Difference between revisions of "Notes:Statistical test results"

Revision as of 09:04, 29 January 2018 (view source) Alec (Talk | contribs) (Created page with "{{ProbMacros}}__TOC__ ==Notes== Let {{M|T:\eq(u,v)}} be a statistical test, and {{M|P\eq 1}} denote the thing being tested for being "true", then: * {{M|\Pcond{T\eq 1}{P\e...")		Latest revision as of 09:17, 29 January 2018 (view source) Alec (Talk | contribs) (Adding more stuff)
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	*** Thus we observe we must have {{M|1-u\eq 0}} or {{M|u\eq 1}}		*** Thus we observe we must have {{M|1-u\eq 0}} or {{M|u\eq 1}}
	**** 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'''		**** 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'''
		+	**** we could also say that '''''[[false negatives]]'' are an impossibility'''
		+	* a [[corollary]] to this is that if {{M|u\eq 1}} then '''testing ''negative'' means you can be completely certain that the subject ''does not'' have the property'''
		+	===Under the conditions of false negatives being an impossibility===
		+	Then:
		+	* {{MM|\Pcond{P\eq 1}{T\eq 1}\eq\frac{p}{p+(1-p)(1-v)} }}
		+	* {{MM|\Pcond{P\eq 0}{T\eq 1}\eq\frac{(1-p)(1-v)}{p+(1-p)(1-v) } }}
		+	* {{MM|\Pcond{P\eq 1}{T\eq 0}\eq 0}} - as discussed above
		+	* {{MM|\Pcond{P\eq 0}{T\eq 0}\eq 1}}
		+	==Analysis==
		+	: Here I document a form of analysis I like to apply in some areas of statistics and probability, it's not named but extremely useful
		+	To study tests I like to make the following definitions:
		+	* {{M|p\eq 10^{-k} }}
		+	** This means that "1 in {{M|10^k}} subjects have the property", for {{M|k\eq 0}} it's 1 in 1 (certainty), for {{M|k\eq 1}} it's 1 in 10, for {{M|k\eq 2}} it's 1 in 100, so on so forth
		+	** Notice how after {{M|k\eq 0}} everything is quite "rare" (as 1 in 10 is certainly not common)
		+	** This is the ''rarity'' convention, as larger k means the property is rarer.
		+	* Conversely we could want "9 in 10" or "99 out of 100", in this case let {{M|q:\eq 1-p}} now:
		+	** {{M|q\eq 1-10^{-k} }} is how we'd define it, so {{M|k\eq 0}} is 0 in 1 (impossibility), for {{M|k\eq 1}} it's 9 in 10, for {{M|k\eq 2}} it's 99 in 100, so on so forth
		+	** This is the ''commonality'' convention
		+	These are also very useful when plotted, compared to a plot that shows {{M|p}} directly on the range {{M|[0,1]}} as it'll get very steep - by using {{M|k}} for {{M|k\in\mathbb{R}_{\ge 0} }} a lot of the situation is visible.
	==Notes==		==Notes==
	<references group="Note"/>		<references group="Note"/>

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Latest revision as of 09:17, 29 January 2018

$\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } }$$\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } }$$\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } }$$\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} }$

Notes

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

• $\Pcond{T\eq 1}{P\eq 1}\eq u$ and
• $\Pcond{T\eq 0}{P\eq 0}\eq v$

Here we will investigate $\P{T\eq 1}$, $\Pcond{P\eq 1}{T\eq 1}$, $\Pcond{P\eq 0}{T\eq 1}$ and so forth

Observations

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

Findings

• $$\P{T\eq j}\eq \P{P\eq 0}\Pcond{T\eq j}{P\eq 0}+ \P{P\eq 1}\Pcond{T\eq j}{P\eq 1}$$
• $$\Pcond{P\eq i}{T\eq j}:\eq\frac{\P{P\eq i\cap T\eq j} }{\P{T\eq j} }$$ $$\eq \frac{\Pcond{T\eq j}{P\eq i}\P{P\eq i} }{\P{T\eq j} }$$
  • Which we develop:

    $$\eq \frac{\Pcond{T\eq j}{P\eq i}\P{P\eq i} } { \P{P\eq 0}\Pcond{T\eq j}{P\eq 0} + \P{P\eq 1}\Pcond{T\eq j}{P\eq 1} }$$ - notice the denominator only depends on $j$ - the value of $T$

  • Notice:
    • We can find $\Pcond{T\eq j}{P\eq i}$ from the definition of $T$
    • $\P{P\eq i}$ must come from somewhere
    • $\P{T\eq j}$ - we will find below

We make the following definitions:

• Let $\P{P\eq 1}:\eq p$

Then:

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

The result $\Pcond{P\eq 1}{T\eq 0}$ 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 $\Pcond{P\eq 1}{T\eq 0} \eq 0$ - no chance of having the property if your test was negative - that we require $p(1-u)\eq 0$^{[Note 1]}
  • if $p\eq 0$ (i.e. $\P{P\eq 1} \eq 0$) then this is a pointless test.
    • Thus we observe we must have $1-u\eq 0$ or $u\eq 1$
      • 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
      • we could also say that false negatives are an impossibility
• a corollary to this is that if $u\eq 1$ then testing negative means you can be completely certain that the subject does not have the property

Under the conditions of false negatives being an impossibility

Then:

• $$\Pcond{P\eq 1}{T\eq 1}\eq\frac{p}{p+(1-p)(1-v)}$$
• $$\Pcond{P\eq 0}{T\eq 1}\eq\frac{(1-p)(1-v)}{p+(1-p)(1-v) }$$
• $$\Pcond{P\eq 1}{T\eq 0}\eq 0$$ - as discussed above
• $$\Pcond{P\eq 0}{T\eq 0}\eq 1$$

Analysis

Here I document a form of analysis I like to apply in some areas of statistics and probability, it's not named but extremely useful

To study tests I like to make the following definitions:

• $p\eq 10^{-k}$
  • This means that "1 in $10^k$ subjects have the property", for $k\eq 0$ it's 1 in 1 (certainty), for $k\eq 1$ it's 1 in 10, for $k\eq 2$ it's 1 in 100, so on so forth
  • Notice how after $k\eq 0$ everything is quite "rare" (as 1 in 10 is certainly not common)
  • This is the rarity convention, as larger k means the property is rarer.
• Conversely we could want "9 in 10" or "99 out of 100", in this case let $q:\eq 1-p$ now:
  • $q\eq 1-10^{-k}$ is how we'd define it, so $k\eq 0$ is 0 in 1 (impossibility), for $k\eq 1$ it's 9 in 10, for $k\eq 2$ it's 99 in 100, so on so forth
  • This is the commonality convention

These are also very useful when plotted, compared to a plot that shows $p$ directly on the range $[0,1]$ as it'll get very steep - by using $k$ for $k\in\mathbb{R}_{\ge 0}$ a lot of the situation is visible.

Notes

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