Consider a bi-exponential function defined as $B(x)=C(e^\frac{-x}{\tau_1}-e^\frac{-x}{\tau_2})+b_0$, where $C$ is the scaling factor, $b_0$ is the offset, and $\tau_1$ and $\tau_2$ are the time constants of the two exponential.
Now consider an alpha function which is defined as: $\alpha(x)=A\frac{x}{\tau}e^{(1-\frac{x}{\tau})}H(x)$, where $A$ is the scaling factor, $H(x)$ is the Heaviside function, and $\tau$ is the "time constant" (time it takes to reach the peak value of A from the onset).
Ignoring the peak values and offset, is there a relation between the time constants of the bi-exponential and the alpha function, for instance: $\tau = f(\tau_1, \tau_2)$, such that shapes of the two curves are roughly identical?
I approached this by doing the following:
$B(x)=\alpha(x)$
$...$
$(e^\frac{-x}{\tau_1}-e^\frac{-x}{\tau_2})\approx\frac{x}{\tau}e^{1-\frac{x}{\tau}}$
Is this the right approach to solve this? What is the expression for $f(\tau_1, \tau_2)$? $f(\tau_1, \tau_2)$ will likely be degenrate, right?
I am trying this in context of electrophysiology where experimentalists like to fit bi-exponentials to the recorded signals, whereas theorists like to use the alpha function in simulations. And I'm trying to theoretically bridge the gap.
I'll try to answer my own question.
Since, $\tau$ for the alpha function corresponds to the time it takes for the peak to occur. One could simply calculate the time to peak of the bi-exponential function. Some googling landed me on to this blog about some miscellaneous functions. In this blog, the time to peak for a bi-exponential is described as (I'm rearranging the bi-exponential equation in the blog to fit the $B(x)$ formula mentioned in my original question):
time to peak $= \frac{\tau_1 \tau_2}{\tau_1 - \tau_2}log_e(\frac{\tau_1}{\tau_2})$.
This is very straight forward to derive; just find the maxima of $B(x)$ by using $\frac{d}{dx}B(x)=0$. Which yields: $x = \frac{\tau_1 \tau_2}{\tau_1 - \tau_2}log_e(\frac{\tau_1}{\tau_2})$. And this is exactly what $\tau$ is for $\alpha(x)$.
Therefore, $\tau = \frac{\tau_1 \tau_2}{\tau_1 - \tau_2}log_e(\frac{\tau_1}{\tau_2})$
I've run a few simulations and this seems to be true. In the figure below, the solid cyan line is the bi-exponential fitted to electrophysiology data (this means $\tau_1$ and $\tau_2$ are given). The time constant for the alpha function was calculated by the aforementioned formula, and an alpha-shaped curve with the estimated $\tau$ was plotted as the dashed black line. The time to peak of the fitted bi-exponential (duration marked by blue arrows) is fairly close to the theoretical estimate (red). And although the two curves appear different at the late stages of decay, the overall shape is replicated. Note that the amplitude of the alpha-shaped curve was adjusted to match.
For my use case, this will suffice. But if you can figure out how to perfectly match the two curves, please add your comments/answers. But I wonder if the alpha function can fully approximate the electrophysiological trace. Thanks for stopping by :)
Edit 1 (25.04.2023):Here's a caveat! Since, $\tau$ depends upon $\tau_1$ and $\tau_2$,and, of course, if $\tau_1 = \tau_2$, the formula does not work. But what's also important to note is that if $\tau_1<<\tau_2$ then the shape of the alpha function will not be able to capture the shape of the bi-exponential (think of an event with dirac-like rise and an exponential decay). In electrophysiology, this does not usually occur (if you are sampling fast enough). But theoretically, the formula has its limits. Bi-exponential and alpha function are part of the same family of probability distributions, and alpha function is a special case of the bi-exponential.