sta_theory_cheatsheet.tex

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     \Large{\textbf{Statistical Theory Cheatsheet}} \\
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\scriptsize

Compiled by Gejun Zhu (zhug3@miamioh.edu) in preparation for the theory comprehensive exam by using William Chen's \href{http://wzchen.com/probability-cheatsheet}{formula sheet template}. 

\begin{center}
    Last Updated \today
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\section{Probability and Distributions} \smallskip \hrule height 1pt \smallskip
\begin{description}
	\item[Section 1] - probability and distributions \\
	Let $Y=g(X)$, where $g(x)$ is a one-to-one differentiable function. $f_Y(y) = f_X(g^{-1}(y)) |\frac{dx}{dy}|$ for $y \in S_Y$ . \\


If $X$ and $Y$ are independent random variables, then $M_{X+Y} (t) = M_{X}(t)M_{Y}(t)$.\\
If $X$ and $Y$ are independent random variables, then $\rho(X, Y) = 0$. \\
If $X$ is a discrete random variable,  then mgf of $X$ is $M_{X}(t) = \sum P(X = k)e^{kt}$. \\
If $X \sim Gamma(a_1,b)$ and $Y \sim Gamma(a_2,b)$, $X+Y \sim Gamma(a_1+a_2,b)$ if $X$ and $Y$ are independent. \\
If $X \sim Gamma(a_1,b)$, then $cX\sim Gamma(a_1, cb)$. \\
$X \sim Gamma(a,b)$, if $a=1$ then $X\sim exponential(b)$; if $b=2$ then $X\sim \chi_{(2a)}^2$. \\
If $X_i \sim Laplace(\mu, b)$ then $\frac{2 \sum_{i=1}^n |X_i-\mu|}{b} \sim \chi^2(2n) $.\\
If $X \sim Laplace(0, b)$ then $|X| ~ Exponential(b^{-1})$.\\
If $X \sim Exp(\lambda)$ then $X \sim Gamma(1, \lambda)$.\\
If $X \sim Gamma(k, \theta)$ where $k$ is the shape parameter and $\theta$ is a scale parameter, then $E(X) = k\theta$, $var(X) = k\theta^2$;  if $X \sim Gamma(\alpha, \beta)$ where $\alpha$ is the shape parameter. $\beta$ is the rate parameter, then $E(X) = \frac{\alpha}{\beta} $, $var(X) = \frac{\alpha}{\beta^2}$.\\

\end{description}

%%%%%%%%%%%%%%%    
    \begin{description}
    		\item[Section 2] - Multivariate Distributions\\ %  \boldsymbol{}
    		Covariance: $EX = \mu_1$, $EY = \mu_2$, $Cov(X, Y) = E(XY) - \mu_1\mu_2$, \\
    		Correlation coefficient: $\rho = \frac{cov(X, Y)}{\sigma_1\sigma_2}$ \\
    		$E(X_2) = E[E(X_2|X_1)]$, $Var(X_2) = E[Var(X_2|X_1)] + Var(E(X_2|X_1)) \geq Var(E(X_2|X_1))$.\\
		Moment generate function: 	$M = E(e^{tX})$, $\mu = E(X) = M'(0)$, $\sigma^2 = E(X^2) - (EX)^2 = M''(0) - [M'(0)]^2$\\
		$Z_n \sim \chi^2(n)$, $M_{Z_n} = (1 - 2t)^{-\frac{n}{2}}, t< \frac{1}{2}$, $w_n = \frac{Z_n}{n^2}$, $M_{w_n}(t) = E[e^{tw_n}] = E[e^{t\frac{Z_n}{n^2}}] = M_{Z_n}(\frac{t}{n^2}) = (1-\frac{2t}{n^2})^{-\frac{n}{2}}$ for $\frac{t}{n^2} < \frac{1}{2}$\\
		Negative binomial distribution: $y = \#$ of failures before the $r^{th}$ success. $p(y) = {{y+r-1}\choose{r-1}}p^r(1-p)^y$ \\
		Poisson distribution: $y = \#$ of successes in a fixed length of time, $p(y) = \frac{\lambda^x e^{-\lambda}}{x!}$;\\
		Gamma distribution: $y =$ waiting time required until the $\alpha^{th}$ success. $f(y) = \frac{1}{\Gamma(\alpha) \beta^\alpha} x^{\alpha - 1} e^{\frac{-x}{\beta}}$. Special cases: (1) exponential distribution - $\alpha = 1, \beta	= \frac{1}{\lambda}$; (2) Chi-square distribution - $\alpha = \frac{r}{2}$, $\beta = 2$.\\
				
