I am not a psychology major, so forgive me if I'm misusing terms.

So on an exam, we had a question about signal detection, and it defined a conservative person as someone who always tend to reply negatively ("no"-s), rather than agree ("yes"-s).

I believe that this isn't correct. Consider the TSA for example, they are extremely liberal in the sense that they would screen you even when they have minimal suspicions about you. However, the percentage of people they screen is still very low, because the probability of the signal itself is very low.

Do you know any source in literature that could point out why that is wrong? Or maybe could you come up with a counter-argument that would make sense.


1 Answer 1


As I said in this answer, for these types of questions I really like Detection Theory: A User's Guide by Macmillan and Creelman. They consider 3 types of bias ($c$, $c'$, and $\beta$) that differ in how they behave when the index of sensitivity $d'$ changes, but all three of their definitions agree with your professor that a conservative observer always replies no more often than yes, regardless of the probability of a signal being present. All three definitions also lead to the conclusion that the ideal (maximum likelihood) observer is biased when the probability of a signal being present is not equal to 0.5 (with the standard Gaussian distributions with equal variance assumptions). While it may be uncomfortable to you that the ideal observer is biased, this happens all the time in estimation. That said, Macmillan and Creelman defined three different biases, there is no reason we cannot define a fourth that behaves as you want it to ...

I suggest starting with Macmillan and Creelman's definition of $\beta$. They define $\beta$ to be equal to $p(x|S_2)/p(x|S_1)$ where $p(x|S_1)$ is the probability of observing $x$ given stimulus $S_1$. Within their framework, an unbiased observer has a $\beta$ of one. If we define $\beta'$ to be $p(S_2|x)/p(S_1|x)$ and an unbiased observer as having a $\beta'$ equal to 1, then the ideal observer is unbiased. That seems like a nice result. In the case where $p(S_1)$ is equal to $p(S_2)$, $\beta$ is equal to $\beta'$, so that is good too. When $p(S_1)$ is less than $p(S_2)$ a conservative observer could say yes more than no, which I don't like, but that is the case when we force the ideal observer to be unbiased.

  • $\begingroup$ I am sorry, I guess I wasn't clear in stating my view. please consider this example: if we have 100 individuals to pass by the TSA inspection. An unbiased officer would usually detect a malicious object in 5 individuals. A liberal one would detect 10 or 20. So even though he is liberal, he tends to say no much more often, because the signal itself is weak. $\endgroup$
    – user534055
    Commented Mar 8, 2019 at 8:43
  • $\begingroup$ Oh now I get it. so the TSA officer who detects 5 only is considered a conservative by the definition. This is really weird to accept because they aren't taking into consideration the differences in signal strengths. But at least I understand it now. Thank you $\endgroup$
    – user534055
    Commented Mar 8, 2019 at 8:51
  • $\begingroup$ @user534055 they are conservative according to c, c', and beta. With the beta' I defined, they would be liberal. Signal strength, is related to d' and how reliably the TSA agent can discriminate between the classes. You are talking about the probability of the signal, which is ignored in c, c', and beta, but not in my beta'. $\endgroup$
    – StrongBad
    Commented Mar 8, 2019 at 12:28
  • $\begingroup$ doesn't d' represent the salience of the signal and how easy it is for the person to distinguish between a signal and noise? is it related to the probability of signal? Your beta' is the greatest among all, and makes sense the most. $\endgroup$
    – user534055
    Commented Mar 9, 2019 at 13:47
  • $\begingroup$ I am interested in knowing if the lack of such ability in beta, c, and c', would draw researchers to conclusions that are not accurate ( like considering the TSA to be conservative) because they don't take into consideration the probability of occurrences of signals. $\endgroup$
    – user534055
    Commented Mar 9, 2019 at 13:52

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