I would like to obtain $g$ by solving the following integral equation

$$\int_s^T R(u) dg(u) + f(s,T)\int_s^T g(u)du =0$$

where $f,R:\mathbb R _+ ^*\rightarrow \mathbb R _+$and $g: \mathbb R _+ \rightarrow[0,1]$ is

• non-decreasing continuous function,
• $g(0)=1$,
• $g(T)<1$
• $\lim _{s\to \infty}g(s)=0$.

One can think about g as being such that $g= 1- F$ where $F$ is the law of an absolute continuous random variable.

If we assume that $g$ is differentiable then we have

$$\int_s^T \left(f(s,T)g(u)+R(u) g'(u)\right) ~du =0, \quad \forall s \in [0,T]$$

So I am tempted to conclude that

$$f(s,T)g(u)+R(u) g'(u)=0, \quad \forall u \in [s,T]$$ therefore $$g(s) = \exp\left(-\int _0^s \frac{f(\tau,T)}{R(\tau)}~d\tau\right), \quad \forall u \in [0,T]$$

Is my approach correct or have I made a mistake when I assumed that the integrand is zero as the integral is zero for each $s$ ?

I have no restrictions on $f$ and $R$ for the moment so we can assume any necessary condition about $f$ as necessary to solve it.

Edit

A friend wisely advised me to take a look at Volterra integral equation which is exactly what we have here after integrating by parts the first integral and inverting the time as follows ( after that point I use the notation abuse $R(u): =R(-u), g(u):=g(-u) \text{ and } f(t,T) = f(-t,T)$):

$$g(t) = \alpha + \int_{-T }^t K_T(t,u))g(u) du$$

where $\alpha := (Rg)(-T)$ and $$K_T(t,u):=\frac{R(u) + f(t,T)}{R(t)}$$

Many thanks

Since this question haven't received any answer or comment but some upvotes I posted it at mathoverflow too

$$f(s,T)g(u) + R(u)g'(u) = 0$$

you have two arguments: $u$ and $s$ and you cannot just take $u=s$ as $u$ is the integration variable ranging over $[s,T]$ for a given $s$.

To derive an equation for $g$ start by taking the derivative of

$$\int_s^T g(u)du + f(s,T)\int_s^T R(u)g'(u)du = 0$$

we obtain

$$g(s) - \frac{df}{ds}\int_s^T R(u)g'(u)du + fRg'(u) = 0$$

which equals (using the first equation to simplify)

$$g(s) + fRg'(u) = \frac{d\log f}{ds}\int_s^T g(u)du$$

Now taking the derivative again gives

$$g'(s) + \frac{dfR}{ds}g'(u) + fR g''(s) = \frac{d^2\log f}{ds^2}\int_s^T g(u)du - \frac{d\log f}{ds}g(s)$$

Which simplifies (by using the equation above) to

$$g''(s) + g'(s)\left(\frac{ff' + 2ff'^2R + f^2 f'R' - f^2f''R}{ff'R}\right) + \left(\frac{2f'^2 - f''f}{f^2f'R}\right)g(s) = 0$$

I highly doubt this equation has a general closed form solution, it just too general.

• Thanks a lot for you answer. I just realised I permeated the integral on the equation which does not change much for the solution. However it's better to adapt it – Paul Nov 18 '14 at 1:46
• I haven't assumed $u= s$ for deducing that the integrand is zero. After that if the integrand do is zero for all u between $s$ and $T$ I can look to the equation for $u =s$. The mistake is in deducting that the integrand is zero if it is. And I particularly think it is wrong. The question is "why is it wrong?" – Paul Nov 20 '14 at 1:02
• @Paul What I try to say is that the integrand is a function of both $u$ and $s$ thats why it does not work: $\int_s^T [g(u) + f(s,T) R(u)g'(u)] du$ so if $g(u) + f(s,T) R(u)g'(u) = 0$ for all $u$ then $f(s,T)$ must be a constant (i.e. does not depend on $s$). To see why write it as $\frac{g(u)}{R(u)g'(u)} = - f(s,T)$ and its clear that the left hand side is not a function of $s$ so the right hand side cannot be either. – Winther Nov 20 '14 at 1:12
• Another way to see it. Take the derivative of your $g(s)$ solution and you get: $g(s) + f(s,T)R(s) g'(s) = 0$. Integrate this to get: $\int_s^T g(u) du + \int_s^T f(u,T) R(u) g'(u)du = 0$. This is not the same equation as you started with since $\int_s^T f(u,T) R(u) g'(u)du \not= f(s,T)\int_s^T R(u) g'(u)du$\$ – Winther Nov 20 '14 at 1:18