\end{description}

\section{Statistical Inference}\smallskip \hrule height 1pt \smallskip

%\subsection{Simple Linear Regression}

\begin{description}
\item[Inequalities] - Important Inequalities \\
	\underline{Markov's Inequality}: $u(X)$ non-negative, $E(u(X))$ exists, $P[u(X)\geq c] \leq \frac{E(u(X))}{c}$.\\ 
	\underline{Chebyshev's Inequality}: $P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}$
\end{description}

    \begin{description}
        \item[Distributions]: \textbf{Some facts }\\
        Sample mean: $\bar{X} = \frac{\sum X_i}{n}$ ,  Sample variance: $S^2 = \frac{\sum X_i^2 - n\bar{X}^2}{n-1}$, $E(S^2) = \sigma^2$; another unbiased estimator for $\sigma^2$ in normal distribution is $(\bar{x}^2 - \dfrac{S^2}{n})$ [also think about $Y_n$]\\
        $E(\bar{X}) = \mu$, $var(\bar{X}) = \frac{\sigma^2}{n}$, $\bar{X} \sim N(\mu, \sigma^2/n)$, $\frac{(n-1)S^2}{\sigma^2} \sim	\chi_{(n-1)}^2$, $\bar{X}\ \&\ S^2$ are indep.; \\
        $\sum a_iX_i$ and $\sum b_iX_i$  are indep. iff $\sum a_ib_i = 0$;\\
        $\frac{\sum (X_i - \mu )^2}{\sigma^2} = \frac{(n-1)S^2}{\sigma^2} + \frac{(\bar{X} - \mu)^2}{\sigma^2/n}$\\
        $\dfrac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim N(0, 1)$ by CLT, $\dfrac{\bar{X} - \mu}{s/\sqrt{n}} \sim T_{n-1}$\\

        %$\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim N(0, 1)$, $\frac{\bar{X} - \mu}{S/\sqrt{n}} \sim t(n-1)$, 
         
        %\item[Properties of SLR]
    \end{description}
 

    
    \begin{description}
    		\item[Order Statistics] - \textbf{Order Statistics}\\ %  \boldsymbol{}
    		$g_k(y_k) = \dfrac{n!}{(k-1)!(n-k)!}[F(y_k)]^{k-1}[1-F(y_k)]^{n-k}f(y_k)$, $a< y_k <b$; 0 elsewhere.
		
