6
$\begingroup$

I know that sensitivity (d') is typically calculated as follows:

d' = z(Hit Rates) - z (False Alarm Rates)

I understand this as the idea is to standardize both distributions, and to then calculate the z-score, which allows to calculate a distance within the same scale so to say.

What is however very confusing to me is the visualization of the process. I am using the following plots from these slides:

enter image description here

My questions are as follows:

(1) As in the plot above, d' is always depicted as the difference of the means of the two distributions (of signal and noise, for instance). Say however that I find in my data that the Hit rate (or Hit probability) is 0.8 - I would think that now, I would need to find z(Hit = 0.8), would this not be to the right of the mean:

enter image description here

Is this correct?

(2) In addition, I am am uncertain about the arrows associated with Z(CR) and Z(H) on panel A above.

For one, in my opinion, CR includes the entire area before the criterion. The arrow, however, associates Z(CR) with only the distance mean-criterion - why?

Then, if Z(X) refers to the cumulative probability density of variable X, Z(X) should be the entire density up to X. In the case of Z(CR) in panel A,however, it seems that it is only the part from the mean of the noise distribution to criterion. So what does then the Z(..) notation mean in this context?

$\endgroup$
  • $\begingroup$ It was not mean as a point - but as a "up to here". If you read my questions, this is also what I am asking as wrt to Z(CR), according to the source this seems not to denote the entire probability distribution, but just the part associated with the arrow. Also, how can Z(H) be the entire probability distribution, if hits are only what is above criterion? $\endgroup$ – user21198 Jan 2 at 20:57
  • 1
    $\begingroup$ Er, sorry, I think I need to look at this figure closer. If I were you I would focus on the probability distributions p instead. $\endgroup$ – Bryan Krause Jan 2 at 21:03
  • $\begingroup$ Welcome to psych.SE. I'm just wondering if this question would be a better fit for CrossValidated (stats.SE)? Does it have a particular connection to cognitive science? $\endgroup$ – Arnon Weinberg Jan 3 at 23:31
  • 1
    $\begingroup$ Thank you. Yes, sensitivity and criterion are two classical and possibly most used behavioral measures in cognitive neuroscience/psychophysics. For instance in 2AFC (two alternative forced-choice) tasks,stimulus absence is equated to noise, allowing to calculate sensitivity. If you mean my specific formal question - I would like to understand the concepts better when I use them in experiments. $\endgroup$ – user21198 Jan 3 at 23:44
2
$\begingroup$

First lets write down some math for the above figures assuming Gaussian distributions with means $\mu_{n}$ and $\mu_{s}$ both with a standard deviation of unity. In this case the probability of a detection is

$$P_D=1/(\sqrt{2\pi})\int_{c}^{\infty}e^{-(x-\mu_{s})^2/2}dx$$

and the probability of a false alarm is

$$P_F=1/(\sqrt{2\pi})\int_{c}^{\infty}e^{-(x-\mu_{n})^2/2}dx$$

and the normal deviates are

$$Z_D=\Phi^\prime(P_D)$$

and

$$Z_F=\Phi^\prime(P_F)$$

where $\Phi^\prime(x)$ is the inverse of $\Phi(x)$

$$\Phi(a)=1/(\sqrt{2\pi})\int_{-\infty}^{a}e^{-x^2/2}dx.$$

We also need to note that

$$1/(\sqrt{2\pi})\int_{-\infty}^{c}e^{-(x)^2/2}dx+1/(\sqrt{2\pi})\int_{c}^{\infty}e^{-(x)^2/2}dx=1$$

and that

$$\Phi(-x) = 1-\Phi(x).$$

Now using integration by substitution on $P_D$ and the above equalities gives us

$$P_D=1/(\sqrt{2\pi})\int_{c-\mu_{s}}^{\infty}e^{-(x)^2/2}dx=1-1/(\sqrt{2\pi})\int_{-\infty}^{c-\mu_{s}}e^{-(x)^2/2}dx=1-\Phi(c-\mu_{s})=\Phi(\mu_{s}-c).$$

You can play the same mathematical games and get that $P_F=\Phi(\mu_{n}-c)$ or that $Z_D=\mu_{s}-c$ and $Z_{CR}=c-\mu_{n}$ (this one is a little trickier ...).

Now on to your questions

(1) As in the plot above, d' is always depicted as the difference of the means of the two distributions (of signal and noise, for instance). Say however that I find in my data that the Hit rate (or Hit probability) is 0.8 - I would think that now, I would need to find z(Hit = 0.8), would this not be to the right of the mean:

This depends on what the means are. If the signal distribution has a mean of $-100\sigma$, then $Z_{hit}$ will likely be to the right of the mean, but if the signal distribution mean is $100\sigma$ it will likely be to the left.

In addition, I am am uncertain about the arrows associated with Z(CR) and Z(H) on panel A above.

For one, in my opinion, CR includes the entire area before the criterion. The arrow, however, associates Z(CR) with only the distance mean-criterion - why?

This is a direct result of the magic of the math above.

$\endgroup$
  • $\begingroup$ I think that is quite helpful. Assuming we have signal ~ N(1,1), noise ~ N(0,1). If I understand you correctly, then hit probability = 1 - normcdf (criterion, 1,1) <--> phi(mu_signal - C) <--> phi(mu_signal) - phi(C). With the same logic we get for False alarm probability: phi(c) - phi(noise), and when we subtract Hit rate - False alarms we get phi(mu_signal) - phi (mu_noise) --- is this correct? $\endgroup$ – user21198 Jan 2 at 23:23
  • 1
    $\begingroup$ @TestGuest I don't think your previous comment was correct. I will try and provide a less rushed answer later today. $\endgroup$ – StrongBad Jan 3 at 16:51
  • 1
    $\begingroup$ @TestGuest does the expanded math help? $\endgroup$ – StrongBad Jan 3 at 18:51
  • 1
    $\begingroup$ And, a very last question, related to (3): Are the distributions standardized (also with respect to their mean)? Because if so, then I understand that if for example criterion = -0.2, then p(D) = 1-phi(-0.2) = phi(0.2) = phi(mu-criterion). But if we would have a distribution N(1,1) and a criterion = 0.8, then phi(1.2) = p(D), which is obviously not equal to phi(mu-criterion) = phi(1-0.8)...many thanks again! $\endgroup$ – user21198 Jan 4 at 21:01
  • 1
    $\begingroup$ @TestGuest When you calculate z-scores by X-$\mu$/$\sigma$, the z-score corresponding to the mean of your distribution will always equal 0. This should answer (3) and (2). (1) is correct, as far as I can see. $\endgroup$ – Pegah Jan 4 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy