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• kelly0303
kelly0303
Hello! I am confused about the results I am getting for an apparently simple situation. I have 2 measurements (counts), call them ##S_+## and ##S_-##. Based on these I build an asymmetry defined as:

$$A = \frac{S_+-S_-}{S_++S_-}$$

The parameter I need to extract experimentally, call it ##x## behaves like ##\frac{dx}{x} = \frac{dA}{A}## where ##dx## and ##dA## are the uncertainties on ##x## and ##A## (ignore systematic uncertainties for now). ##x## is fixed (given by the physics process I am studying) and let's say I want to extract ##x## with ##10\%## relative uncertainty i.e. ##\frac{dx}{x} = \frac{dA}{A} = \frac{1}{10}##. I have 2 situations (I will give the actual numbers I get). In the first one I have:

$$S_+ = 0.0484 N$$
$$S_- = 0.0324 N$$
where ##N## is the number of initial events and ##S_+## and ##S_-## are the events I am actually measuring. In the second case I have:

$$S_+ = 0.0085 N$$
$$S_- = 0.0027 N$$

Using the formula above, in the first case I am getting ##A_1 = 0.198## and in the second case I am getting ##A_2 = 0.519##. If I do an error propagation, I end up with the formula:

$$dA = \frac{2}{(S_++S_-)^2}\sqrt{S_+S_-^2+S_+^2S_-}$$
from which I get ##dA_1 = \frac{3.448}{\sqrt{N}}## and ##dA_2 = \frac{8.083}{\sqrt{N}}##. So I get ##\frac{dA_1}{A_1} = \frac{17.4}{\sqrt{N}}## and ##\frac{dA_2}{A_2} = \frac{15.6}{\sqrt{N}}##. Which means that in the first case I need about ##N_1 = 30276## events and in the second case I need ##N_2 = 24336## events. But this doesn't make sense to me. For a fixed ##N##, in the first case the number of events I am actually measuring are about an order of magnitude bigger than in the second case. Given that I am only looking at the statistical uncertainty, I would expect to need ~100 times more events in the second case, to reach the same uncertainty on the parameter of interest i.e. ##N_2 \sim 100N_1##. What am I doing wrong? Shouldn't I use that error propagation on ##A##? What should I do such that the uncertainty on ##x## reflects that fact that in the second case I have much lower statistics? Thank you!

If S = k ⋅N then σS = k⋅√N. It is ≠ √(k⋅N).

I think you missed this.

BvU
gleem said:
If S = k ⋅N then σS = k⋅√N. It is ≠ √(k⋅N).

I think you missed this.
Thank you for this. I am not sure I get that. For example, for the first case, for ##N = 30276## I have ##S_+ = 1465##. Shouldn't the uncertainty be given by the events I am actually measuring i.e. ##\sqrt{S_+} = 38##. Just to clarify a bit, in my case ##N## is well known (i.e. there is no uncertainty on ##N##). The coefficient k in this case is actually a binomial distribution coefficient, so for each event out of the N ones, I have k probability to get a count. On average I have ##k\cdot N## events.

kelly0303 said:
Just to clarify a bit, in my case N is well known (i.e. there is no uncertainty on N)
I don't understand. How is N determined if it is events so if you repeat the experiment you get the same number?

Edit: Also how is the coefficient of N determined?

gleem said:
I don't understand. How is N determined if it is events so if you repeat the experiment you get the same number?

Edit: Also how is the coefficient of N determined?
I am sorry for the confusion. I will try to give a bit more details (I should have done it from before). We have a two level quantum system (an atom in this case), which we prepare in a given state. After a fixed time we measure the probability of the system to be in the other state. This transition can be described by a binomial distribution with probability ##p<<1## (in practice this is done by detecting an ion after a given amount of time). We do this one atom at a time, so we prepare an atom in the initial state (this is done with basically 100% efficiency), then wait, then try to detect an ion (if the transition didn't happen we detect nothing). We repeat this ##N## times, so N is exactly known (i.e. there is no uncertainty associated to N), as that is given simply by how many time we repeat this initial state preparation. Then we do the measurement and extract ##S_+## as the number of detected events. We then change some experimental parameters and redo the experiment N other times and define ##S_-## as the number of events in this experimental configuration. Depending on the experimental setup, we can increase or decrease the values of ##S_+## and ##S_-##. From this step, I do the analysis in the original post.

Just for completeness, I have the following formula:

$$S_{\pm} = N(a^2 \pm ax)$$
where a is an experimental parameter, and we can assume ##a>>x##.

So you want the uncertainty in S/N?

gleem said:
So you want the uncertainty in S/N?
What I need is the uncertainty on x.

How do you define x?

gleem said:
How do you define x?
I added a formula to my post above

a is fixed?

gleem said:
a is fixed?
For a given experimental run (i.e. in order to perform N measurements), yes (and we can assume it doesn't have any uncertainty associated to it).