    		
    \end{description}   
    
    \begin{description}
    		\item[Section 4] - \textbf{Confidence Interval Estimation}\\ %  \boldsymbol{}
    		\underline{Error types}: Type I - reject $H_0$ while $H_0$ is true, $P_{\theta}(X\in R)$; Type II - fail to reject $H_0$ while $H_0$ is false. $\alpha = P(Type\ I\ error),\ \beta =  P(Type\ II\ error)$, $Power = 1 - \beta$. \\
    		\underline{Level of significance}: the probability of making type I error - reject $H_0$ when $H_0$ is true. \\ 
    		\underline{Power function}: $K(\theta) = P_{\theta}[(X_1, X_2, ..., X_n) \in C | \theta]$,  $\theta \in \omega_1$. The \textit{power function} of a hypothesis test with rejection region $R$ is the function of $\theta$ defined by $\beta(\theta) = P_{\theta}(X\in R)$. In other words,  the power of a hypothesis test is the probability of rejecting $H_0$ when $H_a$  is true. $Power = P(X\in R|\theta = \theta_a) = 1 - \beta$.\\
    		\underline{Level of significance} : $\alpha =  max_{\theta	\in \omega_0} K(\theta)$ or $\alpha = P(Type\ I\ error) = P(rejecting\ H_0\ when\ H_0\ is\ true) = P(X\in R | \theta = \theta_0)$\\
    		\underline{CI for difference in Means}: $X_i \overset{iid} \sim N(\mu_1, \sigma^2)$, $Y_i \overset{iid} \sim N(\mu_2, \sigma^2)$, where $\sigma^2$ is unknown. $S_p^2 = \frac{(n-1)S_1^2 + (m-1)S_2^2}{n+m-2}$, $\frac{(\bar{X} - \bar{Y}) - (\mu_1 - \mu_2)}{\sqrt{S_p^2(\frac{1}{n} + \frac{1}{m})}} \sim T(n+m-2)$. \\
    		Note: $x > \theta	\Rightarrow I(x - \theta)$
    		
    		
    \end{description}   
    
   \begin{description}  %%%  \overset{P}{\rightarrow}   %%%
	\item[Section 5] - \textbf{Consistency and Limiting Distributions} \\
	\underline{Convergence in Probability}: $X_n \overset{P}{\rightarrow} X$ if for every $\epsilon > 0$, $lim_{n\rightarrow\infty} P(|X_n - X| \geq \epsilon) = 0$  or ($lim_{n\rightarrow\infty} P(|X_n - X| < \epsilon) = 1$ ).\\
	\underline{Degenerate r.v.}: $p(x) = 1, if\ x=a; p(x) = 0,\ if\ x\neq a$ and $F(x) = 0\ if\ x<a; F(x) = 1\ if\ x\geq a$. We write $X_b \overset{P}{\rightarrow}  a$. \\
	\underline{Consistency}: The statistic $T_n$ is a consistent estimator for $\theta$ iff $T_n \overset{P}{\rightarrow} \theta$. [find limit of the estimator $E(\hat{\theta}) \rightarrow \theta$ and $var(\hat{\theta}) \rightarrow 0$]\\
	\underline{Convergence in Distribution}: $X_n \overset{D}{\rightarrow} X$ iff $lim_{n\rightarrow \infty} F_n(x) = F(x)$. F(x) is said to be the limiting distribution or asymptotic distribution of X; \\
	\underline{Limiting distribution}: $X_n \rightarrow X$ if and only if $lim_{n\rightarrow \infty} F_n(x) = F(x)$ where $F_n(x)$ is the cdf of $X_n$. $F(x)$ is said to be the limiting distribution of $X_n$. \\
	\underline{Theorem 5.2.10}: Suppose $X_n$ has m.g.f. $M_{X_n}(t)$ that exists for $-h\leq x \leq h$ for all $n$. Let $X$ has m.g.f. $M(t)$ which exists for $|t|\leq h_1 \leq h$. If $lim_{n\rightarrow \infty} M_{X_n}(t) = M(t)$ for $|t| \leq h_1$, then $X_n \overset{D}{\rightarrow} X$. \\
	\underline{m.g.f technique}: (1). $lim_{n\rightarrow \infty} (1 + \frac{b}{n} + \frac{\phi (n)}{n})^{cn} = lim_{n\rightarrow \infty} (1+\frac{b}{n})^{cn} = e^{bc}$ where $b$ and $c$ are constants and $lim_{n\rightarrow \infty} \phi(n) = 0$. (2). $e^x = 1 + x + \frac{x^2}{2} + ... + \frac{x^m}{m!} + ...$; \\
	\underline{CLT}: $X_1, X_2, ..., X_n \overset{iid}{\sim} f(x)$ with mean $\mu$ and variance $\sigma^2$, $y_n = \frac{\bar{X} - \mu }{\sigma / \sqrt{n}} \overset{D}{\rightarrow} z$, where $z \sim N(0, 1)$. \\
	
