1 2015-Spring

1.2 Problem 1

1.2.1 Part 1(a)

\[ \begin{align*} P(\sqrt{n}(\widetilde{X_n}-\theta) \leq a) &= P(\sum_{i=1}^n \geq \frac{n+1}{2})\\ &= P(\frac{\sum_iY_i - np}{\sqrt{np(1-p)}} \geq \frac{\frac{n+1}{2}-np}{\sqrt{np(1-p)}})\\ p &= P(X_i \leq \theta + \frac{a}{\sqrt{n}})\\ &= \frac{\pi+2 \tan^{-1}(\frac{a}{\sqrt{n}})}{2\pi}\\ &= {\frac{1}{2} + \frac{a}{\pi\sqrt{n}}}\\ 1-p &= {\frac{1}{2} - \frac{a}{\pi\sqrt{n}}}\\ \lim_{n \rightarrow \infty} \frac{\frac{n+1}{2}-np}{\sqrt{np(1-p)}} &=\lim_{n \rightarrow \infty} \frac{\frac{1}{2}-\frac{a\sqrt{n}}{\pi}}{\sqrt{\frac{n}{4} - \frac{a^2}{\pi}}}\\ &= \lim_{n \rightarrow \infty} \frac{\frac{1}{2\sqrt{n}}-\frac{a}{\pi}}{\sqrt{\frac{1}{4} - \frac{a^2}{\pi\sqrt{n}}}}\\ P(\sqrt{n}(\widetilde{X_n}-\theta) \leq a) &\rightarrow \frac{-2a}{\pi} \end{align*} \]

1.2.2 Part 1(b)

\(H_\theta: \theta=\theta_0\) ; \(H_{\theta_1} = \theta \neq \theta_0\)

$| -| z_{}

1.3 Problem 2

\(Z_i | \theta_i \sim \mathcal{N}(\theta_1, 1)\)

\(\theta_i \sim \mathcal{N}(\mathcal{\epsilon}, 1)\)

1.3.1 Part 2(a)

\[ \begin{align*} f_Z_i(z_i) &\propto exp(-\frac{1}{2}\big( (z_i-\theta_i)^2 + \frac{(\theta-\epsilon)^2}{\sigma^2} \big)) \\ & \propto exp(-\frac{1}{2(\sigma^2+1)} (z_i-\epsilon_i)) \end{align*} \]
Skipped details

1.3.2 Part 2(b)

Define \(p = P(Z_i > u) = P( \frac{Z-\mathcal{\epsilon}}{\sqrt{\sigma^2+1}} > \frac{u-\mathcal{\epsilon_i}}{\sqrt{\sigma^2+1}}) = \phi(\frac{\mathcal{\epsilon_i}-u}{\sigma{\sigma^2+1}})\)

The incomplete data log likelihood: \[ \begin{align*} L(\theta | X) &= \prod_{i=1}^n p^{x_i}(1-p)^{1-x_i}\\ &= p^{\sum_i x_i}(1-p)^{n-\sum_i x_i} \end{align*} \]

Complete data log likehood:

\[ \begin{align*} f(Z|\mathcal{\epsilon}, \sigma^2) &= (2\pi(\sigma^2+1))^{-\frac{n}{2}} \exp(-\sum_i\frac{(z_i-\mathcal{\epsilon})^2}{2(\sigma^2+1)})\\ \log L(\mathcal{\epsilon} | Z,\sigma^2) & = -\frac{n}{2} \log(2\pi(\sigma^2+1)) -\sum_i\frac{(z_i-\mathcal{\epsilon})^2}{2(\sigma^2+1)}\\ E[\log L(\mathcal{\epsilon} | Z,\sigma^2) ] &= -\frac{n}{2} \log(2\pi(\sigma^2+1)) -\frac{1}{2(\sigma^2+1)} E[\sum_i z_i^2 -2\mathcal{\epsilon}\sum_i z_i + n\mathcal{\epsilon^2}] \end{align*} \]

where the expectation is with repsect top the conidiotnal distribution of \(Z\) having observed \(X\), \(k(z|x,\theta) = h(x,z|\theta)/g(x | \theta)\)

\[ \begin{align*} Q(\mathcal{\epsilon}| \mathcal{\epsilon_0}, \sigma^2, X) &= -\frac{n}{2} \log(2\pi(\sigma^2+1)) -\frac{1}{2(\sigma^2+1)} E[\sum_i z_i^2 -2\mathcal{\epsilon}\sum_i z_i + n\mathcal{\epsilon^2}]\\ \frac{\partial Q}{\partial \mathcal{\epsilon}} &= -\frac{1}{2(\sigma^2+1)} \big( -2 \sum_iE[ z_i | x_i] + 2n\mathcal{\epsilon} \big) = 0\\ \implies \mathcal{\epsilon^{(i)}} = \frac{1}{n} E[z_i|x_i,\mathcal{\epsilon^{i-1},\sigma^2}] \end{align*} \]