Difference between revisions of "Notes:Statistical test results"

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Revision as of 09:04, 29 January 2018

[ilmath]\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }[/ilmath]
[ilmath]\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } } [/ilmath][ilmath]\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} } [/ilmath]

Notes

Let [ilmath]T:\eq(u,v)[/ilmath] be a statistical test, and [ilmath]P\eq 1[/ilmath] denote the thing being tested for being "true", then:

  • [ilmath]\Pcond{T\eq 1}{P\eq 1}\eq u[/ilmath] and
  • [ilmath]\Pcond{T\eq 0}{P\eq 0}\eq v[/ilmath]

Here we will investigate [ilmath]\P{T\eq 1} [/ilmath], [ilmath]\Pcond{P\eq 1}{T\eq 1} [/ilmath], [ilmath]\Pcond{P\eq 0}{T\eq 1} [/ilmath] and so forth

Observations

To find say [ilmath]\P{P\eq i} [/ilmath] you'd have to have [ilmath]\P{T\eq j} [/ilmath] for [ilmath]j\eq 0[/ilmath] and [ilmath]j\eq 1[/ilmath] already known - these both require some knowledge about the population

Findings

  • [math]\P{T\eq j}\eq \P{P\eq 0}\Pcond{T\eq j}{P\eq 0}+ \P{P\eq 1}\Pcond{T\eq j}{P\eq 1} [/math]
  • [math]\Pcond{P\eq i}{T\eq j}:\eq\frac{\P{P\eq i\cap T\eq j} }{\P{T\eq j} } [/math][math]\eq \frac{\Pcond{T\eq j}{P\eq i}\P{P\eq i} }{\P{T\eq j} } [/math]
    • Which we develop:
      [math]\eq \frac{\Pcond{T\eq j}{P\eq i}\P{P\eq i} } { \P{P\eq 0}\Pcond{T\eq j}{P\eq 0} + \P{P\eq 1}\Pcond{T\eq j}{P\eq 1} } [/math] - notice the denominator only depends on [ilmath]j[/ilmath] - the value of [ilmath]T[/ilmath]
    • Notice:
      • We can find [ilmath]\Pcond{T\eq j}{P\eq i} [/ilmath] from the definition of [ilmath]T[/ilmath]
      • [ilmath]\P{P\eq i} [/ilmath] must come from somewhere
      • [ilmath]\P{T\eq j} [/ilmath] - we will find below

We make the following definitions:

  • Let [ilmath]\P{P\eq 1}:\eq p[/ilmath]

Then:

  • Results given the test evaluates to positive
    • [math]\Pcond{P\eq 1}{T\eq 1}\eq\frac{pu}{(1-p)(1-v)+pu} [/math]
      • Notice that next we could find [ilmath]\Pcond{P\eq 0}{T\eq 1} [/ilmath] as [ilmath]1-\Pcond{P\eq 1}{T\eq 1} [/ilmath]
    • [math]\Pcond{P\eq 0}{T\eq 1}\eq\frac{(1-p)(1-v)}{(1-p)(1-v)+pu} [/math]
  • Results given the test evaluates to negative
    • [math]\Pcond{P\eq 1}{T\eq 0}\eq\frac{p(1-u)}{(1-p)v+p(1-u)} [/math]
      • Notice that next we could find [ilmath]\Pcond{P\eq 0}{T\eq 0} [/ilmath] as [ilmath]1-\Pcond{P\eq 1}{T\eq 0} [/ilmath]
    • [math]\Pcond{P\eq 0}{T\eq 0}\eq\frac{(1-p)v}{(1-p)v+p(1-u)} [/math]

The result [ilmath]\Pcond{P\eq 1}{T\eq 0} [/ilmath] is very important in diagnostic tests as this would be a subject that has the property but failed the test, usually the function of a (preliminary at least) test is to not miss any possible subjects - usually at the costs of more false positives - which are cases where the test was positive, but the property is absent.


Specifically:

  • Notice that to have [ilmath]\Pcond{P\eq 1}{T\eq 0} \eq 0[/ilmath] - no chance of having the property if your test was negative - that we require [ilmath]p(1-u)\eq 0[/ilmath][Note 1]
    • if [ilmath]p\eq 0[/ilmath] (i.e. [ilmath]\P{P\eq 1} \eq 0[/ilmath]) then this is a pointless test.
      • Thus we observe we must have [ilmath]1-u\eq 0[/ilmath] or [ilmath]u\eq 1[/ilmath]
        • this is to say in order to have a subject with the property failing the test being an impossibility we require the probability of the test being positive given the subject has the property is complete certainty

Notes

  1. The reader should convince himself that a limit where the denominator tends to positive infinity cannot happen