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,



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:

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

  • $\begingroup$ (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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