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:

\begin{equation} \frac{f(t)}{1-F(t)} = p + q F(t) \end{equation}

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

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

My solution

Using Chain Rule, we rewrite this as:

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

Notice that:

\begin{equation} \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), \end{equation}

substituting this to the above equation, implies:

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

integrating yields:

\begin{equation} \log (p + qF(t)) - \log (1 - F(t)) = (p + q) t \end{equation}

using log properties:

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


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

cross producting the fraction yields:

\begin{equation} p + qF(t) = e^{(p+q)t} - e^{(p+q)t} F(t) \end{equation}


\begin{equation} (q + e^{(p+q)t}) F(t) = (e^{(p+q)t} - p) \end{equation}


\begin{equation} F(t) = \frac{e^{(p+q)t} - p}{e^{(p+q)t} + q} \end{equation}

The problem:

But Bass, somehow found the following: enter image description here

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

Can you please help me with this inconsistency?

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

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

  • $\begingroup$ Thanks, that solves it. But why would Bass want this particular constant? Is it special somehow? Anyway, thanks again. $\endgroup$ – Ravshan S.K. Dec 3 '18 at 19:23
  • 1
    $\begingroup$ @RavshanS.K It just makes more convenient the definition of the initial state $F(t = 0)$ $\endgroup$ – caverac Dec 3 '18 at 19:24
  • $\begingroup$ @caverac: Your answer to the original question was beautiful but can you explain your answer to the comment in more detail. Thanks. $\endgroup$ – mark leeds Dec 3 '18 at 20:22
  • $\begingroup$ @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 $$ F(-C) = \frac{q - p}{2q} $$ $\endgroup$ – caverac Dec 3 '18 at 20:27
  • 1
    $\begingroup$ @caverac: I get it better now. thanks. Ravshan: I followed caverac's derivation except for the value of C. $\endgroup$ – mark leeds Dec 4 '18 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.