I'm sitting on a rather complicated task and I'm not sure which method to use here.

For a university class, we are consulting a psychologist who has shown people images like this:

enter image description here

The task was to

  • a) Find which of the diamonds has a different color than the others. This takes an amount of time called "selection time"

  • b) Find which side of the diamond is chipped off. This takes an amount of time called "discrimination time"

  • c) Report the side of the missing edge by clicking the left or right mouse button.

Now, we were only able to measure the total reaction time, i.e., selection plus discrimination time, because we don't know how much of a person's total reaction time is spent searching or, respectively, discriminating.

But we are actually interested in only the selection time, part a) of the above list. This is why we additionally performed the following experiment: Before showing people the picture, we let a little square (called a 'cue') flash up in the spot where the target diamond will appear. The person will know at what location the target diamond appears. By this we hoped to eliminate the selection time, thus only measuring part b), the discrimination time. But note that we are still interested in selection time, not discrimination time.

So our data set consists of 16 persons, with reaction times of around 360 shown pictures per person. About half of them were presented without the little flashing 'cue' square (which means the total reaction time includes the selection time), and the other half had the flashing square (measuring only discrimination time).

Our primary interest is the distribution of the selection times.

We heard a few keywords for which method we could use to extract that selection time. One was "deconvolution of distributions", then we heard about a "fast Fourier transform", which we're not sure how to use yet. Simple subtraction of reaction times without cue from the reaction times with cue might work, but we don't know how to pair those reaction times.

I know this sounds confusing, but I hope I could make our task somewhat clear. If something is not explained properly, I can of course clarify.

If you had any suggestions on which method might be appropriate here, we would be most grateful.

  • 1
    $\begingroup$ It's a very interesting question. Seems like they knew that this can be done or has been implemented in the literature. This one is for a psycometric guy I guess... I'm curious to read the answers they would give you. $\endgroup$
    – JDav
    Aug 5, 2012 at 12:45

1 Answer 1


If $X$ has density $f$ and $Y$ has density $g$ and $X$ and $Y$ are independent then $Z=X+Y$ has a density called the convolution of $f$ and $g$.

$$H(z)=P(Z\le z) = \int \int f(x) g(y)dy dx$$

where $x$ runs from $-\infty$ to $\infty$ and $y$ from $-\infty$ to $z-x$. Then $h(z)=H^\prime(z)$.

So the question becomes if you know $H$ and $f$ can you deconvolve to get $g$?

Here is a wikipedia article that describes deconvolution and gives you an idea of how the FFT enters into it: http://en.wikipedia.org/wiki/Deconvolution

  • $\begingroup$ We have around 3600 samples from H and another 3600 samples from f. We could estimate those densities with the samples - would that suffice? However, I'm not sure if X and Y are independent. They are two different kinds of time measured for one person. I suspect those are dependent. Would there be a way out of this? $\endgroup$ Aug 5, 2012 at 15:55
  • $\begingroup$ if there is dependence they you need to use conditional densities rather than unconditional in the formula. That would probably be more difficult to specify. $\endgroup$ Aug 5, 2012 at 16:37
  • $\begingroup$ So your g(y) would become a g(y|x), which is independent from f(x), right? $\endgroup$ Aug 5, 2012 at 16:43
  • $\begingroup$ Yes you can replace g(y) by g(y|x) but don't call it independent of x. It will be a function of x. $\endgroup$ Aug 5, 2012 at 17:04
  • $\begingroup$ Oops, of course. But the important thing is that f(x)*g(y|x) will be h(x,y), right? Just for independent X and Y, g(y|x) == g(y). $\endgroup$ Aug 5, 2012 at 18:28

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