Difference between revisions of "Notes:Statistical test results"

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Revision as of 09:04, 29 January 2018

[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