<p>You are missing an integration constant</p>

<p><span class="math-container">$$
\log\left(\frac{p + qF(t)}{1 - F(t)}\right) = (p + q)t + \color{red}{\tilde{C}}
$$</span></p>

<p>This constant you can name it whatever you want, I'm going to name it as</p>

<p><span class="math-container">$$
\color{red}{\tilde{C}} = \color{blue}{C}(p + q) + \ln q
$$</span></p>

<p>where <span class="math-container">$C$</span> is just another constant. So I basically changed one constant for  another one (completely allowed). Now the problem becomes</p>

<p><span class="math-container">\begin{eqnarray}
\ln\left(\frac{p + qF(t)}{1 - F(t)}\right) &amp;=&amp; (p + q)t + \color{blue}{C}(p + q) + \ln q = (p + q)(t + C) + \ln q\\
p + qF(t) &amp;=&amp; 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)&amp;=&amp; qe^{(p + q)(t + C)} - p \\
F(t) &amp;=&amp; \frac{1}{q} \frac{qe^{(p + q)(t + C)} - p}{1 + e^{(p + q)(t + C)}}
\end{eqnarray}</span></p>

<p>Rearranging a bit the terms</p>

<p><span class="math-container">$$\bbox[5px,border:2px solid blue]
{
F(t) = \frac{1}{q}\frac{q - pe^{-(p + q)(t + C)}}{1 + e^{-(p + q)(t + C)}}
}
$$</span></p>

CC BY-SA 4.0
2018-12-03T19:23:51.517Z
Thanks, that solves it. But why would Bass want this particular constant? Is it special somehow? Anyway, thanks again.
Ravshan S.K.
CC BY-SA 4.0
2018-12-03T19:24:55.150Z
@RavshanS.K It just makes more convenient the definition of the initial state $F(t = 0)$
caverac
CC BY-SA 4.0
2018-12-03T20:22:14.177Z
@caverac: Your answer to the original question was beautiful but can you explain your answer to the comment in more detail. Thanks.
mark leeds
CC BY-SA 4.0
2018-12-03T20:27:59.010Z
@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
CC BY-SA 4.0
2018-12-03T22:32:53.780Z
@markleeds
If we disregard "C" (I don't really need it), then it turns out that I missed the bounary value. This all comes from the boundary value, given by previous assumptions.
Ravshan S.K.
CC BY-SA 4.0
2018-12-04T07:57:48.000Z
@caverac: I get it better now. thanks. Ravshan: I followed caverac's derivation except for the value of C.
mark leeds
CC BY-SA 4.0
2018-12-03T19:19:05.097Z
2018-12-03T19:19:05.097Z
caverac
<p>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.</p>

<p><strong>The premise:</strong></p>

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

<p><span class="math-container">\begin{equation}
 \frac{f(t)}{1-F(t)} = p + q F(t)
\end{equation}</span></p>

<p>where <span class="math-container">$f(t) = \frac{dF(t)}{dt}$</span>. This gives the following differential equation:</p>

<p><span class="math-container">\begin{equation}
 \frac{dF(t)}{dt} = (1 - F(t)) (p + qF(t))
\end{equation}</span></p>

<p><strong>My solution</strong></p>

<p>Using Chain Rule, we rewrite this as:</p>

<p><span class="math-container">\begin{equation}
 \int \frac{dF(t)}{(1-F(t))(p+qF(t))} = \int dt = t
 \label{eq:diff}
\end{equation}</span></p>

<p>Notice that:</p>

<p><span class="math-container">\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}</span></p>

<p>substituting this to the above equation, implies:</p>

<p><span class="math-container">\begin{equation}
 \int \frac{q}{p+qF(t)} dF(t) - \int \frac{-1}{1-F(t)} dF(t) = (p + q) t
\end{equation}</span></p>

<p>integrating yields:</p>

<p><span class="math-container">\begin{equation}
 \log (p + qF(t)) - \log (1 - F(t)) = (p + q) t
\end{equation}</span></p>

<p>using log properties:</p>

<p><span class="math-container">\begin{equation}
 \log \left( \frac{p + qF(t)}{1 - F(t)} \right) = (p + q) t
\end{equation}</span></p>

<p>or:</p>

<p><span class="math-container">\begin{equation}
 \frac{p + qF(t)}{1 - F(t)} = e^{(p+q)t}
\end{equation}</span></p>

<p>cross producting the fraction yields:</p>

<p><span class="math-container">\begin{equation}
 p + qF(t) = e^{(p+q)t} - e^{(p+q)t} F(t)
\end{equation}</span></p>

<p>finally:</p>

<p><span class="math-container">\begin{equation}
 (q + e^{(p+q)t}) F(t) = (e^{(p+q)t} - p) 
\end{equation}</span></p>

<p>or:</p>

<p><span class="math-container">\begin{equation}
 F(t) = \frac{e^{(p+q)t} - p}{e^{(p+q)t} + q}
\end{equation}</span></p>

<p><strong>The problem:</strong></p>

<p>But Bass, somehow found the following:
<a href="https://i.sstatic.net/zgdcM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/zgdcM.png" alt="enter image description here"></a></p>

<p>cancelling out my answer does not yield Bass' answer. The unnecessary <span class="math-container">$q$</span> is ruining everything.</p>

<p>Can you please help me with this inconsistency?</p>

<p>Thank you.</p>

<value><marketing><difference-equations><theory-of-the-firm>
What am I doing wrong in the derivation of Bass diffusion model?
Ravshan S.K.
# What am I doing wrong in the derivation of Bass diffusion model?

(Asked by: username_0 on 2018-12-03)

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}

or:

\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}

finally:

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

or:

\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](https://i.sstatic.net/zgdcM.png)](https://i.sstatic.net/zgdcM.png)

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.

**Tags:** value, marketing, difference-equations, theory-of-the-firm

---

## 1 answers

### Answer by username_1 on 2018-12-03. Score: 3 | Accepted answer

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

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

> **Comments:**
> **username_0**: Thanks, that solves it. But why would Bass want this particular constant? Is it special somehow? Anyway, thanks again.  
> **username_1**: @RavshanS.K It just makes more convenient the definition of the initial state $F(t = 0)$  
> **username_2**: @caverac: Your answer to the original question was beautiful but can you explain your answer to the comment in more detail. Thanks.  
> **username_1**: @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   $$ F(-C) = \frac{q - p}{2q} $$  
> **username_0**: @markleeds If we disregard "C" (I don't really need it), then it turns out that I missed the bounary value. This all comes from the boundary value, given by previous assumptions.  
> **username_2**: @caverac: I get it better now. thanks. Ravshan: I followed caverac's derivation except for the value of C.  

---