	
   \end{description} 
   
   	\begin{description}  %%% \overset{iid}{\sim} 
   		\item[Section 6] - \textbf{Maximum Likelihood Estimation }\\
   		\underline{Consistent}: If there is a unique solution to the likelihood equation $\frac{\partial}{\partial \theta}L(\theta) = 0$, then $\hat{\theta} \overset{P}{\rightarrow} \theta	$ ($\hat{\theta}$ is consistent for $\theta$).\\
   		\underline{Score function}: $\frac{\partial ln(f(x; \theta))}{\partial \theta}$;\\
   		\underline{Fisher Information}: $I(\theta)= var(\frac{\partial ln(f(x; \theta))}{\partial \theta}) = -E(\frac{\partial^2 ln(f(x; \theta))}{\partial \theta^2})$;\\
   		\underline{Efficient}: $y$ is unbiased for $\theta$, $y$ is efficient for $\theta$ iff $var(y) = [nI(\theta)]^{-1}$. In general, $var(y) \geq [nI(\theta)]^{-1}$ where $y$ is unbiased for $\theta$. \\
   		\underline{Efficiency}: The efficiency of an unbiased estimator is given by the ratio $\frac{RCLB}{var(\hat{\theta})}$, where $RCLB = [nI(\theta)]^{-1}$. \\
   		\underline{Relative Efficiency}: Relative efficiency of unbiased estimators $\hat{\theta}_1$ to $\hat{\theta}_2$ is $\frac{var(\hat{\theta}_2)}{var(\hat{\theta}_1)}$. If $\theta_2$ is biased, then relative efficiency is $\frac{var(\theta_2) + [bias(\theta_2)]^2}{var(\theta_1)}$ where $bias(\theta_2) = E(\theta_2) - \theta$.\\
   		\underline{Theorem 6.1.2}: Suppose $\hat{\theta}$ is the MLE of $\theta$ and $g(\theta)$ is a function of $\theta$. Then MLE of $g(\theta)$ is $\hat{g(\theta)} = g(\hat{\theta})$. \\
   		\underline{Theorem 6.2.1 (Rao-Cramer Lower Bound)}: $X_1, X-2, ..., X_n \overset{iid}{\sim} f(x; \theta)$ for $\theta \in \Omega$. Let $y = u(X_1,X_2, ..., X_n)$ be a statistic with mean $E(Y) = k(\theta)$. Then $var(Y) \geq \frac{[k'(\theta)]^2}{nI(\theta)}$. \\
   		\underline{MVUE}: $\hat{\theta} = u(X_1, X_2, ..., X_n)$ is a minimum variance unbiased estimator for $\theta$  iff $E(\hat{\theta})$ and $var(\hat{\theta})$ is less than or equal to the variance of every other unbiased  estimator. \\
   		\underline{Theorem}: If $\hat{\theta}$ is asymptotically unbiased for $\theta$ and $var(\hat{\theta}) \rightarrow 0$ as $n\rightarrow \infty$, then $\hat{\theta} \overset{P}{\rightarrow} \theta$. 
	\end{description}      

