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Asymptotic CI may be challenging. Besides the nagging issue of finite-sample bias, the non-iid feature of stock returns is likely to make the asymptotic variance difficult to compute. Personally I would bootstrap the CI. Just be careful if a block bootstrap is needed to handle serial correlations in stock returns.


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If $\sigma$ is known, you use Gaussian distribution. When it is unknown, to account for the uncertainty over possible $\sigma$'s, you use student-t distribution. So, the issue is the unknown variance. When the student-t distribution is well approximated by Gaussian with large samples, the problem with uncertainty over $\sigma$ decreases and the two ...


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Generate different 10% validation 90% training splits. Run model generation on each training set. Determine model performance on accompanying testing set. Determine confidence intervals from resulting vector of metrics. Formore indepth look see https://www.sisostds.org/DesktopModules/Bring2mind/DMX/API/Entries/Download?Command=Core_Download&EntryId=36208&...


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I have exactly the same problem. Rubin's rule tells us to average the estimates (and something a little more complex to calculate the pooled variance). The estimates in a logistic model are the betas, not the exponent of the betas (odds ratios), although these are the ones we are usually interested in. My conclusion, therefore, is that I should average the ...


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I had a similar problem. R code demonstration. t.test(c(1.1,2),conf.level = .9)$co t.test(c(2.1,2),conf.level = .9)$co t.test(c(3.1,2),conf.level = .9)$co It is caused by the standard deviation shrinking. I offset the problem by using 75th percentile of standard deviations of all items linearly combined with actual sd. However my goals were a bit different ...


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In this answer I aim to describe the difference between confidence intervals and credible intervals in an intuitive way. I hope that this may help to understand: why/how credible intervals are better than confidence intervals. on which conditions the credible interval depends and when they are not always better. Credible intervals and confidence ...


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A possible explanation for the lack of symmetry in the prediction interval is that the dependent variable could have subjected to a data transformation. Classically economic data is subject to percentage errors. As such, a traditional transformation is the log function. In medicine, similarly, drug dosing likely varies as a function of percent differences ...


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As @whuber notes in a comment, you do need to deal first with what seem to be incorrect premises in your approach. Most important, if your samples aren't truly random then "you can't use any of these methods to make inferences," as he put it. Fix that first. In terms of mean versus median as a measure of central tendency, the choice is yours based on your ...


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In general the confidence interval gives a range where the value is expected to be in a certain fraction of repeated experiments. But this can mean different things, e.g. based on the distribution of the underlying data. Few examples: Normal Distribution ("your definition of CI"): Upper limits and lower limits $\pm$ exactly $Z \sigma / \sqrt{n}$ around the ...


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The paper indicates that data was put into Epi Info 2000. The Epi Info manual indicates that they use Wilson 95% Confidence Limits. These are described here. These intervals are asymmetric by design.


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Framing the Question The model supposes that when $n$ subjects are given a dose $x \gt 0$ and independently develop toxic responses, the count of those responses $Y(x)$ has a Binomial distribution with count parameter $n$ and probability parameter $$p(x;a,b) = \frac{1}{1 + 10^{(\log(a) - \log(x))/b}} = \frac{1}{1 + \exp\left(\beta_0 + \beta_1 \log(x)\...


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For simplicity, assume that there is one focal continuous predictor $x$ and a continous outcome $y$. Standardization doesn't really make a lot of sense with categorical predictors, imo. The regression model could include more predictors but the following answer focuses only on one of them. Then, we have four possibilities: Both $y$ and $x$ are standardized (...


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Asymptotic confidence intervals on 'small' samples can have wider than nominal coverage, narrower than nominal coverage, or a mixture of both… sometimes a complicated mixture of both wider and narrower coverage. (This is a broad question, so this answer is similarly broad.) The particulars of wider vs narrower vs a mixture of both will depend on the ...


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I think you might make progress by asking your audience to assume that these values are distributed on the range [0,5] in the set {(0:10)/2} with a beta-binomial distribution. The beta-binomial distribution arose from a different process than your situation but it is an ordered discrete distribution. Ben Bolker has a nice discussion of simulation using the ...


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One approach is to use bootstrapping: library(simpleboot) library(boot) set1 <- as.data.frame(c(3,3,2.5,2.5,4.5,3,2,4,3,3.5,3.5,2.5,3,3,3.5,3,3,4,3.5,3.5,4,3.5,3.5,4,3.5)) colnames(set1) <- "numbers" set1.boot = one.boot(set1$numbers, mean, R=10^4) ## hist(set1.boot) boot.ci(set1.boot, type="bca") ## BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS ## ...


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You have two options: Option A) Try to figure out a confidence band for each observation in y, given its corresponding value of x (or x's, in case of multiple regressors).Keep on doing this independently for each observation, as if the other predictions did not exist, and you get your red-band. Note, the band has some non-zero width since the predicted ...


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It depends on what the error estimates represent: errors in the measurements going into the model, or expected errors in predictions from a fitted model. For terminology, it's simplest to discuss in terms of the error variance estimates (which for a given study size bear a one-to-one relationship with the confidence-interval widths). Standard linear ...


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Since I feel you like to go more advanced, I edit my answer. Let's assume you have data for one week $\{y_1, y_2, \ldots, y_7\}$. As you pointed out, each data-point consist of two components: the "true" traveled distance per day $d_i$ and measurement error $m_i$. Thus, our model is $Y = D + M$, where $D$ and $M$ are random variables and $d_i$ and $m_i$ ...


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