Notes:Statistical test results

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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 $p(1-u)\eq 0$
  • 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