Understanding the concept behind d-prime

Question

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:

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:

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?

Answer 1

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.