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This question is inspired by Confidence Interval on a random quantity?. That question introduces an interesting concept for a type of interval that is neither a prediction nor a confidence interval (possibly one could see it as a tolerance interval although I believe it is neither that).


A frequentist interval estimate

In short: For pairs of (possibly multidimensional) variables $x_i,y_i$, which are both distributed according to a distribution parameterized by $a$, and where $x_i|a \not\!\perp\!\!\!\perp y_i|a$, we wish to perform interval estimation for the value of $x_i$ as function of $y_i$, where $a$ is unknown.

Given the following:

  • Let $X,Y$ be random variables that are paired.
  • The random variables $X$ and $Y$ follow a distribution function that is parameterized by $a$ $$f_{Y|a}(y|a) \equiv g_Y(y,a)$$ $$f_{X|a}(y|a) \equiv g_X(y,a)$$
  • There is a known relationship between $X$ and $Y$ and $a$, that defines a conditional distribution for $X$ $$f_{X|y,a}(x|y,a) \equiv h(x,y,a)$$
  • There is a sample of measured values $y_i$

We wish to compute:

for each $x_i$ a one-sided interval bound $c(y_i,\alpha)$ such that: $$\forall a : P(X<c(Y,\alpha)) = \alpha$$ or less strong $$\sup \lbrace P(X<c(Y,\alpha)):a \rbrace = \alpha$$

That is, probability in a frequentist sense. If we would have a large sample with pairs $x_i,y_i$ (where we only measure $y_i$ and do not know $a$) then the frequency/fraction of 'failures' of the interval, $x_i<c(y_i,\alpha)$, should be around $\alpha$ independent from the true value of $a$ (or the smallest upper bound is $\alpha$).


How do/should we call that sort of interval?

This is not a confidence interval, because the estimate is for $X$, which is not a (fixed) population parameter, but a random variable.

This is neither a prediction interval, because $c(y_i,\alpha)$ is only a region for the $x_i$ that is paired with $y_i$ and it is not a region for future values of $X$.

What is it?


Example case problems

  • (this one was mentioned by shabbychef in the comments and relates to the before mentioned question)

    You observe returns from $p$ stocks in vector $\vec{y}_i$. Then from a sample of $n$ such observations, you form the Markowitz Portfolio, based on the sample mean and covariance. Then you wish to estimate the Sharpe Ratio of that sample Markowitz Portfolio.

  • Say I have a batch of films for which I want to predict the strength $X$ of each film. Let the strength be a function of two parameters, say film thickness $Y$ and film density $a$.

    Say I can not measure $X$ directly (would damage the film), and I do not know $a$ for every film, nor do I wish to measure it (say it is a costly measurement). I can, however, measure $Y$ for each film and I know that $Y$ is distributed according to some pdf that is parameterized by $a$.

    So now the idea is to use measurements of film thickness $Y$, which carries information of $a$ to compute some confidence/prediction/tolerance/whatever interval for $X$ which I know depends on $Y$ and $a$. I want this interval to fail only $\alpha$ percent of the time.

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  • $\begingroup$ I think it's not useful to make the stipulation in the fourth (final) bullet, due to the dependence on the unknown parameter $a.$ You need to consider either the supremum or the infimum of the left hand side over the set of posited distributions of $Y,$ depending on your objective. $\endgroup$ – whuber Jan 28 at 22:37
  • $\begingroup$ I agree. That is what I did in my answer here. Beyond that one may wonder whether there ain't better approaches for the problem in practice (but that is beyond the point of the question which is about the principle). $\endgroup$ – Martijn Weterings Jan 28 at 22:48
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    $\begingroup$ Another example would be: you observe returns from $p$ stocks in vector $\vec{y_i}$. Then from a sample of $n$ such observations, you form the Markowitz Portfolio, based on the sample mean and covariance. Then you wish to estimate the Sharpe Ratio of that sample Markowitz Portfolio. $\endgroup$ – shabbychef Jan 29 at 5:42
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We could describe the distribution of $Y$, conditional on $X$ and $a$, as a distribution parameterized by $X$ and $a$:

$$f_{Y|x,a}(y,x,a) = \frac{f_{X|y,a}(x,y,a)f_{Y,a}(y,a)}{f_{X,a}(x,a)}$$

In this view the random variable $X$ is a parameter in the (conditional) distribution of $Y$, and we could see the interval estimation of $X$ as a confidence interval for the parameter $X$.

Complications are that the estimate of $X$ is dependent on the value of the parameter $a$ which acts as a nuisance parameter, and in addition $X$ itselve is distributed according to distribution parameterized by $a$. So one may not tackle the interval estimation as a 'regular' confidence interval estimation.

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