# What am I doing wrong in the derivation of Bass diffusion model?

I've been deriving Bass diffusion model and keep consistently finding a different result than Bass' original answer. To make things worse, every single link in the Google results page just copies the Bass original solution, while I am finding the different one. You don't have to know the model, just let me show you the math part, so you can check my solution.

The premise:

We have two formulations of the hazard rate --- as Bayesian conditional probability and as a linear function:

$$$$\frac{f(t)}{1-F(t)} = p + q F(t)$$$$

where $$f(t) = \frac{dF(t)}{dt}$$. This gives the following differential equation:

$$$$\frac{dF(t)}{dt} = (1 - F(t)) (p + qF(t))$$$$

My solution

Using Chain Rule, we rewrite this as:

$$$$\int \frac{dF(t)}{(1-F(t))(p+qF(t))} = \int dt = t \label{eq:diff}$$$$

Notice that:

$$$$\frac{1}{(1-F(t))(p+qF(t))} = \left( \frac{1}{p+q} \right) \left( \frac{q}{p+qF(t)} + \frac{1}{1-F(t)} \right),$$$$

substituting this to the above equation, implies:

$$$$\int \frac{q}{p+qF(t)} dF(t) - \int \frac{-1}{1-F(t)} dF(t) = (p + q) t$$$$

integrating yields:

$$$$\log (p + qF(t)) - \log (1 - F(t)) = (p + q) t$$$$

using log properties:

$$$$\log \left( \frac{p + qF(t)}{1 - F(t)} \right) = (p + q) t$$$$

or:

$$$$\frac{p + qF(t)}{1 - F(t)} = e^{(p+q)t}$$$$

cross producting the fraction yields:

$$$$p + qF(t) = e^{(p+q)t} - e^{(p+q)t} F(t)$$$$

finally:

$$$$(q + e^{(p+q)t}) F(t) = (e^{(p+q)t} - p)$$$$

or:

$$$$F(t) = \frac{e^{(p+q)t} - p}{e^{(p+q)t} + q}$$$$

The problem:

But Bass, somehow found the following:

cancelling out my answer does not yield Bass' answer. The unnecessary $$q$$ is ruining everything.

Thank you.

You are missing an integration constant

$$\log\left(\frac{p + qF(t)}{1 - F(t)}\right) = (p + q)t + \color{red}{\tilde{C}}$$

This constant you can name it whatever you want, I'm going to name it as

$$\color{red}{\tilde{C}} = \color{blue}{C}(p + q) + \ln q$$

where $$C$$ is just another constant. So I basically changed one constant for another one (completely allowed). Now the problem becomes

$$\begin{eqnarray} \ln\left(\frac{p + qF(t)}{1 - F(t)}\right) &=& (p + q)t + \color{blue}{C}(p + q) + \ln q = (p + q)(t + C) + \ln q\\ p + qF(t) &=& e^{(p + q)(t + C) + \ln q}(1 - F(t)) = qe^{(p + q)(t + C)}(1 - F(t)) \\ [q + qe^{(p + q)(t + C)}]F(t)&=& qe^{(p + q)(t + C)} - p \\ F(t) &=& \frac{1}{q} \frac{qe^{(p + q)(t + C)} - p}{1 + e^{(p + q)(t + C)}} \end{eqnarray}$$

Rearranging a bit the terms

$$\bbox[5px,border:2px solid blue] { F(t) = \frac{1}{q}\frac{q - pe^{-(p + q)(t + C)}}{1 + e^{-(p + q)(t + C)}} }$$

• Thanks, that solves it. But why would Bass want this particular constant? Is it special somehow? Anyway, thanks again. – Ravshan S.K. Dec 3 '18 at 19:23
• @RavshanS.K It just makes more convenient the definition of the initial state $F(t = 0)$ – caverac Dec 3 '18 at 19:24
• @caverac: Your answer to the original question was beautiful but can you explain your answer to the comment in more detail. Thanks. – mark leeds Dec 3 '18 at 20:22
• @markleeds Hi Mark! Sure, unfortunately I don't have access to the original reference and couldn't tell you what the author is after with this choice. All I can tell you is that $C$ is uniquely defined once you set a point through which the function $F = F(t)$ needs to go through. For instance, in this case the choice is $$F(-C) = \frac{q - p}{2q}$$ – caverac Dec 3 '18 at 20:27
• @caverac: I get it better now. thanks. Ravshan: I followed caverac's derivation except for the value of C. – mark leeds Dec 4 '18 at 7:57