The relationship is as follows: $l(\beta) = \sum_i L(z_i)$. Define a logistic function as $f(z) = \frac{e^{z}}{1 + e^{z}} = \frac{1}{1+e^{-z}}$. They possess the property that $f(-z) = 1-f(z)$. Or ...
Observe that $$\frac{e^{x_i}}{\sum_j e^{x_j}} = \frac{e^{-m}}{e^{-m}}\frac{e^{x_i}}{\sum_j e^{x_j}}= \frac{e^{x_i-m}}{\sum_j e^{x_j-m}}$$ for any constant $m$. Obviously it is not true that $e^{... View answer 21 votes They're the same. Here's how... Doing a Regression Say you fit the model $$y_t = \sum_{j=1}^n A_j \cos(2 \pi t [j/N] + \phi_j)$$ where$t=1,\ldots,N$and$n = \text{floor}(N/2)$. This isn't ... View answer Accepted answer 18 votes Let$X$be a standard normal random variable, and let$Y = -X$(pointwise). Then both are a.c., but$X+Y$is$0everywhere. View answer 18 votes \begin{align*} \frac{P(A,B|C)}{P(B|C)} &= \frac{P(A,B,C)}{P(C)}\frac{P(C)}{P(B,C)} \\ &= \frac{P(A,B,C)}{P(B,C)} \\ &= P(A|B,C) \end{align*} View answer Accepted answer 14 votes It's a common trick. IfX = \min(Y_1,Y_2)$and$F$,$F_X$are the CDFs of the$Y_i$s and$X, respectively, then \begin{align*} F_X(x) &= 1 - P(X > x) \\ &= 1- P(Y_1 > x, Y_2 > x) \\... View answer Accepted answer 14 votes There are a few reasons. For one, your ARMA model doesn't include a mean/intercept. For another, the ARMA by default uses sum of squares only to find starting points for an iterative maximum ... View answer 11 votes SIR uses two ideas. The first idea is importance sampling. The main idea is that you draw from one probability distribution (in your case, it's the uniform), in order to get information about another. ... View answer Accepted answer 11 votes In Bayesian statistics, a maximum a posteriori probability (MAP) estimate is an estimate of an unknown quantity, that equals the mode of the posterior distribution. This is correct. As far as ... View answer Accepted answer 11 votes The second is a special case of the first. Your first reference discusses the case where eachy_i$is distributed as a Binomial distribution with sample size$n_i$, while the second reference assumes ... View answer Accepted answer 10 votes A few things: The BF gives you evidence in favor of a hypothesis, while a frequentist hypothesis test gives you evidence against a (null) hypothesis. So it's kind of "apples to oranges." These two ... View answer 10 votes I assume you can evaluate$f$and$g$up to a normalizing constant. Denote$f(x) = f_u(x)/c_f$and$g(x) = g_u(x)/c_g. A consistent estimator that may be used is $$\widehat{D_{KL}}(f || g) = \left[... View answer 10 votes This is the same as above, but I thought I would provide a shorter, more concise answer. Again, this is Hamilton's representation for a causal ARMA(p,q) process, where r=\max(p,q+1). This r ... View answer Accepted answer 9 votes You can understand the shape of the ellipsoid better if you look at the spectral/eigen decomposition of the precision matrix (inverse of the covariance matrix). You want to look at the eigenvalues of ... View answer 9 votes "IID" stands for independent and identically distributed. An IID sample is also known as a "random sample." It is referring to the distribution of all of your random variables: X_1, \ldots, X_n. In ... View answer 8 votes Notice that this is the same model as$$ X_t - 10 = .5(X_{t-1} - 10) + Z_t . 10 is the mean, while 5 was the intercept. This means we can add ten to the mean zero series. In other words, if ... View answer Accepted answer 8 votes Yeah, that sounds right. If you have parameters \mu and \Sigma and data point x, then the set of all data points that are less likely than x are the ones that have less density, or in other ... View answer Accepted answer 8 votes When you have an independent prior on X and Y, then the posterior might not factor into X and Y pieces just because the likelihood doesn't factor into X and Y pieces. It's easy to see ... View answer Accepted answer 8 votes Yule Walker (for parameter estimation) is usually only used for AR models, but this method you're using is still a valid technique for finding the autocovariance function. I'm assuming that's what you'... View answer Accepted answer 8 votes From the preface to the first edition of "Markov Chains and Stochastic Stability" by Meyn and Tweedie: We deal here with Markov Chains. Despite the initial attempts by Doob and Chung [99,71] to ... View answer Accepted answer 8 votes This is S. Catterall's hint: \begin{align*} E\{ZΦ(Z)\} &= \int_{\mathbb{R}} z \Phi(z) \phi(z) dz. \end{align*} And an extra hint: let u=\Phi(z) and v' = z\phi(z). One more hint: v = \int_{-\... View answer Accepted answer 7 votes If X = (X_1,\ldots, X_n) are jointly normal, too, then yes. Otherwise, no. In this case \Sigma = \text{diag}(\sigma_1^2,\ldots, \sigma_2^2) and \mu = (\mu_1,\ldots,\mu_n)' \begin{align*}f_X(x) ... View answer Accepted answer 7 votes The model is true if the data are generated according to the model you are doing inference with. In other words, the unobserved parameter is generated by the prior, and then, using that parameter draw,... View answer Accepted answer 7 votes \begin{align*} P(\theta \mid y,x) &= \frac{P(\theta, y,x)}{P(y,x)} \tag{defn. condtl. prob} \\ &= \frac{P(y \mid \theta, x)P(\theta \mid x)P(x)}{P(y \mid x)P(x)} \tag{defn. condtl. prob}\\ &... View answer Accepted answer 7 votes If you target f with g, and you know f(x) \le g(x)c, then \begin{align*} P(\text{accept proposal}) &= P\left( U \le \frac{f(X)}{g(X)c} \right) \\ &= E\left( \mathbf{1}\left[U \le \frac{f(... View answer Accepted answer 7 votes Deterministic Trend If your drift intercept is c, you can just add the function c t to the zero mean process. Code: xt <- arima.sim(n=50, list(order=c(1,0,1), ar = c(.9), ma = -.2)) ... View answer Accepted answer 7 votes It's easy if you write \hat\Sigma_n= \frac{1}{n-1}\sum_{i=1}^nX_i X_i^T - \frac{n}{n-1}\hat{\mu}_n\hat{\mu}_n^T. $$Split up the sum over n elements into two parts. One will involve the first n-... View answer Accepted answer 7 votes Pretty close. Remember that exponentials have support on the positive reals.$$ \frac{1}{N}\sum_i f(x_i)/g(x_i) \to E_g[f(x)/g(x)] = \int_0^{\infty}f(x)/g(x) g(x)dx = \int_0^{\infty}f(x) dx \neq \... View answer Accepted answer 7 votes Assuming absolute summability of the autocovariance function (i.e.\sum_{h=-\infty}^{\infty}|\gamma(h)| < \infty) \begin{align*} \lim_n n \text{Var}(\bar{X}_n) &= \lim_n n^{-1} \sum_i \sum_j \... View answer 6 votes The KL divergence $$KL\left(P \middle\| Q\right) = \int \log \frac{d P}{d Q }dP$$ is only defined if the Radon-Nikodym derivative exists, which is whenP$is dominated by$Q$(written$P \ll Q\$). ...