I'm trying to find a closed form formula for the price of an asset $D$ that has the following properties:

  1. The asset grows by some interest rate $\mu$ at every instant.
  2. Another asset's ($B$) interest rate $\sigma$ is also added to the asset at every instant.

I came up with the following differential equation to express this (up to debate whether this is correct):

$$ \Delta D = D \mu \Delta t + B \sigma \Delta t $$

Solving this for $D(t)$ without the $B \sigma \Delta t$ part would be easy (some exponential function) but this additive factor trips me up (sorry if trivial, uni has been a while). I have a feeling this can be done with the product rule just by looking at the shape of the formula.

Also, I know that $B$ grows according to some interest rate $\sigma$ so $B(t)=B_1 e^{\sigma t}$.


Here's my (almost certainly incorrect) attempt using the product rule:

We want to solve:

$$ \frac{dD}{dt} = D \mu + B \sigma \tag{1}\label{1} $$

Let's assume $D(t)=u(t)*v(t)$.

From the product rule, we have:

$$ \frac{dD}{dt} = v \frac{du}{dt} + u \frac{dv}{dt} \tag{2}\label{2} $$

Equating the first terms of $\ref{1}$ and $\ref{2}$, we can set:

$$ D \mu = v \frac{du}{dt} $$

Substituting $D=uv$ we have:

$$ uv \mu = v \frac{du}{dt} \\ \frac{du}{dt} = u \mu $$ $$ u = u_1 e^{\mu t} \tag{3}\label{3} $$ Equating the second terms of $\ref{1}$ and $\ref{2}$, we can set:

$$ B \sigma = u \frac{dv}{dt} $$

Rearranging, we have:

$$ \frac{dv}{dt} = \frac{B}{u} \sigma $$

Substituting $B(t)=B_1 e^{\sigma t}$ (known from the original problem statement) and $\ref{3}$, we have:

$$ \frac{dv}{dt} = \frac{B_1 e^{\sigma t}}{u_1 e^{\mu t}} \sigma \\ \frac{dv}{dt} = \frac{B_1 \sigma}{u_1} e^{ \left( \sigma - \mu \right) t} $$ $$ v = \frac{B_1 \sigma}{ \left( \sigma - \mu \right) u_1} e^{ \left( \sigma - \mu \right) t} \tag{4}\label{4} $$

Combining $\ref{3}$ and $\ref{4}$, we have:

$$ D = uv = \frac{B_1 \sigma}{ \sigma - \mu} e^{ \sigma t} $$

I don't like this for many reasons though. If $\mu > \sigma$ this will grow in the negative direction, which doesn't make sense. More interest should surely increase the growth of $D$, not decrease it. Also, intuitively I expect $\mu$ to appear in an exponential somewhere (it is interest, after all). Where am I making a mistake?


1 Answer 1


Your system of differential equations is: \begin{align} \text{d}B_t&=\sigma B_t\text{d}t \\[2pt] \text{d}D_t&=(\mu D_t+\sigma B_t)\text{d}t \end{align} Define the process $X_t:=e^{-\mu t}D_t$ and differentiate: \begin{align} \text{d}X_t&=-\mu e^{-\mu t}D_t\text{d}t+e^{-\mu t}\text{d}D_t \\ &=\sigma e^{-\mu t}B_t\text{d}t \\ &=\sigma e^{(\sigma-\mu)t}B_0\text{d}t \end{align} By integration, we readily obtain: \begin{align} X_t&=D_0+\sigma B_0\left(\frac{e^{(\sigma-\mu)t}-1}{\sigma-\mu}\right) \end{align} That is: \begin{align} D_t=e^{\mu t}D_0+\frac{\sigma}{\sigma-\mu}(e^{\sigma t}-e^{\mu t})B_0 \end{align} An interesting observation to understand the dynamics is that: $$\lim_{\mu\rightarrow\sigma}D_t=(D_0+\sigma t B_0)e^{\sigma t}$$ This last equation is consistent with your initial dynamics: the process $D_t$ has a growth component equal to the continuous return generated by $B_t$, which for each infinitesimal unit of time generates a return equal to $\sigma B_t$. The "simple interest" earned between $0$ and $t$ will therefore be equal to: $$\text{Rate}\times\text{Time}\times\text{Initial wealth}=\sigma\times t\times B_0$$ But this quantity is also continuously being compounded at rate $\sigma$, so that the final wealth earned from the second term is $\sigma t B_0e^{\sigma t}$.

  • $\begingroup$ Thank you! 2 questions: $$ $$ 1. How did you know to define $X_t$ with the $e^{- \mu t}$ term? Is this a common technique?$$ $$ 2. Why did my initial attempt at using the product rule not work? Is it because I assumed $D$ to have a form that it doesn't? $$ $$ I think there's a typo when you're calculating the limit: $e^{\mu t}$ should be $e^{\sigma t}$ on the RHS. The site doesn't allow me to make me <6 character edits though (silly rule IMO). $\endgroup$
    – Attila Kun
    Jan 22, 2022 at 15:46
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    $\begingroup$ Thank you @AttilaKun. For 1, check this answer, it might help. This is known as the method of integrating factors. For 2. I guess that that is the reason, namely you assumed a functional form that is not the actual one. $\endgroup$ Jan 22, 2022 at 16:18
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    $\begingroup$ I'm noticing now that my solution is the special case of your solution in case of $\frac{D_0}{B_0}=\frac{\sigma}{\sigma - \mu}$. Bit surprising to me. $\endgroup$
    – Attila Kun
    Jan 22, 2022 at 17:09
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    $\begingroup$ Actually, I shouldn't be surprised. Your solution can indeed be written as a product of 2 functions of $t$ if $\frac{D_0}{B_0}=\frac{\sigma}{\sigma - \mu}$. $\endgroup$
    – Attila Kun
    Jan 22, 2022 at 17:18

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