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You begin with a prior distribution. You combine this with the likelihood to create a posterior distribution. From the posterior distribution, you can create a posterior predictive distribution. A prior predictive distribution also exists if you do not collect data. In your case, you are beginning with the beta distribution as your prior distribution and ...


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You’re asking how to estimate the maximum likelihood parameters of a normal distribution from data. That is—you want the parameters with highest likelihood, given the data. How you find these is up to you; there are several optimization procedures you could turn to. You could start at some initial guess of $\mu$ and $\sigma$, then climb the gradient of the ...


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I'll use $\mu_i = \eta_1 - 2\theta\eta_2x_i + \eta_2 x_i^2$ for convenience. If we're thinking of $\mu_i$ as a function of $\theta$, so only $\eta_1$ and $\eta_2$ are parameters, then we can write this as $$ \mu_i = \eta_1 + \eta_2(-2\theta x_i + x_i^2) = \eta_1 + \eta_2 z_i $$ for $z_i = -2\theta x_i + x_i^2$. This is just a simple linear regression now so $...


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It depends on which parameters you are trying to estimate. In the particulare case of exponential kernels, the likelihood functional can be computed in linear time using the trick mentionned, which has to do with the fact that $(N,\lambda)$ is a Markovian system in this case. But despite this increase of speed in the computation of the objective, the ...


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You have a Poisson point process with intensity (rate) function $\lambda(t)$, say. Assume the observation window is contained in the interval $[0, T]$ and the observed points $t_1, t_2, \dotsc, t_n$. Let $N(T)$ be the total count of points. Then we can show that $$ N(T) \sim \mathcal{Poisson}\left( \int_0^T \lambda(t)\; dt \right) $$ and there is the ...


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The way I understand the Tobit model, you either observe $y_i$ when $y_i>0$ or the fact that $y_i<0$ but not the precise value of $y_i\sim\mathcal N(x_i'\beta,\sigma^2)$. (This means $y_L=0$ in the Wikipedia page.) Thus, the distribution of the data, made of the $$y_i\mathbb I_{y_i>0}\qquad i=1,\ldots,n$$is mixed, with a continuous component when $y&...


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As far as I know there are no special tricks for handling this problem. When I ran into a similar problem before, I ended up using the method of steepest descent to find approximate solutions. If $P(x \vert \theta)$ can sensibly be written as $P(x \vert \theta) = e^{f(x; \theta)}$, then $$ E\left[ \log P \right] = \int f(x; \theta) e^{f(x; \theta)} dx $$ ...


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This problem has been extensively covered in this forum, to wit: Why is computing the Bayesian Evidence difficult? Bayesian MCMC methods that need to calculate the evidence / normalizing factor Normalizing constant irrelevant in Bayes theorem? Why Normalizing Factor is Required in Bayes Theorem? What does it mean intuitively to know a pdf “up to a constant”? ...


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The numerator is typically easy to compute. I.e. $p(x | \theta, \phi)$ is just a likelihood function multiplied by a prior. The problem here is integration. For a lot of functions, integration is straightforward. For example, $\int_0^1 x^2 \mathrm{d}x = \frac{1}{3}$. However, in general, it's not easy to compute an integral in an analytic form (i.e. I can't &...


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The likelihood function $L(\theta|X)$ is indeed a random function (since $X$ is random) in the set of likelihood functions $$\{L(\cdot|x);\ x\in\mathsf X\} $$ Its distribution depends (obviously) on the statistical model. For instance if $L(\theta|X)$ is the likelihood function attached to a Normal $\mathcal N(\theta,1)$ $n$-sample, and if $\theta^*$ is the ...


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