	\begin{description}
		\item[Section 7] - \textbf{Measure of Quality of Estimators}\\
		\underline{MVUE}: $\hat{\theta} = u(X_1, X_2, ..., X_n)$ is a minimum variance unbiased estimator for $\theta$ iff $E(\hat{\theta}) = \theta$ and $var(\hat{\theta})$ is less than or equal to the variance of every other unbiased estimator. Relative efficiency of MVUE to any other unbiased estimator must be $\geq 1$. MUVE would be consistent if $var(\hat{\theta}) \rightarrow 0$ as $n\rightarrow\infty$. \\
		\underline{Sufficient Statistics}: Let $X_1, X_2, ..., X_n$ be a random sample from $f(x;\theta)$, $y_1 = u(X_1,X_2, ..., X_n)$ is sufficient for $\theta$ iff $\frac{\Pi(f(x_i;\theta))}{g_1(y_1;\theta)} = H(x_1, x_2, ..., x_n)$ where $g_1$ is a marginal pdf for $y_1$.\\
		\underline{Factorization Theorem}: $y_1 = u(x_1, x_2,..., x_n)$ is sufficient for $\theta$ iff $\Pi_{i=1}^n f(x_i; \theta) = k_1(u(x_1, x_2, ..., x_n); \theta) k_2(x_1, x_2, ..., x_n)$ where $k_2(x_1, x_2, ..., x_n)$ does not depend on $\theta$. \\
		\underline{Theorem 7.3.2}: $X_1, X_2, ..., X_n \overset{iid}{\sim} f(x; \theta)$. If a sufficient statistic $y_1 = u(X_1,X_2, ..., X_n)$ for $\theta$ exists and if an MLE $\hat{\theta}$ exists uniquely, then $\hat{\theta}$ is a function of $y_1$. \\
		\underline{Theorem}: Suppose $y_1 = u(X_1,X_2, ..., X_n)$ is a sufficient statistic for $\theta$. Let $z = u(y_1)$ be a 1-to-1 transformation not involving $\theta$. $z$ is also sufficient for $\theta$. \\
		\underline{\textbf{Rao-Blackwell Theorem}}: (1). $y_1 = u_1(x_1, x_2, ..., x_n)$ be sufficient for $\theta$; (2) $y_2 = u_2(x_1, x_2, ..., x_n)$ be unbiased for $\theta$; (3). $\phi(y_1) = E(y_2|y_1)$. Then, (1) $\phi(y_1)$ is a statistic; (2) $\phi(y_1)$ is a function of $y_1$ alone; (3) $\phi(y_1)$ is unbiased for $\theta$; (4). $\phi(y_1)$  has variance $< \sigma_{y_2}^2$. \\
		\underline{Completeness}:  Suppose $Z \sim h(z; \theta)$, a member of a family of p.d.f's (p.m.f's): $\{h(z;\theta), \theta	\in \Omega \}$. If $E(u(z)) = 0, \forall \theta \in \Omega$ implies that $u(z) = 0$ except on a set of points that has probability 0 for each $h(z; \theta), \theta	\in \Omega$, then the family $\{h(z;\theta), \theta	\in \Omega \}$ is called a complete family of density (mass) functions. \\
		\underline{Lehmann and Scheffe (MVUE)}: Let $Y_1 = u_1(X_1,X_2, ..., X_n)$ be sufficient for $\theta$ and $\{g_1(y_1, \theta): \theta	\in \Omega \}$ be a complete family of densities (or p.m.f's). If there is a function if $Y_1$ which is unbiased for $\theta	$, then this function of $Y_1$ is the unique MVUE for $\theta$. \\
		\underline{Find MVUE}: (1) Find a sufficient statistic $t$; (2) show that the family of distributions of $t$ is complete [shortcut - theorem 7.5.2]; (3) Find a crude unbiased estimator; (4) evaluate \\
		\underline{Exponential Family}: $f(x; \theta) = exp\{p(\theta)k(x) + \delta (x) + q(x)\}$ where $\delta(x)$ dose not depend on $\theta$, $p(\theta)$ is nontrivial continuous, $k'(x) \neq 0$.\\
		$X_1,X_2, ..., X_n \overset{iid}{\sim} f(x; \theta)$, a regular case of exponential class with $y_1 = \sum k(x_i)$. Then, (1) $g_1(y_1; \theta) = R(y_1)exp\{p(\theta)y_1 + nq(\theta)\}$; (2) $E(y_1) = -n\frac{q'(\theta)}{p'(\theta)}$; (3) $var(y_1) = n\frac{1}{[p'(\theta)]^3}\{p"(\theta)q'(\theta) - q"(\theta)p'(\theta)\}$ \\
		\underline{\textbf{Theorem 7.5.2}}: $X_1,X_2, ..., X_n \overset{iid}{\sim} f(x; \theta)$, a regular case of exponential class with $\Omega = \{\theta: \gamma < \theta < \delta \}$, $y_1 = \sum	k(x_i)$ is \textbf{sufficient} for $\theta$ and the family $\{g_1(y_1; \theta): \gamma < \theta< \delta \}$ is \textbf{complete}. \\
		\underline{Theorem 7.4.1}: $X_1, X_2, ..., X_n$, a random sample from $f(x;\theta)$, $y_1 = u_1(X_1, X_2, ..., X_n)$ complete sufficient for $\theta$ and  $E(\phi(y_1)) = \alpha(\theta)$, then $\phi(y_1)$ is unique MVUE for $\alpha(\theta)$.\\
		unique MVUE = sufficient + complete (Theorem 7.5.2) + unbiased \\
		 
