I have a confusion on this model. I get how to solve the equation using the integrating factor and find out ru in terms of r0 plus other. Here we start with an equation, apply Taylor formula , and come up with dst. We then integrate and find a solution for ru. I can then see at some places drt is taken and integrated directly to come up with integrated r(u) between t and T. In this we substitute solved ru to come up with another equation. When do we do that. This has made me all confused on what else can be asked on this question. And which way to solve it. One such question is Q 10 April 2008. At another place we take solved ru and integrate it . Example of this is question 5 , April 2014. I would appreciate any help in this regard. Many thanks Sarika
To start off, r(u) and the integral of r(u) are two different things, used for two different purposes. r(u) gives the equation of the short interest rate at time u under the Vasicek model. It can be used to tell us the statistical distribution of the short rate in the future. The integral of r(u) would be needed if we wished to use the Vasicek model to calculate zero-coupon bond prices, as a zero-coupon bond price is given by a formula of the form exp(- integral of r(u) over the term of the bond). Some questions just ask for r(u), some ask for the integral of r(u), some ask for both. Both April 2008 Q10 and April 2014 Q5 ask for the integral of r(u). The answers to these questions are the same (except for different letters). As with lots of maths, there are different ways of getting to the same result. You've not said precisely what solutions you are looking at, so I cannot comment specifically, but there are two ways to derive the integral of r(u). I'll give an overview below. 1. Start with the SDE for dr(t). Solve this using integrating factor to get an expression for r(u). Integrate this expression for r(u) between the limits of t and T and simplify to get the integral of r(u). The simplification here involves dealing with a double integral. 2. Start with the SDE for dr(u). Integrate this directly between the limits of t and T (without using an integrating factor) and rearrange to get an expression for the integral of r(u). This will be in terms of r(T). You then need to substitute in an expression for r(T) to get the same answer as 1. This avoids the double integral, but you need to separately obtain the expression for r(T) to substitute in (by solving the SDE for dr(t) using the integrating factor). You can choose which approach you prefer.
