# Tag Info

## Hot answers tagged estimation

5

Let $Z \sim N_n(μ_Z, Σ_Z)$ and $Z= \left(\begin{array}\\Y\\X\end{array}\right)$, where $X$ has been observed. What you want is to generate from the distribution of $Y|X$. If $μ_Z$ and $Σ_Z$ are partitioned as follows $$\mu_Z = \begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}$$ corresponding to the means of the $Y$ and $X$ components respectively, where ...

5

As stated in comment, the prior distribution represents your prior beliefs about the distribution of the parameters. When you have prior beliefs you can: convert your belief in terms of moments (e.g. mean and variance) to fit a common distribution to these moments (e.g. Gaussian if your parameter lies to the real line) use your intuitive understanding ...

3

A Bayesian approach will work as follows: Chose a prior distribution on the number $n$ of balls in the urn. If you don't have any prior idea, you can chose the uniform distribution : $\def\P{\mathbb P}$ $$\P_0(n = i) = {1\over 1001} \ \ (i = 0, \dots, 1000).$$ For a sample of 100 balls, denote $X$ the number of red balls. For each sample with $X=x$ ...

3

Use Caterpillar Plots showing Mean +/- squared error. Some examples are shown here: http://stackoverflow.com/questions/13847936/in-r-plotting-random-effects-from-lmer-lme4-package-using-qqmath-or-dotplot

2

The difference between the confidence interval for the mean response and the prediction interval is subtle but important. I'll explain it first and then provide you with a graphical intuition which helped me a lot when I learned this. Obviously, there is an error component to our prediction. Under the normal probability model, the prediction is normally ...

2

Grid search at appropriate resolution will always produce excellent results. The problem arises when getting to that appropriate resolution is intractable. Numerical approximations become infeasible very quickly as the dimension of the data increases, but it's entirely possible to have an intractably large space to explore in just two dimensions. Consider a ...

2

Consider $\lambda =\pi_1\lambda_1 + \pi_2\lambda_2$ where $\pi_1 = \exp(1\beta_{11} + a_1\beta_{21})$ $\pi_2 = \exp(1\beta_{12} + a_1\beta_{22})$ So $\lambda =\exp(\beta_{11}) \exp(a_1\beta_{21})\lambda_1 + \exp(\beta_{12}) \exp(a_1\beta_{22})\lambda_2$, or $\lambda =\exp(a_1\beta_{21})[\exp(\beta_{11}) \lambda_1] + \exp(a_1\beta_{22})[\exp(\beta_{12}) ... 2 The question asks about how the estimation variance of the parameters in least squares fitting changes when an additional observation is included in the data. Letting$A$be the model matrix ($A = (a)$or$A=(a,b)'$in the examples in the question),$X$be the parameter vector ($X = (x)$in the question), and$Y$be the response vector ($Y = (y_1)$or ... 1 I run a similar script in R, but it seems that OK, setting prior of the b How is your prior of b setting? Has your MCMC chain already converged? PS: I tried something as below as you set b as fixed. But I also tried some other prior of b, it is OK. N <- 100 x <- rgamma(N, 5, 1) write(" model { for (i in 1:length(x)) { x[i] ~ dgamma(a, ... 1 First of all, I'd start by restating your hypothesized relationship in the more "conventional" format, i.e., with the dependent variable$N$appearing on one side of the equation, and all of the independent variables and unknown parameters on the other side, like so: $$N = g(x,y;\alpha,\beta,\gamma) = \frac{f(x^{\beta} (y-\gamma))}{x^{\alpha}}$$ Restated in ... 1 Oversized comment warning You could phrase your problems in terms of link prediction. There is a lot of literature on this for things like social networks and whatnot. The problem that is going to make all (almost all) of of that standard techniques I'm aware of inapplicable is that you've assumed you don't know any of the links for the node you're trying ... 1 Bias in estimating$p$-quantiles is investigated in a distribution-free way in http://www.sciencedirect.com/science/article/pii/S016771520000242X (a pdf can be found on the same page). The authors focus on the quantile estimator based on ECDF inversion. No assumptions on the underlying distribution is made (except finite second moment), thus also discrete ... 1 Plots like this are usually shown in classes to explain the meaning of a$(1-\alpha)\cdot 100\%$confidence interval for a population parameter$\mu$. Under repeated sampling, they cover$\mu$in about$(1-\alpha)\cdot 100\%$of all hypothetical random samples. In your case, only three of 50 intervals are 'wrong', i.e. they lead to wrong conclusions about ... 1 Are these the "not very nice" formulas that you found? (Note that$\hat{b}_1$and$\hat{b}_2$have the same denominator.)$\hat{b}_1 = (s_{zz}\,s_{xy} - s_{xz}\,s_{zy})\,/\,(s_{xx}\,s_{zz} - s_{xz}^2)\hat{b}_2 = (s_{xx}\,s_{zy} - s_{xz}\,s_{xy})\,/\,(s_{xx}\,s_{zz} - s_{xz}^2)\hat{b}_0 = \overline{y}\$

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