Dear SPM list,

Suppose you find an activation cluster that you believe
may originate from a small midbrain nucleus, and you want to test
whether it is significant at a p-value that is properly
corrected for multiple comparisons.

One way would be to contrive a small-volume correction, after
dredging PubMed to find suitable previous papers to justify your ROI.
Hey presto, the small-volume correction will show that your cluster
is significant. This doesn't feel quite right.
(By the way, I'm not suggesting that all small-volume corrections
are bad in this way, I'm just saying that it is unsatisfying here).

Because the cluster is small, other multiple-comparisons correction
methods will probably not count it as significant:

1. The voxels' intensity will probably not be very high,
as tissue from outside the nucleus will get smoothed in with
the active tissue from the nucleus. So, voxelwise correction will
tend not to count the nucleus's activity as significant.

2. Cluster-based correction judges clusters by their size.
So, small clusters won't make the cut.

3. FDR requires a lot of low p-value voxels in order for the sorted
p-value curve to be able to sneak under the alpha-sloped line.
If only small brain areas are active, there won't be many low p-values.

4. Non-parametric multiple-comparisons correction in SnPM
works by recording the maximum suprathreshold cluster size.
Small nuclei will produce small clusters, and hence small maximum sizes.
Again, the small cluster won't make the cut.

Given this, it seems to me that even if a small nucleus genuinely
is active, the resulting fMRI cluster will fail to be counted as 
significant by existing multiple-comparisons correction methods.

However, I'm not at all sure that this reasoning is correct.
I'd be very interested to hear people's thoughts on this issue.

Many thanks,

Raj
P.S. Of course, given that various small midbrain nuclei are all nestled
right next to each other, even a significant cluster still presents us
with the tricky problem of figuring out which nucleus it actually is.
But that's a separate question. I'm wondering whether it can even get
past the corrected p-value significance hurdle.