Normalise by vanadium in (d, 2theta)

[jl-wynen]

          From today's discussion, it sounds like we need to compute 2d maps (2theta - dspacing) for sample and vanadium, divide them, and then sum along 2theta?

Originally posted by @nvaytet in #155 (comment)

[jokasimr]

Is this issue related to this one?

[jl-wynen]

Not directly. This issue is about whether we histogram vanadium in dspacing or in (2theta, dspacing). We normalise in event mode either way.

[jokasimr]

It would be good to have a description of the model that motivates why we do the normalization one way (normalize in $d$) or another (normalize in $d, 2\theta$). Is there such a motivation we can link to here? I think that could help clarify what needs to be done.

[jl-wynen]

This was brought up by @celinedurniak. I don't have any real insights here. Céline, do you?

[jokasimr]

There was a similar discussion in essreflectometry a while back, and there the conclusion was that normalizing in 1D or 2D produced the same result in the limit of infinite counts, but for realistic count rates normalizing in 2D leads to worse statistics and errors in the final result.

The context here might be different in some way that makes normalizing in 2D necessary.

To clarify why the errors might be worse when we normalize in 2D:

For the sake of this discussion assume the relationship between the measured intensity $I_s$ and the quantity of interest $S$ is described by the simple model

$$ I_s(d, 2\theta) = I(d, 2\theta) S(d) $$ $$ I_v(d, 2\theta) = I(d, 2\theta) $$

where $I$ is some sort of intrinsic instrument intensity and $I_v$ is the measured intensity from the vanadium measurement. Substituting the intrinsic instrument intensity for the vanadium measurement we get

$$ I_s(d, 2\theta) = I_v(d, 2\theta) S(d). $$

Normalizing in 1D is the same as first integrating both sides over slices $\Delta_i$ of $(d, 2\theta)$ where $d$ is approximately constant and equal to $d_i$ in each slice

$$ \int_{\Delta_i} I_s(d, 2\theta) \ \delta d \delta 2\theta \approx \int_{\Delta_i} I_v(d, 2\theta) \ \delta d \delta 2\theta \ S_i. $$

and then isolating $S_i$

$$ \frac{\int_{\Delta_i} I_s(d, 2\theta) \ \delta d \delta 2\theta}{\int_{\Delta_i} I_v(d, 2\theta) \ \delta d \delta 2\theta} \approx S_i \approx S(d_i). $$

Normalizing in 2D is the same as first integrating both sides over rectangles $\Delta_{ij}$ of $(d, 2\theta)$ where $(d, 2\theta)$ is approximately constant and equal to $(d_i, 2\theta_j)$ in each rectangle

$$ \int_{\Delta_{ij}} I_s(d, 2\theta) \ \delta d \delta 2\theta \approx \int_{\Delta_{ij}} I_v(d, 2\theta) \ \delta d \delta 2\theta \ S_{ij}, $$

then isolating $S_{ij}$

$$ \frac{\int_{\Delta_{ij}} I_s(d, 2\theta) \ \delta d \delta 2\theta}{\int_{\Delta_{ij}} I_v(d, 2\theta) \ \delta d \delta 2\theta} \approx S_{ij} \approx S(d_i) $$

and then averaging $\{S_{ij}\}$ over $j$ to estimate $S(d_i)$.

In theory they are the same
For large enough number of counts both methods give the same result, the estimates $S_{ij}$ for one $i$ ought to all converge to the same value. But in practice the number of counts is finite and the error of our final estimate of $S(d_i)$ is going to depend on the distribution of the errors in $S_{ij}$ and how we average them.

If the vanadium measurement is excellent they are the same
If the vanadium measurement has a large number of counts, so that the error of $\int_{\Delta_{ij}} I_v(d, 2\theta) \ \delta d \delta 2\theta$ is negligible then the error of $S_{ij}$ only has contribution from the sample measurement $\int_{\Delta_{ij}} I_s(d, 2\theta) \ \delta d \delta 2\theta$ and can be assumed to be normal distributed.
In that case a weighted average of $S_{ij}$ over $j$ (taking into account the estimated uncertainty of each $S_{ij}$) is going to estimate $S(d_i)$ with about the same error as the method that normalizes in 1D.

In practice the 2D normalization probably leads to larger errors
If we can't make the assumption that the statistics of the vanadium measurement are great everywhere in the $d, 2\theta$ binning then the average of $S_{ij}$ over $j$ might be a worse estimate of $S(d_i)$ than the estimate obtained using the 1D normalization.

[jokasimr]

To move forward on this issue I think we need to figure out what the original motivation for normalizing in 2D instead of in 1D was. @nvaytet or @jl-wynen do you remember? I couldn't figure it out from the conversation that was linked in the issue description.

[SimonHeybrock]

Note that after #176 we always go via a 2-D (d, 2theta) intermediate, even if final result is only 1-D.