Bartlett's theorem is useful for this. If $\{\mathbf{X}_{t}\}$ is a bivariate time series whose components are defined by $$X_{t1} = \sum_{k=-\infty}^{\infty} \alpha_k Z_{t-k,1}, \hspace{10mm} \{Z_{... View answer Accepted answer 3 votes \displaystyle {\boldsymbol {T}}(X) is a vector of estimators and \displaystyle {\boldsymbol{\psi}}({\boldsymbol {\theta}}) is its expectation. That is$$ \mathbb{E}\left[\displaystyle {\boldsymbol ...

Maybe one autocovariance function is theoretical, but we estimate it with the sample autocovariance function. Yes, that's correct. One is based on random data, and the other is theoretical and based ...

Can the log-likelihood actually decrease when adding a parameter, is there something special about how the TSA package calculates AIC, am I doing something wrong in R, or is R pulling my leg ...

It looks like you're having trouble with the derivative of a moment generating function. The reason they have the name they have is because they generate (via differentiation) moments (as long as you ...

This book chapter is a pretty good explanation of MCMC techniques for general state space models; you might want to start with this. For your particular model here, instead of looking at a review ...

This proof is easier if you iterate expectations. You want to show that $\mathbb{E}(y|x)$ is in a sense the best predictor of $t$, so you want to show that \mathbb{E}[(y(x) - t)^2] \ge \mathbb{E}[\{... View answer Accepted answer 3 votes X^* is not discrete; it's continuous. The X_i's are continuous, too. It is X^*|X_{1},\ldots,X_n that's discrete. I'll copy and paste your \TeX and and write over the stuff that's incorrect. I ... View answer Accepted answer 3 votes You're pretty much there. Here are all your steps, with the intended answer added at the end. \begin{align*} P(t(\boldsymbol{y}) = \boldsymbol{t}) &= \sum_{t(\boldsymbol{y}) =\boldsymbol{t}} P(t(\... View answer Accepted answer 3 votes Going from the first line to the beginning of the second line, you aren't pulling z out of the right summation. You are simply substituting P(h(X,Y) = z) = \sum_{\{(x,y): h(x,y) = z\}}P(X=x,Y=y)....

Is it correct to swap these two and put the measurement update first and then the prediction so at each time step have a prediction for the next time step? Depends what you mean by "swap." Let $... View answer 3 votes Say the parameter to each distribution is$\lambda$. First, $$E\left[\frac{\bar{X}}{6}\right] = [6\lambda]^{-1}$$ by linearity. Then, because$Y = \sum_i X_i \sim \text{Gamma}(n,1/\lambda)$\begin{... View answer 3 votes Hint: $$P(|X_n - 1|> \epsilon) = P(nY_n > \epsilon) = P(Y_n > \epsilon/n) = 1-P(Y_n\le \epsilon/n).$$ View answer Accepted answer 3 votes If you mean$X_1, \ldots, X_n \overset{iid}{\sim} \text{Uniform}(0,n)$, then$1-\prod_{k = 1}^n(1-F_k(y))$turns into $$1-\prod_{k =1}^n(1-y/n) = 1 - (1-y/n)^n \to 1 - e^{-y}.$$ This is the CDF of ... View answer Accepted answer 3 votes The Gaussian assumption is used in the predict and update steps of the Kalman Filter. They are the reason you only have to keep track of means and variances. First,$Z_t|X_t$is Normal. Second,$X_t|...

Statement: The Pythagorean theorem says, for any elements $T_1$ and $T_2$ of an inner-product space with finite norms such that $\langle T_1,T_2\rangle = 0$, $$||T_1+T_2||^2 = ||T_1||^2 + ||T_2||^2 \... View answer 3 votes Your OLS estimator is$$ \hat{\beta}_{ols} = (X'X)^{-1}X'y, $$while your ridge regression estimator is$$ \hat{\beta}_{ridge} = (X'X + \lambda I)^{-1}X'y. Take the expectation and variance of ... View answer 3 votes I'm not sure you can. Unless \phi \in \mathbb{N} is known. You can see the problem in the last line. \begin{align*} &f(y_i, \mu, \phi) \\ &= \frac{\Gamma (y+ \phi)}{\Gamma(\phi) \Gamma(y+1)}... View answer Accepted answer 3 votes It is still Gaussian, but the state is affecting the scale, not the location of the observaions. Here x_k represents the standard deviation of y_k. So y_k|x_k \sim \text{Gaussian}(y_k;0, x_k^2)...
Let $Z=\max(X,Y)$. It's always true by independence that $$F_Z(z) = P(\max(X,Y) \le z) = P(X \le z)P(Y \le z).$$ If $0 \le z \le 1$, then $P(\max(X,Y) \le z) = z^2/2$. If $1 < z \le 2$ then P(\... View answer 3 votes Just make sure you resample from your data with replacement, and you give every data point the same chance of being chosen. Then you should be good. Here's a quick R example: num_samples <- 3 ... View answer Accepted answer 2 votes \begin{align*} \partial_x D[p|q] &= \int_{-\infty}^{\infty} \bigg (\partial_xp(x)\log\frac{p(x)}{q(x)}+ p(x)\partial_x \log\frac{p(x)}{q(x)}\bigg)dx \\ &= \int_{-\infty}^{\infty}\partial_xp(x)\... View answer Accepted answer 2 votes Must we make an assumption that\mathbf{Q}$is known in order to do what the authors are suggesting? No, this is a matter of identifiability, and things like this are commonly done to make the model ... View answer Accepted answer 2 votes Finding the first moment Whenever I look up the Black-Scholes formula, I always worry about changing parameterizations and notations, so let's start off by reproducing the result in Hull's text. All ... View answer 2 votes "Is this normal?" I don't use this software, and I'm not sure what you mean by "normal," but nothing jumps out at me. A few thoughts: Asymptotic distributions of MLE estimates ... View answer 2 votes This makes me think of Meyn and Tweedie's book. They use$P$to denote the transition kernel for a Markov chain, and$\mathsf{P}$for the law of the entire chain on$\mathsf{X}^{\infty}$. This answer ... View answer Accepted answer 2 votes The inverse of a cdf$F : \mathbf{R} \mapsto [0,1]$is usually $$F^{-1}(p) = \inf\{ x : F(x) \ge p \}.$$ The way you invert a c.f.$\phi$to get measures of intervals is $$\mu([a,b]) = \lim_{T\... View answer Accepted answer 2 votes Adding to @Jarle Tufto's comment, the likelihood can be written as$$ p(y_{1:t} \mid r, \sigma_x, \mu, p) = \int p(y_{1:t} \mid x_{1:t}, \mu, p) \overbrace{p(x_{1:t} \mid r, \sigma_x)}^{{\text{AR(1)}}}... View answer Accepted answer 2 votes SSMs aren't always identifiable without restrictions on the parameter space. There is a common distinction between centered and uncentered parameterizations. What you have written would be called the ... View answer 2 votes You can use the law of large numbers to approximate your expectation pretty easily. Edit: Analyticaly you can multiply a bunch of normal cdf evaluations together. For$i > 0\$ \begin{align*} \...