# Finding the combined uncertainty of weighted measurements

I am currently doing some simple simulations of radiation dose from several treatment fields.

I am looking at the dose within a certain voxel of a patient receiving radiation therapy, being treated using a total of 3 treatment fields (F1,F2,F3). The dose in a voxel from one treatment field is given as the dose per primary particle. Therefore, to obtain the dose per primary particle from all treatment fields, the dose must be weighted. As a result, the total dose in one voxel is: $$D_v^{tot} = w_1D_v^{F1} + w_2D_v^{F2} + w_3D_v^{F3},$$ where $D^{F1,F2,F3}$ are the doses from each field, and $w_{1,2,3}$ are the weights. These weights are summed up to 1: $$W = w_1+w_2+w_3 = 1.$$ This further leads to $$w_1 = 1-w_2-w_3, \ w_2 = 1-w_1-w_3, \ w_3 = 1-w_1-w_2,$$ and thus $$D_v^{tot} = (1-w_2-w_3)D_v^{F1} + (1-w_1-w_3)D_v^{F2} + (1-w_1-w_2)D_v^{F3}$$

The dose in a voxel from each treatment field has its own uncertainty $\Delta D_v^{F_1,F_2,F_3}$.

I was first thinking of calculating the overall uncertainty in one voxel by the following formula: $$\Delta D_v = \sqrt{\bigg(\frac{\partial}{\partial D_v^{F1}}D_v^{tot}\Delta D_v^{F1}\bigg)^2+\bigg(\frac{\partial}{\partial D_v^{F2}}D_v^{tot}\Delta D_v^{F2}\bigg)^2+\bigg(\frac{\partial}{\partial D_v^{F3}}D_v^{tot}\Delta D_v^{F3}\bigg)^2}\\ = \sqrt{\big(w_1\Delta D_v^{F1}\big)^2+\big(w_2\Delta D_v^{F2}\big)^2+\big(w_3\Delta D_v^{F3}\big)^2}$$

However, as my doses are weighted and the weights are dependent on each other, is this the right approach? As far as I have understood, the measurements themselves are still independent of each other, even though the weights are not.

Note: I say measurements, but really mean Monte Carlo simulations.

• Your approach seems correct to me, but I wonder if you've forgotten to square the items under the square root ? It should be sqrt((w1 ..)^2 + (w2...)^2 + (w3...)^2 ), I think. Dec 13, 2016 at 10:44
• Also, what is the physical interpretation of $w$? If $D$ is the dose per primary particle, then $w$ is the number of particles? Dec 13, 2016 at 14:23
• Yes, or, the weight for each field is defined by the fraction of the total number of particles. So in a way, the weight is the number of particles, yes. The normalization to obtain the total dose from all particles in e.g. Gy, is done later. Dec 13, 2016 at 15:52
• So do the weights have uncertainties too, then? Or do you know the fractions of particles precisely? Dec 13, 2016 at 16:12
• No uncertainties for the weights. They are indeed precisely known. Dec 13, 2016 at 17:06

Your approach is correct, it doesn't matter that the $w_i$ values are not independent, just that the measurements are independent.