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Hi: Notice that the initial equation you wrote [ except that it should be $f(x_1, \ldots x_{n}) = f(x_{n} | x_1, \ldots x_{n-1}) * f(x_1, \ldots x_{n-1})$] is a product of a full and a conditional. This is the technique used in building likelihoods for time series models. Harvey calls it the prediction error decomposition but it's really just the full ...


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No, it is called the posterior. For completion, $p(o|hi)$ is called the likelihood $p(hi)$ is called the prior $p(hi|o)$ is called the posterior


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A full understanding of this issue requires a theory of integration over probability distributions, not just functions. However, even in such an abstract theory it's possible to visualize the integrals as areas under curves. The universal principle is that in any "reasonable" theory of integration, it should be possible to integrate by parts. Consider the ...


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$X_i$ is continuous a random variable, with pdf $f_{X_i}(x_i;\theta)$, and the expectation requires an integral. The integral limits contain the domain of $X_i$. Not $i$ from $1$ to $n$. The $n$ samples you have are just realizations of $X_i$, i.e. $X_1,X_2,...,X_n$. You're not integrating/summing across these variables. You're integrating for a particular $...


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This proof corresponds to the case of a single data point (so $n=1$ in this context), where the distribution of the random variable $X_i$ is continuous, so it has a probability density function $f$. The proof uses the integral form from the law of the unconscious statistician, which holds that the expected value of the score function is an integral of that ...


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The proof you are examining starts by assuming $f(x_i; θ)$ is "a regular pdf." A pdf, or probability density function, is, by definition a continuous (i.e. not discrete) function. Since $X_i$ is continuous (hence pdf), you would use an integral to obtain the expected value of a function of $X_i$ by the Law of the Unconscious Statistician.


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A couple of notes: Most often the p-values given next to regression coefficients are based on the Wald test statistic, which is the estimated value of the coefficient divided by its standard error. This tests the null hypothesis that the specific coefficient is zero, and all other coefficients are non-zero. You could test the same hypothesis also with a ...


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For independent observables $\{x_{it}\}$ and $\{a_{it}\}$, I would write the joint conditional likelihood as $$ \begin{aligned} P(\{x_{it}\},\{a_{it}\}|\{\lambda_i\},\{\xi_i\},\beta) = \prod_i TN(a_{it}|\lambda_i,\beta) Pr(x_{it}|\lambda_i,\xi_i) \end{aligned} $$ Some comments: Notice that I've only included mention of the parameters whose values are ...


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