A few years later... here's a solution that does not involve invoking Cesaro's Lemma. It follows from the following result: If $v_n \to \ell$ as $n \to \infty$, then $\bar{v} := n^{-1}\sum_i v_i \to \... View answer Accepted answer 1 votes I'm picking up on some confusion about a few things. Linear dependence in the columns of your design matrix$\mathbf{X}$, or equivalently singularity of$\mathbf{X}^\intercal \mathbf{X}$, is bad ... View answer 0 votes I do not know how to use the factorization theorem You have to write the likelihood as a product of two functions. One function is allowed to have parameters in it, and one is not. The one with the ... View answer 1 votes Hint: consider the definition of random variable, then try to write the set$\{\omega: \xi(\omega) = \eta(\omega)\}$in terms of sets you know are in$F$. The definition of sigma-algebra/sigma-field ... View answer 0 votes Here's a model: $$\mathbf{x}(t) = \mathbf{A} \mathbf{x}(t-1) + \mathbf{v}(t) \tag{1}$$ $$\mathbf{x}^o(t) = g(\mathbf{x}(t)) \tag{2}$$ where$g$is the ordering function,$\{\mathbf{v}(s)\}_s$are ... View answer 1 votes Going for the shortest answer here: just look at the first row of$\mathbf{X}^\intercal \mathbf{H} = \mathbf{X}^\intercal$. View answer Accepted answer 3 votes The normal conditional density's normalizing constant is not free of lambda, so Wolfram is integrating the wrong thing. $$(2\pi)^{-1}\int_0^\infty \underbrace{(2\pi\tau^2\lambda^2)^{-1/2}}_{\text{... View answer 1 votes You can reduce this question to this: why is$$ E[XY|X] = XE[Y|X] $$(with probability 1)? If this is true, then you can just take expectations on both sides. The answer is "because of the (... View answer Accepted answer 1 votes There are a few things you can do with a distribution: a. write it down as an unsolved integral, b. write it down as a solved integral, c. evaluate its normalized version on a computer at different ... View answer -1 votes I wrote a paper arxiv.org/abs/1907.09090 that describes how the pseudo-marginal approach can impute missing data. 400 covariates sounds tough, though, to be completely honest. Depends on what kind of ... View answer Accepted answer 2 votes Must we make an assumption that \mathbf{Q} is known in order to do what the authors are suggesting? No, this is a matter of identifiability, and things like this are commonly done to make the model ... View answer 2 votes "Is this normal?" I don't use this software, and I'm not sure what you mean by "normal," but nothing jumps out at me. A few thoughts: Asymptotic distributions of MLE estimates ... View answer Accepted answer 18 votes Let X be a standard normal random variable, and let Y = -X (pointwise). Then both are a.c., but X+Y is 0 everywhere. View answer 2 votes This makes me think of Meyn and Tweedie's book. They use P to denote the transition kernel for a Markov chain, and \mathsf{P} for the law of the entire chain on \mathsf{X}^{\infty}. This answer ... View answer Accepted answer 2 votes The inverse of a cdf F : \mathbf{R} \mapsto [0,1] is usually$$ 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 1 votes A few comments: Regarding @BruceET's comment, MGFs don't uniquely define random variables all the time, though. Practically, yeah you can identify many/most common random variables by their mgf. But ... View answer 1 votes The problem is: It works very poorly. If I deviate a little from the true parameters, the likelihood gets, as expected, smaller. However, there are always alternative constellations which give me a ... View answer Accepted answer 0 votes If you accept$\{\tau < \infty\} = \bigcup_{k\ge 1} \{\tau = k\}, then $$P(\tau < \infty) = \sum_{k=1}^{\infty}P(\tau = k).$$ by countable additivity. To add to the intuition, take ... View answer 0 votes By the triangle-inequality \begin{align*} 0 &\le \left|\frac{1}{nm}\sum_{j=1}^m\sum_{i=1}^nX_i^j - E[\mu]\right| \\ &\le \left|\frac{1}{nm}\sum_{j=1}^m\sum_{i=1}^nX_i^j - m^{-1}\sum_{j=1}^m\mu^... View answer Accepted answer 2 votes Adding to @Jarle Tufto's comment, the likelihood can be written as $$p(y_{1:t} \mid r, \sigma_x, \mu, p) = \int p(y_{1:t} \mid x_{1:t}, \mu, p) \overbrace{p(x_{1:t} \mid r, \sigma_x)}^{{\text{AR(1)}}}... View answer Accepted answer 2 votes SSMs aren't always identifiable without restrictions on the parameter space. There is a common distinction between centered and uncentered parameterizations. What you have written would be called the ... View answer 1 votes The equation$$ \mathbf{w}^T\mathbf{z_1}+b=0 $$defines a (hyper)plane. The vector \mathbf{w} is the normal vector. For a refresher on multivariable calculus, see here. So what happens if you have ... View answer Accepted answer 3 votes Multiplying X by W gives you transformed data. Multiplying X by P gives you X again (P is the identity matrix because the eigenvectors are orthogonal). Observe that$$ \text{Var}\left[W^... View answer Accepted answer 1 votes Define \begin{align*} S &= \{ x : Ax = y, x_2 \le s\} \\ &= \{ (x_1, x_2) : a_1 x_1 + a_2 x_2 = y, x_2 \le s \} . \end{align*} We can say that \begin{align*} p(x \in S \mid y) &= \frac{... View answer Accepted answer 3 votes Hint: ifX \sim \text{HalfNormal}(\sigma^2)$and$Y \sim \text{Normal}(0,\sigma^2)$, then for any symmetric and integrable$f$$E[f(X)] = E[f(|Y|)] = E[f(Y)|Y \ge 0] = 2 E[f(Y) 1(Y \ge 0)].$$ I ... View answer 1 votes Observe that \begin{align*} \log \psi(x) &= \log \sum_{k\in\mathbb Z}\phi(x+k) \\ &= \log \sum_k \exp\left[ \log \phi(x+k) \right] \\ &= m + \log \sum_k \exp\left[ \log \phi(x+k) - m \... View answer Accepted answer 2 votesX_n = O_p(\frac{1}{n})$means it's not terrible to think of$X_n$as something like $$Y / n.$$ This is a single random variable over a changing nonrandom constant. Indeed$Y/n = O_p(\frac{1}{n})$... View answer Accepted answer 1 votes What do the subscripts of the expectations mean here? They are the distribution you are taking the expectation with respect to. They are the "weights" you're using to calculate the weighted average. ... View answer Accepted answer 2 votes In the context of an SIS particle filter with the transitional prior as the proposal density$q(x_k|x^i_{k-1},z_k)= p(x_k | x_{k-1},z_k)$Two things: one, SIS stands for "sequential importance ... View answer Accepted answer 3 votes You're mistaken about a few things (and that's okay!). In this case the optimal proposal density...is [available]. I believe this is only true if$f\$, the state transition is Gaussian. It can be ...