# Tag Info

## New answers tagged nonparametric

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I have a number of questions. Why do you care about the tail of the distribution if you don't know where it is? You have said that you would need to evaluate the density at x = 10 in your example. How many samples do you have? For KDE I am fairly certain you don't have enough for this. Looking at that part of the tail, we have $$P(|X| > 10) = ... 0 It may be a consequence of the way you have presented your example, but it looks like your density function has finite support (e.g, a truncated Gaussian)? If this is the case, why not use a spline density estimator with linear tails: http://cran.r-project.org/web/packages/pendensity/vignettes/pendensity.pdf You could also have a look into wavelet density ... 0 I might be telling you something you already know, but keep in mind that really \hat{f}(x)=\hat{f}(x,\{X_k\}), where \{X_k\} is the set of sample points over which you build your estimate. For most non-parametric estimators, the X_k are assumed independent, and the method is additive, so you can just look at the MSE of \hat{f}(x,X_k) and then ... 0 I think it's important to check for dependence because 1) the statistical properties of the things being estimated depend on that. For example, standard errors will generally be affected by the presence of autocorrelation Autocorrelation is an obvious form of dependence to check for in a time series. 2) as you note, if there's dependence structure in ... 1 The Kruskal-Wallis procedure is the multi-group equivalent of the Wilcoxon rank-sum test, which is akin to one-way ANOVA without the normality assumption. A nonparametric ordered alternative equivalent of the Kruskal-Wallis test is the Jonckheere-Terpstra test. That is, where the Kruskal-Wallis tests against a general alternative "at least one \neq", J-T ... 11 I'd say logistic regression isn't a test at all; however a logistic regression may then lead to no tests or several tests. You're quite correct that labelling something nonparametric because it's not normal is insufficient. I'd call the exponential family explicitly parametric, so I'd usually regard logistic regression (and Poisson regression and Gamma ... 9 Larry Wasserman defines a parametric model as a set of distributions "that can be parameterized by a finite number of parameters." (p.87) In contrast a nonparametric model is a set of distributions that cannot be paramterised by a finite number of parameters. Thus, by that definition standard logistic regression is a parametric model. The logistic ... 2 There's two questions here really. The first is whether you can use a 95% confidence interval for the mean of a non-normal population. And you can, with several options: (1) If the sample size is large the mean will be approximately normally distributed anyway (2) Fit a specific distribution & calculate confidence intervals for the mean based on your ... 1 The box plots seem to tell an interesting story: Assuming they are in order, there are some complex things going on. I don't think any single test is going to tell the whole story. Wilcoxon was significant between the first two, but that seems solely due to higher highs - the medians are almost identical. Therefore, I suggest quantile regression on a range ... 0 As far as I know, no such book exists yet as the area is still quite new. The couple Bayesian nonparametrics books I've seen are basically just a bunch of review papers from various researchers bound together. If you have a PhD in math, applied or not, I'm sure you can get your head around by reading the standard papers. Probably the gentlest yet most ... 0 normality assumption needs to be validated, especially if it has implications on the analysis or method of analysis in you data. however, at times even with big or large data normality will be rejected and that has some meaning in itself about the data set or the random variable in question, hence i would suggest that if normality is a prerequisite for ... 0 Actually you were on the right track. The KDE you wrote there is actually obtained by convolving a distribution, which you get by placing a Dirac-delta function over each observed datapoint, by a Gaussian kernel :). Something like: p_0(x)={1 \over |X|}\sum\limits_{x_i \in X} \delta_{x_i}(x) p(x) = p_0(x) * \mathcal{N}(x| 0,\sigma)  0 The Guassian assumption is just that -- an assumption. If it holds, great, and you'll have a low-complexity estimate of the joint pdf. But if the assumption does not hold, you will have to use a more complex estimator, like the nonparametric one. For example, suppose that you want to approximate a joint pdf over two variables (X,Y), i.e., p(x,y). Now, if ... 0 Bayes, Actually, you don't need to estimate separate KDEs for f(x,y,z) and f(z) to get f(x,y,z). You can just estimate the joint f(x,y,z) by KDE that uses the Gaussain kernels and then analytically compute f(x,y|z) for any selected value of z. Then, as Pigeon already explained, just integrate over the x. But, if you are using the Gaussian kernels, you ... 0 I imagine that your dependent variable will be a composite of some or all of the 25 items. If this is the case, your composite variable can take$$kc - c + 1 categories, where $k$ is the number of items, and $c$ is the number of categories. Thus, if you formed a composite based on the 25 items, your scale would take on $25 \times 4 - 4 + 1 = 97$ different ...

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One assumption of the two sample t-test is that both samples come from a normal population. If you are convinced your samples both do not come from normal populations, you could use the Wilcoxon rank-sum test to test for difference in medians.

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