So x= (S/N-a2)/a

I would say that

σx = σS/(aN)

gleem said:
So x= (S/N-a2)/a

I would say that

σx = σS/(aN)
what would ##\sigma_S## be in this case?

S is the number of detected events with a binomial distribution thus σs = √S

Got to go will be back is about an hour.

gleem said:
S is the number of detected events with a binomial distribution thus σs = √S
Thanks a lot for help and no worries! So doing what you suggested, given that a>>x, we can assume ##S = Na^2## so ##\sigma_S = \sqrt{N}a##. Then, using the formula you provided we get ##\sigma_x = \sigma_S/(aN) = \sqrt{N}a/(aN) = \frac{1}{\sqrt{N}}##. This is basically consistent with what I am getting, but I am still not sure I understand conceptually why. By increasing a, we can detect more events (for fixed N). So I would expect to have a reduced uncertainty for higher values of a (still with 1>a>>x). Why the final formula for uncertainty on x doesn't depend at all on a?

Hi,

I reproduced your results (with difficulty -- as I'm somewhat rusty).

I think it's just a numerical issue. If I observe that, with ##\eta=S_-/S_+##, we have $$A = {1-\eta \over 1+\eta}$$ Such a two-step calculation (via ##\eta##) can be done for the error calculation also -- with, of course, the same result. It shows that ##\Delta \eta/\eta ## is indeed quite big in situation 2. See below. But then, in the propagation to ##dA\over A##, that is mitigated drastically due to the value of ##\eta##.As a numerical example, let me take ##N = 30000## and calculate the errors in ##\eta## and ##A##:

Situation 1:
##S_+ = 1452 ## and -- assuming Poisson statistics -- ##\ \ \Delta S_+ = \sqrt{1452} =38 ##
##S_- = \ \ 972\pm 31 ##

Situation 2:
##S_+ = \\255\pm 16##
##S_- = \ \ \ \ 81\pm 9 ##

Your one-step yields ##dA_1=0.020, \ \ dA_2 = 0.047## i.e. relative errors 10% and 9%, respectively.

For the two-step I get ##\eta_1 = 0.669 \pm 0.028, \ \ \eta_2 = 0.318 \pm 0.045##, so relative errors 4% and 13% ! (so, as our intuition expected)

But then, with $${dA\over d\eta}={2\over (1+\eta)^2} \Rightarrow {dA\over A} = {2\eta\over 1-\eta^2} {d\eta\over \eta}$$ the relative errors in A are the same 10% and 9% as above.

Note that, for clarity, I show errors with too much accuracy -- the error in the error usually doesn't justify more than one digit accuracy (unless the first digit is a 1)

##\ ##

Be careful σS ≠ a√N

S ≅ aNx

So σS =aN σx

gleem said:
Be careful σS ≠ a√N

S ≅ aNx

So σS =aN σx
Sorry I got lost. You said ##\sigma_S = \sqrt{S}## (which makes sense). Also we have that ##S = N(a^2+ax)## and ##a>>x##, so shouldn't ##S \cong Na^2 ## (as ##a^2>>ax##) and thus ##\sigma_S = a\sqrt{N}##?

S = N(a2 +ax)

N and a are constants

So an incremental change in S is ΔS = a⋅N⋅Δx which we may assume that σs = a⋅N⋅σx

It just occurred to me that the "A" you defined is independent of the number of trials N and the parameter x that you seek or am I missing something?

gleem said:
S = N(a2 +ax)

N and a are constants

So an incremental change in S is ΔS = a⋅N⋅Δx which we may assume that σs = a⋅N⋅σx
I agree with this. What I don't understand is why ##S\cong aNx##?

gleem said:
It just occurred to me that the "A" you defined is independent of the number of trials N and the parameter x that you seek or am I missing something?
So "A" is obtained experimentally as described above. But the formula it has does involve x:

$$A = \frac{S_+-S_-}{S_++S_-} = \frac{N(a^2+ax)-N(a^2-ax)}{N(a^2+ax)+N(a^2-ax)} = \frac{2ax}{2a^2}=\frac{x}{a}$$
So by measuring A and the associated uncertainty I get x and its uncertainty. And yes, A doesn't depend on ##N##, but the uncertainty on A depends on ##N##.

In my above post, I inadvertently assumed a<<x and a2 could be ignored, my error sorry for dragging this out. But even if a>>x you still may have a problem

A depends on a and x where x <<a and dA =dx/a. A is a small number.

Your data are the "S"s S= (a2 ± ax)N so dS = aNdx or dx = dS/aN

So dA = dS/(aN )

therefore dA/A = dS±/ (S± - a2N) which you want to = 0.1

dS± = σS±

N = (S±-10σS±)/a2

Please fill in the missing steps yourself to see if I did not make an error.

Does this work for your data?

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