biochemistry - What does it mean for a distribution to be "consistent with a two rate-limiting stochastic steps"?

WYSIWYG is almost there, but you need one more piece of information to make this explicit.

The distribution cited in the paper is $h(t)\propto te^{-t/\tau}$ (we're going to ignore normalizing constants today). We can recognize this as a particular case of the Gamma distribution, with PDF:

$$f_{\mathrm{Gamma}}(t\,\big|\,k,\theta)\propto t^{k-1}e^{-t/\theta}.$$

In particular, $h(t)$ looks like the PDF of a $\mathrm{Gamma}(2,\tau)$ random variable,

$$h(t)\propto f_{\mathrm{Gamma}}(t\,\big|\,2,\tau)\propto te^{-t/\tau}.$$

So the authors of the paper are claiming that the burst duration $\tau_{\mathrm{burst}}$ is a random variable with a $\mathrm{Gamma}(2,70\,\mathrm{ sec})$ distribution.

How do we get from that to "This distribution was consistent with two rate-limiting stochastic steps?"
If we recall from our stats class (or look up properties of the Gamma distribution on Wikipedia), a Gamma distribution with integer values of $k$ is the distribution of the waiting time for $k$ events to occur in a Poisson process. The setup of a Poisson process is that you are tracking the occurrence of unlikely/infrequent events in time, so the fact that the distribution $h(t)$ of burst times looks like a $\mathrm{Gamma}(2,\tau)$ suggests that the burst duration $\tau_{\mathrm{burst}}$ is determined by the occurrence of 2 stochastic events (with the same "frequency" $1/\tau$). In other words, this distribution is consistent with the following scenario: a nuclear localization occurs, and then will stop after two particular events, where the timing of the events is governed by a Poisson process (which, as WYSIWYG pointed out, is a reasonable model for the occurrence of chemical reactions with reasonably slow kinetics, i.e. you can count individual reactions in time). If this is true, the distribution of $\tau_{\mathrm{burst}}$ will look like $te^{-t/\tau}$.

The authors generalize that statement somewhat by saying that technically there could be more events/reactions required to terminate a localization burst, but all but 2 of those reactions happen extremely rapidly, i.e. there are 2 rate-limiting steps in the decision process (both of which happen to have the same value for $\tau$).

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Edit: I realized that one more mathematical point might make this more clear: The statement about the Gamma distribution and Poisson processes above is equivalent to the statement that the sum of $k$ iid random variables with distribution $\mathrm{Exponential}(\tau)$ is a random variable with a $\mathrm{Gamma}(k,\tau)$ distribution. Thus $h(t)$ is literally (up to normalization) the distribution of a sum of 2 independent exponential random variables. If you read up on the connection between the exponential distribution and waiting times that might make the claim in the paper seem more transparent.

Source: https://en.wikipedia.org/wiki/Gamma_distribution#Special_cases

http://en.wikipedia.org/wiki/Exponential_distribution

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