Talk:Pre- and post-test probability: Difference between revisions

Revision as of 14:27, 23 February 2011

Revision as of 14:57, 23 February 2011

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:<math> P( Cancer = True | Test = True) = \frac{ P(Cancer = True \and Test = True)}{P(Test=True)} = \frac{2}{2+8} = 0.10</math>

That seems '''way''' easier than the complicated 4 multi-step process described in the example. So why would you ever do it that way? <small><span class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Quadari|Quadari]] ([[User talk:Quadari|talk]] • [[Special:Contributions/Quadari|contribs]]) 14:26, 23 February 2011 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

In fact, it's easy to prove mathematically. Let's let the following table be true:

{| class="wikitable"

|-

! !! Disease = True !! Disease = False

|-

| Test = Positive || ''a'' || ''b''

|-

| Test = Negative || ''c'' || ''d''

|}

Where ''a,b,c,d'' are probabilities. I.e. ''a+b+c+d=1''. (This is without loss of generality.)

Then, the definitions as given in the article are:

* Sensitivity = <math>\frac{a}{a+c}</math>

* Specificity = <math>\frac{d}{b+d}</math>

And now we can just follow the algorithm of the article:

1. Likelihood ratio positive = sensitivity / (1 − specificity) = <math>\frac{a(b+d)}{b(a+c)}</math>

2. Pretest probability = <math>a+c</math>

3. Pretest odds = pretest prob / (1 - pretest prob) = <math>\frac{a+c}{1-(a+c)} = \frac{a+c}{b+d}</math>

4. Positive posttest odds = pretest odds * likelihood ratio positive = <math>\frac{a+c}{b+d}\cdot\frac{a(b+d)}{b(a+c)} = \frac{a}{b}</math>

5. Positive posttest probability = positive posttest odds / (1+positive posttest odds) = <math>\frac{a/b}{1+a/b} = \frac{a}{a+b}</math>

Thus we see that it would be '''way''' easier just to calculate the positive predictive value.

Now, all of this is assuming that the pretest probability for the patient in question is the same as the population probability. However, if that is not the case, then the entire chart is invalid. By using the chart you are assuming that the properties of the diagnostic test (i.e. the predictive values, sensitivity, specificity, etc.) are the '''same''' for the population (or the sample group) as they are for the patient in question. There's no reason to think that has to be the case. If we're willing to assume that the ''a+c'' for our patient is different than the sample group, why are we willing to assume that <math>\frac{a}{a+c}</math> is the same?

I just think that perhaps the article should point out some of this.