<p>If <span class="math-container">$ \delta &gt; 0 $</span> is very small, then the interest incurred during the small subperiod <span class="math-container">$[t, t + \delta]$</span> can be approximated using the simple interest formula. More specifically, the interest incurred in <span class="math-container">$[t, t + \delta]$</span> is:</p>
<p><span class="math-container">$$ B(t + \delta) - B(t) = B(t) r(t) \delta + o(\delta^2) $$</span></p>
<p>where <span class="math-container">$ B(t) r(t) \delta $</span> is the interest that would be incurred if we were to use simple interest.</p>
<p>Dividing by <span class="math-container">$\delta$</span>, we have</p>
<p><span class="math-container">$$ \frac{B(t + \delta) - B(t)}{\delta} = B(t)r(t) + o(\delta) $$</span></p>
<p>Taking limit <span class="math-container">$ \delta \to 0 $</span>, we have</p>
<p><span class="math-container">$$ B'(t) = \lim_{\delta \to 0} \frac{B(t + \delta) - B(t)}{\delta} = B(t)r(t) $$</span></p>
<p>The solution of the differential equation <span class="math-container">$ B'(t) = B(t)r(t) $</span> with the initial condition <span class="math-container">$ B(0) = B_0 $</span> is</p>
<p><span class="math-container">$$ B(t) = B_0 \exp\left( \int_0^t r(s) ds \right) $$</span></p>

CC BY-SA 4.0
2021-11-02T18:26:04.713Z
Thank you so much for your detailed answer! It makes much more sense to me now :)
Marc Allan
CC BY-SA 4.0
2021-11-01T04:12:21.910Z
2021-11-01T06:35:28.313Z
2021-11-01T06:35:28.313Z
user141240
<p>I've researching some mathematical finance and I've stumbled upon something I can't seems to find sources on. I'm probably overlooking something, but I hope someone can enlighten me and give me some sources which I can take a look at.
In Björk 2020 on page 95 it is stated that the a risk free asset with price process <span class="math-container">$B$</span> has the following expression:
<span class="math-container">$$B_{t} = B_{0} \exp\left(\int_{0}^{t} r_{s} ds\right).$$</span>
Can someone explain me why it is so or link to some source where they explain why it is given like that. Thanks in advanced.</p>

<finance><bonds><risk><stochastic-processes><stochastic-calculus>
Bond Price expression
Marc Allan
# Bond Price expression

(Asked by: username_0 on 2021-10-31)

I've researching some mathematical finance and I've stumbled upon something I can't seems to find sources on. I'm probably overlooking something, but I hope someone can enlighten me and give me some sources which I can take a look at.
In Björk 2020 on page 95 it is stated that the a risk free asset with price process $B$ has the following expression:
$$B\_{t} = B\_{0} \exp\left(\int\_{0}^{t} r\_{s} ds\right).$$
Can someone explain me why it is so or link to some source where they explain why it is given like that. Thanks in advanced.

**Tags:** finance, bonds, risk, stochastic-processes, stochastic-calculus


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## 1 answers

### Answer by username_1 on 2021-11-01. Score: 3 | Accepted answer

If $ \delta > 0 $ is very small, then the interest incurred during the small subperiod $[t, t + \delta]$ can be approximated using the simple interest formula. More specifically, the interest incurred in $[t, t + \delta]$ is:

$$ B(t + \delta) - B(t) = B(t) r(t) \delta + o(\delta^2) $$

where $ B(t) r(t) \delta $ is the interest that would be incurred if we were to use simple interest.

Dividing by $\delta$, we have

$$ \frac{B(t + \delta) - B(t)}{\delta} = B(t)r(t) + o(\delta) $$

Taking limit $ \delta \to 0 $, we have

$$ B'(t) = \lim\_{\delta \to 0} \frac{B(t + \delta) - B(t)}{\delta} = B(t)r(t) $$

The solution of the differential equation $ B'(t) = B(t)r(t) $ with the initial condition $ B(0) = B\_0 $ is

$$ B(t) = B\_0 \exp\left( \int\_0^t r(s) ds \right) $$

> **Comments:**
> **username_0**: Thank you so much for your detailed answer! It makes much more sense to me now :)  

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