# Uncertainties on Bayes' theorem

• A
Hello! I have a setup consisting of some charged particles each of which is produced at a different position, ##(x_i,y_i,z_i)##. I don't know the exact position, but I know that each of the 3 variables is normally distributed with mean zero and standard deviation of 3 mm. What I measure in the experiment is the position of the particles on a 2D screen perpendicular to the trajectory of motion of the particles, call it x' and y' and the time from the moment they are produced to when they hit the 2D screen, call it t. So what i measure is ##(x',y',t)## with the associated uncertainties, given by the detector, ##(dx',dy',dt)##. What I want to do is, for each ion, from these measurements, to get the most probable starting point (together with the associated uncertainty). I am not sure what is the best way to proceed. What I want to get basically is:

$$p(x_i,y_i,z_i|x_f,y_f,t)$$
and I will write this just as ##p(x_i|x_f)## for simplicity. For example, assuming we had no uncertainty on our measurements, I could use Bayes theorem and have:

$$p(x_i|x_f) = \frac{p(x_f|x_i)p(x_i)}{p(x_f)}$$

I know ##p(x_i)## (which is Gaussian) and for a given event, ##p(x_f)## is constant (I don't know what it is, but it is constant). For ##p(x_f|x_i)## I can get a value by running some simulations. Assuming everything would be ideal, this would be 1 for just one value and 0 for everything else. Given that we can't model the electrodes perfectly, and I will run the simulation several times with slightly different potentials, this will probably have a distribution, too (say Gaussian for simplicity). From here I can get ##p(x_i|x_f)## and the associated standard deviation, which is exactly what I need.

However, given that I have noise on ##(x_f,y_f,t)##, I can't just apply the Bayes theorem directly as above. How can I account for these errors? Thank you!

EDIT: I attached before a diagram of the trajectory of the particle (red line) from the source to the 2D detector.

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## Answers and Replies

Stephen Tashi
I don't know the exact position, but I know that each of the 3 variables is normally distributed with mean zero and standard deviation of 3 mm.

If that is the model, then the final position of a particle is distributed 3-D normally about the point (0,0,0). Is that the case?

Or perhaps you mean that the ##i##th particle is produced at some unknown location ##(x_i,y_i,z_i)## and travels to position ##(x_i+\triangle x, y_i + \triangle y, z_i + \triangle z)## where each of ##\triangle x, \triangle y , \triangle z## is distributed normally with mean zero and standard deviation 3 mm.

If that is the model, then the final position of a particle is distributed 3-D normally about the point (0,0,0). Is that the case?

Or perhaps you mean that the ##i##th particle is produced at some unknown location ##(x_i,y_i,z_i)## and travels to position ##(x_i+\triangle x, y_i + \triangle y, z_i + \triangle z)## where each of ##\triangle x, \triangle y , \triangle z## is distributed normally with mean zero and standard deviation 3 mm.
Hello! The initial position is distributed 3-D normally about the point (0,0,0). The final position is the one I measure.

Office_Shredder
Staff Emeritus
Gold Member
I have a couple questions

1.) Is there any uncertainty about the angle of motion, or is it 100% guaranteed to be in the z direction?

2.) Is the uncertainty in the detector also guassian?

3.) What kind of uncertainty do you have on the speed of the particle?

I have a couple questions

1.) Is there any uncertainty about the angle of motion, or is it 100% guaranteed to be in the z direction?

2.) Is the uncertainty in the detector also guassian?

3.) What kind of uncertainty do you have on the speed of the particle?
1) The motion is not in the z direction. As I mentioned the x' and y' of the detector are in a different plane compared to the starting point (I will add a simulation picture in the original post)

2) Yes

3) The uncertainty on the energy is also gaussian, but for simplicity we can assume that the energy is constant for all particles.

Office_Shredder
Staff Emeritus
Gold Member
Got it. Does the uncertainty in the detector measurement include uncertainty about how literally the particle follows the red line that you've drawn? I'm a little nervous that if the tube is long, very small deflections in the path of the particle could result in it striking a place on the screen at the end that is arbitrarily far from where you would "expect" it to hit.

Got it. Does the uncertainty in the detector measurement include uncertainty about how literally the particle follows the red line that you've drawn? I'm a little nervous that if the tube is long, very small deflections in the path of the particle could result in it striking a place on the screen at the end that is arbitrarily far from where you would "expect" it to hit.
In practice we have a very good control on the voltage on the potentials (I can check the actual relative uncertainty). But of course there will be small variations. This is why I mentioned that in practice I would slightly change the potential in the simulations to account for that. So basically, for a given initial point (that we don't know but it's fixed), the final point on the detector will have a distribution given by this variation in the voltage, even if the detector would have infinite resolution. In practice, on top of that, we also have the resolution of the detector which gives this ##(dx',dy',dt)##.

Twigg