Solving a Riccati Equation for a Short-Rate Bond Coefficient
Summary
The document works through a Riccati differential equation for the coefficient in a short-rate model and compares an attempted derivation with a proposed closed-form expression. The approach shifts the unknown by a constant particular solution, then takes the reciprocal of the residual to convert the nonlinear equation into a first-order linear equation. A terminal condition is meant to determine the integration constant before transforming back to the bond coefficient.
The question provides intermediate algebra and the target expression but does not include a final correction or verification. In particular, the reciprocal substitution’s resulting linear equation and its integration require careful sign, time-variable, and boundary-condition handling. This makes the material useful as a derivation problem and a reminder to check transformed equations against the original Riccati equation and terminal value. The document does not provide empirical evidence or discuss calibration, so it explains a mathematical step in short-rate modeling rather than evaluating a rate strategy.
Key ideas
- A constant particular solution can reduce a Riccati equation to an equation for the residual.
- Taking the reciprocal of that residual can linearize the differential equation.
- The terminal condition determines the integration constant in the solution.
- Sign conventions and time direction must be checked throughout the transformation.
- Substituting the final expression back into the original equation is a useful verification.
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# Differential Equation of Type Ricatti as part of Short Rate Model
# Differential Equation of Type Ricatti as part of Short Rate Model
I currently despair of the following solution of a differiental equation (Ricatti Type) as part of a short rate model:
$$ B_t=\frac{1}{2}aB^2+bB-1 $$ First I am "guessing" a particular solution $$ c \in \mathbb{R} $$ as its derivative is zero. That is straightforward and I get e.g. $$ -\frac{b}{a}+\frac{\sqrt{b^2+2a}}{a}=s $$
I call $$ \sqrt{b^2+2a} = k $$ from now on.
For a general solution y I set $$y=s+u$$ and insert y into the initial equation.
$$ s'+u'=\frac{1}{2}a(s+u)^2+b(s+u)-1 $$ As s'= 0: $$ u'=\frac{1}{2}as^2+\frac{1}{2}au^2+asu+bs+bu-1 $$ We already know that $$ u'=\frac{1}{2}as^2+bs-1=0 $$ thus we are left with $$ u'=\frac{1}{2}au^2+asu+bu $$ We now can set $$ z=\frac{1}{u} $$ Therefore, $$ u'=-\frac{z'}{z^2} $$ Inserting this into the equation delivers: $$ z_t+z(as+b)-\frac{1}{2}a $$
which is a linear DGL with the solution: $$ z=\frac{-\frac{1}{2}a \int\limits_{t}^{T} exp(k(T-t)) d\tau+C}{exp(k(T-t))} $$
Furthermore (see definition of k above): $$ as+b=k $$
Now I have to resubstitute to find y (which is B) with: $$ y=\frac{1}{z}+s $$ So I have: $$ y=\frac{exp(k(T-t))}{-\frac{1}{2}a \int\limits_{t}^{T} exp(k(T-t)) d\tau+C}+s $$ And I now that $$ B(T,T)=y(T,T)=0 $$ in order to find C.
Does anyone already spot any mistake I might have made, because no matter what I do I never obtain the solution proposed, being: $$ B(t,T)=\frac{2(exp(k(T-t))-1)}{(b+k)(exp(k(T-t))-1)+2k} $$ Or if no mistake is spotted, can somebody carry me over the finish line. I tries everything, also by inserting numbers, but still...something is off.
Thank you very much in advance for any help! KevShown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.