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Neyman Construction

Note to the reviewer: This obviously still needs a lot of work. The subject is turning out be more difficult than I originally thought. Please add input/ideas on how we can better. Thank you!

In 1937 Jerzy Neyman proposed a frequentist method to construct an interval at a confidence level $C,\,$ such that if we repeat the experiment many times the interval will contain the true value of some parameter a fraction $C\,$ of the time.

Theory

Assume $X_{1},X_{2},...X_{n}$ are random variables with joint pdf $f(x_{1},x_{2},...x_{N}|\theta _{1},\theta _{2},...,\theta _{k})$, which depends on k unknown parameters. For convenience, let $\Theta$ be the sample space defined by the n random variables and subsequentially define a sample point in the sample space as $X=(X_{1},X_{2},...X_{n})$
Neyman originally proposed defining two functions $L(x)$ and $U(x)$ such that for any sample point,$X$,

• $L(X)\leq U(X)$ $\forall X\in \Theta$
• L and U are single valued and defined.

Given an observation, $X^{'}$, the probability that $\theta _{1}$ lies between $L(X^{'})$ and $U(X^{'})$ is $P(L(X^{'})\leq \theta _{1}\leq U(X^{'})|X^{'})=0$ or $1$. These calculated probabilities fail to provide meaningful information to create an interval estimate of $\theta _{1}$. $\theta _{1}$ is either in the interval or not with probability 1 or 0.

Under the frequentist construct the model parameters are unknown constants and not permitted to be random variables. Considering all the sample points in the sample space as random variables defined the joint pdf above, that is all $X\in \Theta$ it can be shown that $L$ and $U$ are functions of random variables and hence random variables. Therefore one can look at the probability of $L(X)$ and $U(X)$ for some $X\in \Theta$. If $\theta _{1}^{'}$ is the true value of $\theta _{1}$, we can define $L$ and $U$ such that the probability $L(X)\leq \theta _{1}^{'}$ and $\theta _{1}^{'}\leq U(X)$ is equal to pre-specified confidence level$,C$.

That is,$P(L(X)\leq \theta _{1}^{'}\leq U(X)|\theta _{1}^{'})=C$ where $0\leq C\leq 1$ where $L(X)$ and $U(X)$ the upper and lower confidence limits for $\theta _{1}$

Coverage probability

The coverage probability, $C$, for Neyman construction is the frequency of experiments that the confidence interval contains the actual value of interest. Generally, the coverage probability is set to a $95\%$ confidence. For Neyman construction, the coverage probability is set to some value $C$ where $0<C<1$. This value $C$ tells how confidently that the true value is contained in the interval.

Implementation

A Neyman construction can be carried out by performing multiple experiments that construct data sets corresponding to a given value of the parameter. The experiments are fitted with conventional methods, and the space of fitted parameter values constitutes the band which the confidence interval can be selected from.

Classic Example

Suppose $X$~$N(\theta ,\sigma ^{2})$, where $\theta$ and $\sigma ^{2}$ are unknown constants where we wish to estimate $\theta$. We can define (2) single value functions, $L$ and $U$, defined by the process above such that given a pre-specified confidence level ,$C$, and random sample $X^{*}$=($x_{1},x_{2},...x_{n}$)

$L(X^{*})={\bar {x}}-{\frac {ts}{\sqrt {n}}}$

$U(X^{*})={\bar {x}}+{\frac {ts}{\sqrt {n}}}$

where ${\bar {x}}={\frac {1}{n}}\sum _{i=1}^{n}x_{i}={\frac {1}{n}}(x_{1},x_{2},...x_{n})$ , $s={\sqrt {{\frac {1}{n-1}}\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}}}$

and $t$ follows a t distribution with (n-1) degrees of freedom. $t$~t_{$({1-C}/2,n-1)$}

Another Example

$X_{1},X_{2},...,X_{n}$ are iid random variables, and let $T=(X_{1},X_{2},...,XZ_{n})$. Suppose $T\sim N(\mu ,\sigma ^{2})$. Now to construct a confidence interval with $\alpha$ level of confidence. We know ${\bar {x}}$ is sufficient for $\mu$. So,

$p(-Z_{\frac {\alpha }{2}}\leq {\frac {{\bar {x}}-\mu }{\sigma ^{2}}}\leq Z_{\frac {\alpha }{2}})=1-\alpha$

$p(-Z_{\frac {\alpha }{2}}\sigma ^{2}\leq {\bar {x}}-\mu \leq Z_{\frac {\alpha }{2}}\sigma ^{2})=1-\alpha$

$p({\bar {x}}-Z_{\frac {\alpha }{2}}\sigma ^{2}\leq \mu \leq {\bar {x}}+Z_{\frac {\alpha }{2}}\sigma ^{2})=1-\alpha$

This produces a $100(1-\alpha )\%$ confidence interval for $\mu$ where,

$L(T)={\bar {x}}-Z_{\frac {\alpha }{2}}\sigma ^{2}$

$U(T)={\bar {x}}+Z_{\frac {\alpha }{2}}\sigma ^{2}$.