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I am reading a paper and I am struggling to find what this notation means. I have given the link to the paper here (https://arxiv.org/abs/1108.0884).

$$\rho(d \mathbf{y}) = \frac{1}{\mathcal{Z}} \exp\left( -\sum_{k = N_0 + 1}^{N} \frac{\lambda_k}{\sigma^2 q_k^2} y_k^2\right) d \mathbf{y}$$

It is an invariant measure of multiple variables that are designated $y_k$ which are orthogonal. I don't know know what $d \mathbf{y}$ particularly with regard to integrating with respect to this measure. Any help that you could offer would be extremely useful.

Kindest regards,

Catherine

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There are basically two notations commonly used for the "differential" in Lebesgue integration with respect to a variable $y$ and an explicitly defined measure $\rho$. One is $\rho(dy)$. The other is $d\rho(y)$. Neither is really very good, but generally everybody uses one or the other. Anyway, this differential is used in the sense that

$$\int_A \rho(dy) = \rho(A).$$

Thus in your example:

$$\rho(A)=\int_A \frac{1}{\mathcal{Z}} \exp\left( -\sum_{k = N_0 + 1}^{N} \frac{\lambda_k}{\sigma^2 q_k^2} y_k^2\right) d y$$

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  • $\begingroup$ (I replaced $F(y)$ with the formula, which I inserted also in the original question. Please verify and edit as needed.) $\endgroup$ – Nominal Animal Aug 20 '18 at 3:41
  • $\begingroup$ I suppose, but my question really is if $y= (y_1,y_2,...y_N)^T$ then what does $dy$ mean? Sorry, I did not know how to make it bold! $\endgroup$ – Catherine Drysdale Aug 22 '18 at 8:36
  • $\begingroup$ @CatherineDrysdale $dy$ on the right side is just the $N$-dimensional Lebesgue measure. $dy$ on the left side is part of $\rho(dy)$ which is an indivisible unit of syntax. $\endgroup$ – Ian Aug 22 '18 at 15:30
  • $\begingroup$ So the integral represents a multiple integration with respect to each $y_k$? $\endgroup$ – Catherine Drysdale Aug 22 '18 at 16:52
  • $\begingroup$ @CatherineDrysdale Yes. $\endgroup$ – Ian Aug 22 '18 at 16:59

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