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I want to predict a factor and also get an upper bound for my prediction. It seems that what I want is prediction interval. but prediction interval only specifies the bound for one next prediction not all. Recently I found tolerance interval definition. Can I use it like prediction interval for future samples?

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    $\begingroup$ Can you give us some information on your design? Usually, a prediction interval does in fact give predictions for all coming observations (with respect to its error probability, of course) - however, if it is the right approach depends on your problem and design. $\endgroup$
    – Thilo
    Sep 12, 2011 at 14:14
  • $\begingroup$ I have some tasks that have some features and want to predict their duration and find an upper bound for my prediction for example say 95% of future tasks' duration will be in this region (or in tolerance interval language, with 90% degree of assurance 95% of data). I have tested different methods (Linear regression, ridge method, CCA, KCCA, Kernel ridge regression and Gaussian Process). It seems that only linear regression and Gaussian Process have methods to find the prediction std deviation. Now I want to know if I should use this std deviation for Prediction interval or Tolerance interval $\endgroup$ Sep 13, 2011 at 5:22

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Prediction intervals can indeed provide a interval for multiple forthcoming observations. The most common usage of the prediction interval is however when the number of forthcoming observations is one. For the normal distribution this is just a special case of the more general version with multiple forthcoming observations. As is seen in the standard ISO standard 16269.

From the standard follows:

A random sample of n observations has been drawn from a normally distributed population with unknown mean $\mu$ and unknown standard deviation $\sigma$. The sample mean is $\bar{x}$ and the sample standard deviation is $s$.

For given values of $n$, $m$ and $\alpha$ the smallest factor $k$ is required such that one may have at least $100(1-\alpha)%$ confidence that none of $m$ further observations will exceed $\bar{x}+ks$.

$k$ is given by $$\int_0^\infty{g\left(s\right)\int_{-\infty}^\infty{\Phi^m\left(\bar{x}+ks\right)f\left(\bar{x}\right)d\bar{x}ds}}\geq1-\alpha $$

where $$f(\bar{x})=\sqrt{\frac{n}{2\pi}}\exp\left(-\frac{n}{2}\bar{x}^2\right), -\infty\lt\bar{x}\lt\infty $$ $$ g(s)= \frac{\nu^{\nu/2}s^{\nu-1}}{2^{(\nu/2)-1}\Gamma(\frac{\nu}{2})}\exp\left(-\nu s^2/2\right),s\geq0 $$ $$ \Phi(t)=\int_{-\infty}^t{\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}u^2\right)du} $$ and $$ \Gamma\left(\frac{\nu}{2}\right)= \int_0^\infty{x^{\frac{\nu}{2}-1}\exp\left(-x\right)dx}$$ $$ \nu=n-1 $$

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