	\end{description}	  
	
	
	\begin{description}
		\item[Section 8] \textbf{Most powerful tests }\\
		\underline{Neyman-Pearson Theorem}: A \textbf{best critical region} of size $\alpha$ for testing $H_0: \theta = \theta '$ v.s $H_1: \theta = \theta "$ (both simple) is such that \\
		(1) $\dfrac{L(\theta ' ; x_1, x_2, ..., x_n)}{L(\theta " ; x_1, x_2, ..., x_n)}\leq k$ for each $(x_1, x_2, ..., x_n) \in C$\\
		(2) $\dfrac{L(\theta ' ; x_1, x_2, ..., x_n)}{L(\theta " ; x_1, x_2, ..., x_n)}\geq k$ for each $(x_1, x_2, ..., x_n) \in C^c$\\
		\textbf{(3)} $P((x_1, x_2, ..., x_n) \in C; H_0) = \alpha$ \\
		\underline{Uniformly Most Powerful Test}: C is a uniformly most powerful critical region of size $\alpha$ if C is a best critical region of size $\alpha$ for testing $H_0$ against \textbf{each} simple hypothesis in $H_1$. Because the critical region $C$ defines a test that is most powerful against each simple alternative $H_1$, this is a uniformly most powerful test, and $C$ is a uniformly most powerful critical region of size $\alpha$.\\
		\underline{Likelihood Ratio Test}: 
		Let (1) $L(\hat{\omega})$ denote the maximum of the likelihood function with respect to $\theta$ when $\theta$ is in the null parameter space $\omega$. (2) $L(\hat\Omega)$ denote the maximum of the likelihood function with respect to $\theta$ when $\theta$ is in the entire parameter space $\Omega$. Then, the likelihood ratio is the quotient: \\
		\begin{center}		
		$\lambda = \frac{L(\hat{\omega})}{L({\hat{\Omega}})} = \frac{L(\hat{\theta}_0|x)}{L(\hat{\theta}|x)}$.\\  %= \frac{max_{\theta \in \Theta_0}L(\theta)}{max_{\theta \in \Theta } L(\theta)}
		\end{center}		
		p.s: MLE estimated $\mu$ and $\sigma^2$ for normal distribution: $\hat{\mu} = \bar{x}$, $\hat{\sigma}^2 = \frac{1}{n}\sum (x_i - \bar{x})^2$; $\hat{\mu} = (n+m)^{-1}\{\sum x_i + \sum y_i \}$, $\hat{\sigma}^2 = (n+m)^{-1} \{\sum (x_i - \bar{x})^2 + \sum (y_i - \bar{y})^2\}$.\\
		\underline{Neyman-Pearson Lamma}: If a critical value is chosen so that $P_{\theta_0}(\Lambda \leq c) = \alpha$, then the test with decision rule \\
		\begin{center}
		Reject $\theta = \theta_0$ in favor of $\theta = \theta_1$ when $\Lambda \leq c$
		\end{center}
		is a uniformly most powerful test of size $\alpha$.
	\end{description}	  
    
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