数学代写|STAC58 Statistical inference

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STAC58课程简介

The course surveys the various approaches that have been considered for the development of a theory of statistical reasoning. These include likelihood methods, Bayesian methods and frequentism. These are compared with respect to their strengths and weaknesses. To use statistics successfully requires a clear understanding of the meaning of various concepts such as likelihood, confidence, p-value, belief, etc. This is the purpose of the course.

Prerequisites 

Evaluation
A midterm worth 40% and a final worth 60%.

References
The following texts will be used in the course.
Probability and Statistics; The Science of Uncertainty by M. Evans and J. Rosenthal – this is the text used for STAB52F/B57S and we will be focusing on chapters 6, 7, 8 and 9 but only some this will be review. The book is available for download on my website.`
Measuring Statistical Evidence Using Relative Belief by M. Evans – this book gives a general discussion of the different approaches to inference. The book is available electronically from the University of Toronto Library.

STAC58 Statistical inference HELP（EXAM HELP， ONLINE TUTOR）

The confidence interval for $\theta$ is given as $[\overline{\mathbf{x}}-1.96 / \sqrt{n}, \overline{\mathbf{x}}+$ $1.96 / \sqrt{n}]$.

We know that the sample mean $\overline{\mathbf{x}}$ is an unbiased estimator of $\theta$ and follows a normal distribution with mean $\theta$ and variance $1/n$.

Thus, we can standardize $\overline{\mathbf{x}}$ as follows:

$$\frac{\overline{\mathbf{x}} – \theta}{1/\sqrt{n}} \sim N(0,1)$$

Now, let’s consider the random variable $Z$ defined as:

$$Z = \frac{\overline{\mathbf{x}} – \theta}{1/\sqrt{n}} \times \sqrt{n}$$

We can rewrite $Z$ as:

$$Z = \frac{\overline{\mathbf{x}} – \theta}{1/\sqrt{n}} \times \sqrt{n} = \frac{\overline{\mathbf{x}} – \theta}{\sqrt{1/n}}$$

Since $\overline{\mathbf{x}}$ follows a normal distribution, we know that $Z$ follows a standard normal distribution, i.e., $Z \sim N(0,1)$.

Now, the probability that an additional independent sample $X_{n+1}$ will fall within the confidence interval is:

\begin{align*} p &= P\left(\overline{\mathbf{x}} – 1.96/\sqrt{n} < X_{n+1} < \overline{\mathbf{x}} + 1.96/\sqrt{n}\right) \ &= P\left(\frac{\overline{\mathbf{x}} – X_{n+1}}{1/\sqrt{n}} < 1.96\right) – P\left(\frac{X_{n+1} – \overline{\mathbf{x}}}{1/\sqrt{n}} < 1.96\right) \ &= P\left(Z < 1.96\right) – P\left(Z < -1.96\right) \ &= 2 \times P\left(Z < 1.96\right) – 1 \ &\approx 0.95 \end{align*}

where the last step follows from the fact that $P\left(Z < 1.96\right) \approx 0.975$ (from standard normal table).

Therefore, we can conclude that $p \approx 0.95$, which is equal to the confidence level of the interval.

We want to test the null hypothesis $H_0: \lambda=\lambda_0$ versus the alternative hypothesis $H_1: \lambda \neq \lambda_0$, where $\lambda=\sigma_Y^2/\sigma_X^2$.

We can use the likelihood ratio test (LRT) to test this hypothesis. The likelihood function for the joint sample $\left(X_1, \cdots, X_n\right)$ and $\left(Y_1, \cdots, Y_m\right)$ is given by:

$$L(\sigma_X^2, \sigma_Y^2) = \frac{1}{(2\pi\sigma_X^2)^{n/2}} \exp\left{-\frac{1}{2\sigma_X^2}\sum_{i=1}^{n}x_i^2\right} \times \frac{1}{(2\pi\sigma_Y^2)^{m/2}} \exp\left{-\frac{1}{2\sigma_Y^2}\sum_{i=1}^{m}y_i^2\right}$$

The maximum likelihood estimator (MLE) of $\sigma_X^2$ is the sample variance $s_X^2 = \frac{1}{n}\sum_{i=1}^{n}x_i^2$ and the MLE of $\sigma_Y^2$ is the sample variance $s_Y^2 = \frac{1}{m}\sum_{i=1}^{m}y_i^2$.

The null hypothesis $H_0: \lambda=\lambda_0$ can be written as $\sigma_Y^2 = \lambda_0\sigma_X^2$, and the alternative hypothesis $H_1: \lambda \neq \lambda_0$ can be written as $\sigma_Y^2 \neq \lambda_0\sigma_X^2$.

The restricted parameter space for the null hypothesis is given by $\Theta_0 = {\sigma_X^2}$, and the unrestricted parameter space for the alternative hypothesis is given by $\Theta_1 = {(\sigma_X^2, \sigma_Y^2): \sigma_Y^2 > 0}$.

The log-likelihood function for the null hypothesis is:

$$\ell(\sigma_X^2) = -\frac{n}{2}\log(2\pi) – \frac{n}{2}\log(\sigma_X^2) – \frac{1}{2\sigma_X^2}\sum_{i=1}^{n}x_i^2$$

The log-likelihood function for the alternative hypothesis is:

$$\ell(\sigma_X^2, \sigma_Y^2) = -\frac{n+m}{2}\log(2\pi) – \frac{n}{2}\log(\sigma_X^2) – \frac{1}{2\sigma_X^2}\sum_{i=1}^{n}x_i^2 – \frac{m}{2}\log(\sigma_Y^2) – \frac{1}{2\sigma_Y^2}\sum_{i=1}^{m}y_i^2$$

The likelihood ratio test statistic is given by:

$$\Lambda = \frac{\sup_{\Theta_1} L(\sigma_X^2, \sigma_Y^2)}{\sup_{\Theta_0} L(\sigma_X^2)} = \frac{s_X^{2n}s_Y^{2m}}{s_X^{2n}(m+n)\hat{\sigma}_X^2\hat{\sigma}_Y^2}$$

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