Taylor
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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 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 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 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 0 votes Sometimes people are interested in estimating the locations of zeros in the precision matrix for the same reason you describe above. IfM$is your square root matrix, i.e.$M'M = \Sigma^{-1}$, then ... View answer 0 votes Use the polar coordinates transformation if you're good at integrating. Define $$\left[\begin{array}{c} R \\ \theta \end{array} \right] = \left[\begin{array}{c} \sqrt{X^2 + Y^2} \\ \text{arctan}(\... View answer 0 votes Check out this Wiki page and note that$$ -(5 - 6 x) = -(-6)(x - 5/6).$$View answer 0 votes First, I think you mean: I know to [dis]prove stationar[it]y it suffices to prove [that] either [the] mean function or [the] autocovariance function is not independent of time. But yeah that's ... View answer 0 votes Let$X,Y_1,Y_2,\ldots,Y_{1000} \in \{0,1\}$denote the true color, and the responses, respectively. "Blue" is coded as a$1$, and vice versa. Assume that$p(x)$is Bernoulli with parameter$p_x\$. ...