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In calculating sensitivity rates where TP =0 and FN =0 and the formula is TP/(TP+FN) - although it mathematically won't compute, does this equate to 100% sensitivity since it has correctly identified no true-positives?

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You can look at this as binomial proportions. I.e. proportion of correctly identified cases out of the ones that should be identified.

So you could use all the theory for estimates (and confidence intervals) for sparse binomial data. One reasonable approach could be to assume a Beta(1/3, 1/3) (or a Beta(1/2, 1/2) or Beta(1,1) or something based on actual prior knowledge) prior and do a Bayesian analysis.

Admittedly, in your case, you essentially have absolutely no data about sensitivity, so your posterior distribution would just be your prior. Once you have low numbers and just one of them being zero, this approach starts to be more interesting.

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    $\begingroup$ Thanks Björn for taking your time to reply. I might have to include more data then to see what it picks up. $\endgroup$ – cms72 Jun 30 '19 at 13:17
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$TP$ and $FN$ are discrete variables. Let's pretend as if they're continuous, and investigate the limiting case using well-known techniques in multivariable calculus. For simpler notation, let $x=TP, y=FN$; and we want to find the value of the multivariate function $f(x,y)=\frac{x}{x+y}$ at $(x,y)=(0,0)$. Since the value is undefined, we can't substitute directly and need to observe the limiting behaviour: $$L=\lim_{(x,y)\rightarrow(0,0)} f(x,y)$$ In order for this limit to be defined and calculated, as we approach origin from any path, it should give the same limiting value. If limits differ while approaching towards different angles, the limit doesn't exist.

  1. Approach from $x=0$ line: $$L_1=\lim_{y\rightarrow 0} f(0,y)=\lim_{y\rightarrow 0}\frac{0}{0+y}=\lim_{y\rightarrow 0}0=0$$
  2. Appriach from $y=0$ line: $$L_2=\lim_{x\rightarrow 0} f(x,0)=\lim_{x\rightarrow 0}\frac{x}{x+0}=\lim_{x\rightarrow 0}\frac{x}{x}=\lim_{x\rightarrow 0}1=1$$

Since $L_1\neq L_2$, $L$ doesn't exist. If we use similar ideas for the discrete case, i.e. pretend as if the values decrease by $1$ as they approach, since there is no continuity; depending on the direction of approach, we get different result.

Mathematically, there is no reliable definition of this ill case; however, one can define the recall/sensitivity as $1$, since we've found every Positive sample.

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  • $\begingroup$ Thanks gunes for taking the time to explain this. I might have to add more data to my analysis to see if this changes - however, if it remains TP=0 and FN=0, I will try to justify it using your approach on limits. Thank you. $\endgroup$ – cms72 Jun 30 '19 at 13:28

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