Critical exponent relation for neural avalanche dynamics I am trying to understand the origin of equation (4) in this paper. Per the arguments of Touboul and Destexhe, a power law distribution for avalanche size and duration, as well as observed data collapse, do not constitute sufficient evidence to conclude that the system is at the critical point of phase transition. Rather, a more stringent condition that should be met is equation (4) in the above paper, which posits a relation $$\gamma = \frac{\alpha - 1}{\beta -1}$$ where $$f(S) \propto S^{-\alpha}$$ and $$g(T) \propto T^{-\beta}$$ are the probability density functions for the avalanche size and avalanche duration, respectively, and $$\langle S \rangle (T) \propto T^{\gamma}$$ is the average size of an avalanche of duration $T$. I can deduce the exponent relation under the assumption that the conditional probability distribution of $S$ given an observed value of $T$ is $P(S\vert T) = \delta(S-T^{\gamma})$. I am confused by this assumption, however, because it seems extremely stringent. It makes sense to me that the average size of an avalanche may be a power function of time, but this density function seems to suggest that every avalanche of duration $T$ has size $T^{\gamma}$. Am I misinterpreting the delta function here? If anyone could elucidate this, I'd greatly appreciate it.
 A: A quick correction before I get started: you appear to have flipped the exponent relation. It should be
$$\gamma = \frac{\beta - 1}{\alpha - 1},$$
given your definitions $f(S) \sim S^{-\alpha}$, $f(T) \sim T^{-\beta}$, and $\langle S \rangle(T) \sim T^{\gamma}$.
As you intuited, the distribution $P(S|T)$ is not a delta function distribution. While the avalanche literature often focuses on the power law scalings $P(S) \sim S^{-\tau}$, etc., these are not the complete story, because close to the critical point you have a more general Widom-type scaling. In particular, following this paper about avalanche statistics in the random field Ising model, close to the critical point the asymptotic scaling form for the joint distribution of both sizes $S$ and durations $T$ (taken to be large: $S$, $T \gg 1$) is
$$P(S,T) \sim S^{-(\tau+\sigma \nu z)} \mathcal D_{S,T}\left( S r^{1/\sigma}, T r^{\nu z}\right),$$
where I have simplified Sethna et al's Eq. (1) slightly by setting the external magnetic field to its critical value and ignoring non-universal constants. The tuning parameter is $r$, which has a critical value of $0$. Note that although this scaling form is from a paper on the random field Ising model this is the generic scaling form used in essentially all avalanche work (up to names of the exponents and interpretation of the tuning parameter $r$).
In addition to the critical exponents $\tau$, $\sigma$, $\nu$ and $z$ (and combinations of them), the function $\mathcal D_{S,T}$ is also universal (in the sense that data with different $S$, $T$, and $r$ should all fall onto a single curve/surface; there are some non-universal constants that must be determined by fit for specific models/data).
From this scaling form one can derive these exponents relations. For example, integrating out $T$:
\begin{align*}
P(S) &\sim \int dT~S^{-(\tau+\sigma \nu z)} \mathcal D_{S,T}\left( S r^{1/\sigma}, T r^{\nu z}\right)\\
&\sim \int du~r^{-1/\sigma} S^{-(\tau+\sigma \nu z)} \mathcal D_{S,T}\left( S r^{1/\sigma}, u\right)\\
&\sim r^{-1/\sigma} S^{-(\tau+\sigma \nu z)} \mathcal D'_S\left(S r^{1/\sigma} \right)\\
&\sim S^{-\tau} \left(S r^{1/\sigma} \right)^{-\sigma \nu z}\mathcal D'_S\left(S r^{1/\sigma} \right)\\
&\equiv S^{-\tau}\mathcal D_S\left(S r^{1/\sigma} \right).
\end{align*}
In doing this calculation I performed a number of steps: I changed variables to $u = T r^{\nu z}$ and then defined $\mathcal D'_S\left(S r^{1/\sigma} \right) = \int du~ \mathcal D_{S,T}\left( S r^{1/\sigma}, u\right)$, ignoring possible $r$-dependence of the integration limits (assumed to be negligible in the asymptotic limit). I then massaged the prefactors out front of this function $\mathcal D'_S$ to collect on a power of $Sr^{1/\sigma}$, which I then absorbed into $\mathcal D'_S$ to define a new scaling function $\mathcal D_S$. From this we see that your $\alpha = \tau$ in this notation.
You can do the same trick to derive $P(T)$, changing variables to $u = S r^{1/\sigma}$ instead. I find $P(T) \sim T^{-\left(\frac{\tau-1}{\sigma \nu z} - 1 \right)} \mathcal D_T(T r^{\nu z})$, identifying your $\beta = \frac{\tau-1}{\sigma \nu z} - 1$.
You can then compute the average of $S$ conditioned on $T$ by computing the conditional distribution
$$P(S|T) = \frac{P(S,T)}{P(T)} \sim \frac{S^{-(\tau + \sigma \nu z)} \mathcal D_{S,T}(Sr^{1/\sigma},Tr^{\nu z})}{T^{-\left(\frac{\tau-1}{\sigma \nu z} - 1 \right)} \mathcal D_T(T r^{\nu z})} \sim S^{-(\tau + \sigma \nu z)} T^{\left(\frac{\tau-1}{\sigma \nu z} - 1 \right)} \mathcal G'(S r^{1/\sigma}, T r^{\nu z})$$
for some scaling function $\mathcal G'$ that's just a ratio of the two $\mathcal D$'s. If we were interested in the scaling form of this distribution we would massage things so that there is just a power of $S$ out front, but since our interest is the average $S$ we won't bother.
Now, with $P(S|T)$ you can compute
$$\langle S \rangle(T) = \int dS~S P(S|T) \sim \int dS~S \left(S^{-(\tau + \sigma \nu z)} T^{\left(\frac{\tau-1}{\sigma \nu z} - 1 \right)} \mathcal G'(S r^{1/\sigma}, T r^{\nu z}) \right).$$
Changing variables to $u = S r^{1/\sigma}$ and then massaging terms and redefining scaling functions to eliminate any powers of $r$ outside of the scaling function ultimately gives
$$\langle S \rangle(T) \sim T^{\frac{1}{\sigma \nu z}} \mathcal G(T r^{\nu z}).$$
This identifies your $\gamma = \frac{1}{\sigma \nu z}$.
You can now verify that the corrected exponent relation I gave at the top, $\gamma = \frac{\beta -1}{\alpha -1}$, is satisfied by the exponents $\alpha = \tau$, $\beta = \frac{\tau-1}{\sigma \nu z} - 1$, and $\gamma = \frac{1}{\sigma \nu z}$ derived from the general scaling form of $P(S,T)$.
