In this case "marginal" is short for marginal distribution. If you have a distribution for a few random variables, that's usually termined the "joint" (distribution). When you look at individual ...

If $Y \sim \text{Binomial}(m,p)$, then its pmf is $f_Y(y;p) = {m \choose y} p^y(1-p)^{m-y}$. You are transforming $Y$ by applying this pmf to it, and you're using the value $p=1/2$. So $$X = f_Y(Y;.5)... View answer 5 votes Yes to (1) and no to (2). Let me explain. The reasoning is from the transformation theorem. This is it generally. Assume you have two original random variables X_1 and X_2, along with their joint ... View answer 5 votes A highest posterior density [interval] is basically the shortest interval on a posterior density for some given confidence level. A highest density region is probably the same idea applied to any ... View answer Accepted answer 4 votes Just rearrange your \text{AR}(p) polynomial \phi(z) = 1 - \phi_1 z - \cdots - \phi_p z^p. If z' is a root, then 1 - \phi_1 z' - \cdots - \phi_p (z')^p = 0. If z' is a unit root, then you can ... View answer Accepted answer 4 votes I will mention three ways to approximate the following$$ \mathrm{E} \left[ \phi(X_1, \cdots, X_{100}) \right] = Pr\left(\frac{\bar{X} - 0.1}{\sqrt{0.1/100}} < -1.645 \right). $$Use the CLT to ... View answer Accepted answer 4 votes Consider a random walk model, X_t = X_{t-1} + Z_t where Z_t \overset{iid}{\sim} \text{Normal}(0,1). Also, assume that X_1 = Z_1 and has variance 1. Take the random vector X = (X_1,X_2,X_3)', ... View answer 4 votes I'll answer this question as if it had been tagged with the self-study tag. Hint: consider Z = \frac{Y_1 - Y_2}{\sqrt{2}} \sim \text{Normal}(0,1). You are looking for the distribution of Z^2. View answer 4 votes No you are not wrong thinking that. If Y \mid X_1 \sim \alpha f_1(y) + (1-\alpha)f_2(y), then you can also let X_2 \sim \text{Bernoulli}(\alpha) independently and say$$ Y \mid X_1, X_2 = 1 \...

The other answer doesn't really give the full story. It's the transformation theorem. I just had an answer involving it here. Call X(t_1) = Y_1,~X(t_2) = Y_2 + Y_1, ...,~X(t_n) = Y_n + \cdots + ... View answer Accepted answer 4 votes When I asked after class, I was told that events are not subsets of the sample space. No you're correct. Events are subsets of the sample space. There could be a few sources of confusion, though. ... View answer Accepted answer 4 votes If a statistic T is sufficient in the frequentist way, then p(\mathbf{x} \mid \theta, t) = p(\mathbf{x} \mid t), so \begin{align*} p(\theta \mid \mathbf{x}, t) &= \frac{p(\mathbf{x}\mid t,\... View answer Accepted answer 4 votes Just use Bayes' rule. You're assuming these: f_{Y|Z}(y \mid z) \propto z^{p/2}\det(\Sigma)^{-1/2} \exp\left[-\frac{z}{2}(y - \mu)^t\Sigma^{-1}(y-\mu) \right] f_Z(z) \propto z^{\nu/2 - 1}\exp\left[... View answer Accepted answer 4 votes For z \in [0,n] \begin{align*} P(Z_n \le z) &= 1-P\left(U_{(n)} \le 1 - \frac{z}{n}\right) \\ &= 1 - \prod_{i=1}^nF_{U_i}\left( 1 - \frac{z}{n}\right) \\ &= 1 - \left(1 - \frac{z}{n}\... View answer Accepted answer 4 votes Adding a little bit to the hint above: P(X=k) = P(X/n = k/n). View answer 4 votes Intuitively, convergence in probability means the random variables get close to a nonrandom constant, and convergence in distribution means that it gets close to another random variable. Closeness ... View answer Accepted answer 4 votes In chapter 5.5 of this book, they discuss how a lot of these model selection criteria arise. They start with Akaike's FPE criterion for AR models, and then go on to discuss AIC, AICc and BIC. They ... View answer 4 votes To expand on @hxd1011's linked-to answer in the comments, \begin{align*} \text{TSS} &= \sum_i(y_i - \bar{y})^2 \\ &= \sum_{i}(y_i - \hat{y}_i + \hat{y}_i - \bar{y})^2\\ &= \sum_{i}(y_i - \... View answer Accepted answer 4 votes If you predict y by taking the conditional mean of it (with respect to one or both of the x_is), then it will always be true that \text{Var}[E(y \mid x_1)] \le \text{Var}[E(y \mid x_1, x_2)], ... View answer Accepted answer 4 votes First $$E[\mathbf{c}] = E[\mathbf{a}\otimes \mathbf{b}] = E[\mathbf{a}] \otimes E[\mathbf{b}] = \mu_a \otimes \mu_b$$ by independence. Then \begin{align*} \operatorname{Var}[\mathbf{c}] &= E\... View answer 4 votes The null hypothesis for DF test is a type of non-stationarity.H_0: \phi_1 = 1$where$X_t - \mu = \phi_1(X_{t-1}-\mu) + Z_t. The null hypothesis for ADF test is a type of non-stationarity as well, ... View answer Accepted answer 4 votes You look at ACF and PACFs of the differenced series. This is because these are tools for looking at stationary processes. You mention that the undifferenced values have a mean increasing over time...... View answer Accepted answer 4 votes Here are a couple hints: \begin{align*} P([X] = x) &= P(x \le X < x+1) \tag{logic} \\ &= P(x < X \le x+1) \tag{Xis a continuous rv} \end{align*} for any non-negative integerx$. View answer 4 votes A random walk process is defined as $$X_t = X_{t-1} + Z_t,$$ where$\{Z_t\}$is white or IID noise. Using just pencil and paper, you can see that subtracting$X_{t-1}$from both sides tells you the ... View answer 4 votes Yes it is a value, but no it doesn't necessarily have to be realized. A random variable can be realized or unrealized. Just as a house can be built or unfinished. The analogy is meant to emphasize ... View answer 4 votes If you can convince yourself that $$\max(X,Y) \overset{d}{=} -\min(X,Y),$$ then taking the variance on both sides will give you your answer. Regarding the other part, you'll probably have to ... View answer 4 votes The answer by Dougal does a good job explaining the empirical distribution. This answer just highlights more the connection between when expectations are sums, and when they are integrals, addressing "... View answer Accepted answer 4 votes If you Rao-Blackwellize a biased estimator, the bias does not change, and the variance potentially shrinks. The result doesn't change much. Let$U$be the estimator for$h(\theta)$. Let$b = E[U]-h(\...

I'm not quite familiar with statsmodels, but if you have data values, $x_1, \ldots, x_n$, and probabilities of these values $p_1, \ldots, p_n$ given to you by the function distribution.pdf(), then an ...
Hint: "method of moments" means you set sample moments equal to population/theoretical moments. For example, the first sample moment is $\bar{X} = n^{-1}\sum_{i=1}^n X_i$, and the second sample ...