# Re: [AUDITORY] question about c SOLVED! (Pierre Divenyi )

```Subject: Re: [AUDITORY] question about c SOLVED!
From:    Pierre Divenyi  <pdivenyi@xxxxxxxx>
Date:    Thu, 19 Sep 2013 08:35:37 -0700
List-Archive:<http://lists.mcgill.ca/scripts/wa.exe?LIST=AUDITORY>

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Sam,

I would very much like to see a reply from Neil Macmillan or Doug Creelman
to your proposed model. However, beyond the question I have on the
observer's evaluation of H and F (which, themselves, could also have biases
even if the person has a perfect memory for the history of his/her past
responses), I would surmise that the recency effect would give the x2 trace
a larger weight and, therefore, put a second-interval bias on y and,
consequently, also on k.

-Pierre

On 9/18/13 11:54 AM, "Sam Mathias" <smathias@xxxxxxxx> wrote:

Dear list,

Many thanks to everyone that replied regarding my question. Based on these
comments and some simulations I ran, I think I have the solution, which I
thought I'd share with everyone.

In a nutshell, it turns out that c = -0.5 * [z(H) + z(F)] is a perfectly
fine measure of response bias for 2I2AFC. However, it does not yield what I
call the "true criterion", which I explain below.

Imagine that on each trial, the listener generates two observations, x1 and
x2, which are Gaussian random variables with different means but the same
variance. The listener then computes the difference between them, y = x2 -
x1, and compares this value to a criterion. To avoid confusion, I call this
the "true criterion", k. If y > k, the listener responds "2nd", otherwise
responding "1st".

To get k, one needs to calculate c using the eq. above, and then MULTIPLY
the result by sqrt(2). I'm happy to supply some python code to illustrate
this on request.

Thanks again!

--
Dr. Samuel R. Mathias
Center for Computational Neuroscience and Neural Technology
Boston University
677 Beacon St., Boston, MA 02215

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pace; -webkit-line-break: after-white-space; color: rgb(0, 0, 0); font-size:=
16px; font-family: Calibri, sans-serif; "><div>Sam,</div><div><br></div><di=
v>I would very much like to see a reply from Neil Macmillan or Doug Creelman=
to your proposed model. However, beyond the question I have on the observer=
's evaluation of H and F (which, themselves, could also have biases even if =
the person has a perfect memory for the history of his/her past responses), =
I would surmise that the recency effect would give the x2 trace a larger wei=
ght and, therefore, put a second-interval bias on y and, consequently, also =
on k.</div><div><br></div><div>-Pierre</div><div><br></div><span id=3D"OLK_SRC=
_BODY_SECTION"><div><div>On 9/18/13 11:54 AM, "Sam Mathias" &lt;<a href=3D"mai=
lto:smathias@xxxxxxxx">smathias@xxxxxxxx</a>&gt; wrote:</div></div><div><br></di=
v><div dir=3D"ltr">Dear list,<div><br></div><div>Many thanks to everyone that =
replied regarding my question. Based on these comments and some simulations =
I ran, I think I have the solution, which I thought I'd share with everyone.=
</div><div><br></div><div>In a nutshell, it turns out that c =3D -0.5 * [z(H) =
+ z(F)] is a perfectly fine measure of response bias for 2I2AFC. However, it=
does not yield what I call the "true criterion", which I explain below.</di=
v><div><br></div><div>Imagine that on each trial, the listener generates two=
observations, x1 and x2, which are Gaussian random variables with different=
means but the same variance. The listener then computes the difference betw=
een them, y =3D x2 - x1, and compares this value to a criterion. To avoid conf=
usion, I call this the "true criterion", k. If y &gt; k, the listener respon=
ds "2nd", otherwise responding "1st".</div><div><br></div><div>To get k, one=
needs to calculate c using the eq. above, and then MULTIPLY the result by s=
qrt(2). I'm happy to supply some python code to illustrate this on request.<=
/div><div><br></div><div>Thanks again!</div><div><br></div><div><div>--&nbsp=
;<br>Dr. Samuel R. Mathias<br>Center for Computational Neuroscience and Neur=
al Technology<br>Boston University<div>677 Beacon St.,&nbsp;Boston, MA 02215=
<br><div><br></div><div><br></div></div></div><div dir=3D"ltr"><br></div></div=
></div></span></body></html>

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