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. $\endgroup$ 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

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