpha)$ so that $\sqrt{n}(F(\frac{t}{\sqrt {n}}+\alpha)-\alpha) \to f(\alpha)$. Consequently:
$$\prob(\sqrt{n}(X_{(k)}-\alpha) \leq t) 
\to 1-\Phi \biggl( \frac{-f(\alpha)t}{\sqrt{F(\alpha)(1-F(\alpha))}} \biggr)$$
Hence, we find:
$$\sqrt{n}(X_{(k)}-\alpha) 
\overset{d}\to N \biggl(0,\frac{\alpha(1-\alpha)}{f^2(\alpha)} \biggr)
= N \biggl(0,\frac{F(\alpha)(1-F(\alpha))}{f^2(\alpha)} \biggr)
= N (0,\alpha(1-\alpha)). \hspace{5 pt} \square$$

%PART C
\noindent
(c) As for the median, it is apparent that in the framework of (b), here we have $k=\frac{n}{2}$ so that $\alpha = \frac{1}{2}$. Thus:
$$\sqrt{n}(X_{(\frac{n}{2})}-\frac{1}{2}) 
\overset{d}\to N(0,\frac{1}{4}). \hspace{5 pt} \square$$ 

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% PROBLEM 3
%PART A
\noindent
(3) (a) We can utilize the cdf $F(x) = \int_{-\infty}^x f(x)dx$ to create a bijective mapping between $[0,1]$ and $\mathbb{R}$. Thus, if $X_i \overset{iid}\sim f$ then $U_i = F(X_i) \overset{iid}\sim U[0,1]$. Additionally, if we assume that $f(x)$ has no discontinuities, then it follows from the definition of $F$ that $F(x)$ is continuous in F. Consequently, from the Continuous Mapping Theorem we have that if we let $\alpha = \frac{k}{n}$, then:
$$ X_{(k)} 
= F^{-1}( F(X_{(k)})) 
= F^{-1}(U_{(k)}) 
\overset{p}\to F^{-1}(\alpha).
\hspace{5 pt} \square$$

%PART B
\noindent
(b) In the derivation of 2(b), we did not assume that $X$ was from a uniform distribution, until the very last step. Hence, we can apply the results to that problem here:
$$\sqrt{n}(X_{(k)}-F^{-1}(\alpha)) 
\overset{d}\to N \biggl(0,\frac{F(F^{-1}(\alpha))(1-F(F^{-1}(\alpha)))}{f^2(F^{-1}(\alpha))} \biggr)
= N \biggl(0,\frac{\alpha(1-\alpha)}{f^2(F^{-1}(\alpha))} \biggr). \hspace{5 pt} \square$$

%PART C
\noindent
(c) Similar to problem 2, the median is just a special case of part (b) with $\alpha = \frac{1}{2}$. Thus, if $m$ is the value such that $\int_{-\infty}^m f(x) dx = \frac{1}{2}$, then it follows that:
$$\sqrt{n}(X_{(\frac{n}{2})}-m)
= \sqrt{n}(X_{(\frac{n}{2})}-F^{-1}(\frac{1}{2}))
\overset{d}\to N \biggl(0,\frac{\frac{1}{2}(1-\frac{1}{2})}{f^2(F^{-1}(\frac{1}{2}))} \biggr)
= N \biggl(0,\frac{1}{4f^2(m)} \biggr). \hspace{5 pt} \square$$

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% PROBLEM 4  <-- Did Not Attempt

% PROBLEM 5
%PART A
\noindent
(5) (a) First note that:
$$F_{T_n} (t)
= \prod_{i=1}^n F_{X_i} (t)
= F_X(t)^n 
= (1-e^{-t})^n.$$
Hence, it follows that:
$$\prob(T_n - \log n \leq t) 
= \prob(T_n \leq t+\log n)
= (1-e^{-(t+\log n)})^n
= \biggl(1-\frac{e^{-t}}{n} \biggr)^n
\to e^{-e^{-t}}.$$
Thus, $(T_n-\log n) \overset{d}\to Y$ where $F_Y = e^{-e^{-t}}.$ $\square$\\

%PART B
\noindent
(b) First note that:
$$F_{T_n}(t)
= \prod_{i=1}^n F_{X_i}(t)
= F_X(t)^n
= (1-t^{-\alpha})^n.$$
Hence, it follows that:
$$\prob(n^{-1/\alpha}T_n \leq t) 
= \prob(T_n \leq t n^{1/\alpha})
= (1-(tn^{1/\alpha})^{-\alpha})^n
= \biggl(1-\frac{t^{-\alpha}}{n} \biggr)^n
\to e^{-t^{-\alpha}}.$$
Thus, $n^{-1/\alpha}T_n \overset{d}\to Y$ where $F_Y = e^{-t^{-\alpha}}.$ $\square$

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% PROBLEM 6

\noindent
(6) We have from (a) that:
$$G_n(\theta_0)+o_p(1) \overset{p}\to G(\theta) + 0 = G(\theta).$$
Additionally, from $(a)$ it follows that there exists a $\theta^*$ such that:
$$\prob(|G_n(\hat{\theta}_n)-G(\theta^*)|>\varepsilon) \to 0$$
Consequently, from the given constraint on $G_n(\hat{\theta}_n)$ it follows that:
$$G_n(\hat{\theta}_n) \overset{p}\to G(\theta^*) \geq G(\theta)$$
Now, condition (b) implies that $G(\theta_0)$ is the unique maximizer of $G(\theta)$ across all $\theta \in \Theta$. Thus:
$$\theta^* = \theta \Rightarrow \hat{\theta}_n \overset{p}\to \theta_0. \hspace{5 pt} \square$$

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