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I need to estimate the confidence interval for the efficiency η which is a function of 5 parameters $P1-P5$, $η = f(P1,…P5)$, with all parameters $P1-P5$ having normal distributions . I know the mean and SD for each of the parameters. How to determine the confidence interval for the parameter $\eta$ for different possible combinations of parameters $P1-P5$. Will $\eta$ have a normal distribution.

In that case, can I generate random samples from a multivariate normal distribution with mean = [ $\mu(P1)....\mu(P2)$] and the co-variance matrix, compute the $\eta$ for each sample (which is a set of parameters), then compute the mean and std deviation of the all the sample efficiency, and then finally compute the confidence interval by assuming the efficiency will also have a normal distribution.

user2821720
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    Do you have a variance-covariance matrix for these parameters? – gung - Reinstate Monica Jun 21 '16 at 01:45
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    The sense in which any such interval would be a "confidence" interval is unclear. Exactly what properties are you requiring of this interval? As far as the normality of $\eta$ goes, suppose $f$ equals $1$ when the sum of the $P_i$ is negative and otherwise equals $0$. Obviously $\eta$--which can attain at most two values--cannot be Normal, no matter what distribution $(P_1,\ldots,P_5)$ might have. – whuber Jun 21 '16 at 02:51
  • I do not know the variance-co variance matrix. But as i know the mean and variance of individual parameters, I can compute them by drawing $n$ samples. – user2821720 Jun 21 '16 at 03:00
  • I would like to know what is the most likely value for the efficiency $\eta$ if i take a random sample from the multivariate distribution (meaning $P1-P5$ takes some value in the range). So i should be able to say for example mean efficiency is $72$% $\pm$ $3$% with 95% confidence. – user2821720 Jun 21 '16 at 03:08
  • That is much closer in concept to a *prediction interval.* It does not appear to be any form of a confidence interval in the usual statistical sense of that term. – whuber Jun 21 '16 at 13:09
  • I think you have to something called simultaneous cofidence intervals. Refer more about that. – Sam88 Jul 11 '17 at 03:27

1 Answers1

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When you are using simulation to approximate confidence intervals there is no need to assume the normality of your random variable.

By simulating the parameters you can produce a large number $M$ of $\eta$s: $$\{\eta_1,\eta_2...,\eta_M\}$$

From this simulated sample a confidence interval can be constructed by properly choosing the sample quantiles.

Lets suppose you want a $95$% bilateral confidence interval and $M = 1000$. If $\{\eta_{(1)},\eta_{(2)},...,\eta_{(M)}\}$ is your simulated sample written in ascending order notice that $950$ observations fall between $\eta_{(26)}$ and $\eta_{(975)}$. For this reason $[\eta_{(26)},\eta_{(975)}]$ is a $95$% confidence interval.

This can be readily seen since:

$$\mathbb P(\eta_{(26)}\leq \eta \leq \eta_{(975)}) \approx \frac{950}{1000} =0.95$$

Mur1lo
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    Could you explain the sense in which this procedure produces a [confidence interval](http://stats.stackexchange.com/questions/26450) and not some other interval? – whuber Jun 21 '16 at 02:53
  • If I understood the question correctly, the questioner was asking for a way to produce a interval $[a,b]$ such that $\mathbb P(a \leq \eta \leq b) \geq \alpha$, if $\alpha$ is the desired probability. – Mur1lo Jun 21 '16 at 03:05
  • In this case, to compute the M of $\eta$s do i have to pick the random parameters samples $P1-P5$ from individual normal distribution or from a multivariate distribution. – user2821720 Jun 21 '16 at 03:26
  • It depends on the particularities of your problem. If you can assume the independence of P1,...,P5 than you can sample them separately. – Mur1lo Jun 21 '16 at 03:31