# Closed form formula of asset that incorporates another asset's interest rate on top of its own

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 $$B(t)=B_1 e^{\sigma t}$$.

Update:

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?

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 $$\sigma t B_0e^{\sigma t}$$.
• 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). Jan 22, 2022 at 15:46
• 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. Jan 22, 2022 at 17:09
• Actually, I shouldn't be surprised. Your solution can indeed be written as a product of 2 functions of $t$ if $\frac{D_0}{B_0}=\frac{\sigma}{\sigma - \mu}$. Jan 22, 2022 at 17:18