 |
 |
Dear Mark,
Thank you for the helpful comment.
What you say is true if one uses the conditional probability that "the path starting at A, passes through x and B, knowing that A connects to B"which, following Saad is: p(A->x->B | A->B) = p(A->x->B)/waytotal(A->B)
But still: When combining p(A->x->B | A->B) and p(A<-x<-B | B->A), when looking for the probability that either the path from A to B or from B to A passes through x, it means that x should be the same voxels, no? However, when there is minimal overlap, x is actually not the same regions, which finally gives a wrongly inflated tract volume - or do I miss something completely?
Thank you so much!Markus
2014-09-08 6:11 GMT+02:00 Mark Jenkinson <[log in to unmask]>:
Dear Markus,
I think what you are not taking into account is that in case [2] where p(A->x->B) and p(A<-x<-B) are similar then the OR value is twice the value of the individual probabilities, whereas the product (AND) will be less than the individual probabilities (given they are always less than 1) which means that the result of subtracting the AND from the OR gives a value that is actually larger than the original probability and so will not take away from the tract.
I hope this helps.All the best,Mark
On 7 Sep 2014, at 04:35, Markus Gschwind <[log in to unmask]> wrote:
Dear all, and especially Saad,
I come back to this old discussion.
You explained the understanding of the combination of two independent paths from a probabilistic point of view (c.f. below).
Probtrackx calculates the probability that the path of least hindrance to diffusion from A passes through x: P(A->x). Using a waypoint B, the values become now the probability that the path of least hindrance to diffusion from A passes through x and B, i.e. p(A->x->B).
Now combining p(A->x->B) and p(A<-x<-B), what you want is the probability that either the path from A to B or from B to A passes through x, which means:
p( [A->x->B] or [A<-x<-B] ) = p( [A->x->B] ) + p( [A<-x<-B] ) - p( [A->x->B] and [A<-x<-B] )
Assuming the two events are independent (i.e. passing through x coming from A to B, or from B to A), the last probability is the product of both maps.
And Cherif detailed it later on:
p = [p( [A->x->B] )]/waytotal(a->b) + [p( [A<-x<-B] )]/waytotal(b->a) - ([p( [A->x->B] )]/waytotal(a->b)) * ([p( [A<-x<-B] )]/waytotal(b->a))
What I understand:The sum p(A->x->B)+p(A<-x<-B) describes the combination of both distributions, where the overlap is counted double (OR condition)
The product p(A->x->B)*p(A<-x<-B) describes the intersection, i.e. the voxels where both direction pass (AND condition)
However, I do not get the reason of the substraction of the product from the sum.
For exemple, in a case [1] where p(A->x->B) and p(A<-x<-B) travel in very different voxels with only very few overlap, the OR-region (p(A->x->B)+p(A<-x<-B) ) will be very large, and the AND-region (p(A->x->B)*p(A<-x<-B) ) will be very small.
However, in a case [2] where p(A->x->B) and p(A<-x<-B) travel in exactly the same voxels with maximal overlap, the OR-region (p(A->x->B)+p(A<-x<-B) ) will be (nearly) the same as the AND-region (p(A->x->B)*p(A<-x<-B) ) and the substraction of the product will thus take away a lot of the whole tract, although it was much a stronger tract with a clear definition than in case [1].
I am asking this because in one of our projects, we observed a much bigger volume of the OR-regions in one group compared to another group (potentially interesting), however, the AND-region has comparable size between both groups.
Could you please explain a little on this?
Thank you very much an advance!Markus
2008-10-01 10:55 GMT+02:00 Saad Jbabdi <[log in to unmask]>:
Hi -It is important to keep in mind what is the probability that you want to calculate, it will tell you whether you want to add, substract, multiply or divide.
Probtrackx calculates the probability that the path of least hindrance to diffusion from A passes through x: P(A->x). Using a waypoint B, the values become now the probability that the path of least hindrance to diffusion from A passes through x and B, i.e. p(A->x->B).
Now combining p(A->x->B) and p(A<-x<-B), what you want is the probability that either the path from A to B or from B to A passes through x, which means:
p( [A->x->B] or [A<-x<-B] ) = p( [A->x->B] ) + p( [A<-x<-B] ) - p( [A->x->B] and [A<-x<-B] )
Assuming the two events are independent (i.e. passing through x coming from A to B, or from B to A), the last probability is the product of both maps.
So in order to calculate the probability that you want, you would need to add both maps, and substract their product.
Cheers,Saad.
On 30 Sep 2008, at 18:25, Cherif Sahyoun wrote:
Hi Markus,
Matt and Saad will correct me if I'm wrong, but I'd say the way you're proposing will cause you to lose any meaningful relationship between the waytotal and your image. i.e. if you add, then essentially the statement that waytotal is the total number of streamlines that "made it" between your masks is preserved, but once you multiply your images, then it would seem like your waytotal is not as helpful...
(i like the idea of multiplying to get rid of outliers though... i've been thresholding as per previous posts).
Cherif.
------------------------------------------------------------------------------------------
Cherif P. Sahyoun HST-MEMP
Developmental Neuroimaging of Cognitive Functions
C: 617 688 8048
H: 617 424 6956
[log in to unmask]
"Live as if this were your last day. Learn as if you'll live forever"
Ghandi
-------------------------------------------------------------------------------------------
On Tue, Sep 30, 2008 at 3:41 AM, Markus Gschwind <[log in to unmask]> wrote:
Hi!
I am following your discussion with a lot of interest!
Here I have a question:
Why wouldn't you MULTIPLY (instead of add) the paths AB and BA?
I thought it is a more conservative measure of what is in common and trancking outliers will dorp out. (And take the mean of waytotals).
Thanks,
Markus
--
Dr Markus Gschwind
Functional Brain Mapping Laboratory | Campus Biotech - Neuroscience Department | phone +41.22.37.90.886
postal adress: 9 Chemin des Mines | Case postale 60 | 1211 Genève 20, Switzerland