How to prove that if $\int_t^\infty(s-t-\frac{1}{\lambda})\,f(s)\ ds =0$ for all $t\ge 0$ then $f(s)=\lambda\, \mathrm{e}^{-\lambda s}$

Question

The problem is motivated by my probability text which states that if the expectation of time to wait conditioned on time already spent waiting $(t)$ is constant (equals $\frac{1}{\lambda}$) then the pdf is $f(s)=\lambda e^{-\lambda s}$

I've tried writing it down:

$$\frac{\int_t^\infty(s-t)f(s)\ ds}{\int_t^\infty f(s)\ ds}=\frac{1}{\lambda} \ \ \ \text{for all } t\in \mathbb R_{\geq 0}$$

which becomes $$\int_t^\infty(s-t-\frac{1}{\lambda})f(s)\ ds =0$$ after rearrangement. Yet I have no idea where to start and how to arrive at $f(s)=\lambda e^{-\lambda s}$.

Answer 1

Take derivatives, and use uniqueness of solutions to first order differential equations. More precisely: $$ \int_t^\infty(s-t-\frac{1}{\lambda})f(s)\ ds \\ = \int_t^\infty(s-\frac{1}{\lambda})f(s)\ ds - t \int_t^\infty f(s)\ ds $$ So taking a derivative, and after some simplification: $$ \frac{d}{dt} \left( \int_t^\infty(s-\frac{1}{\lambda})f(s)\ ds - t \int_t^\infty f(s)\ ds \right) \\ = \frac{1}{\lambda} f(t) - \int_t^\infty f(s) ds = 0 $$ Differentiating the above we obtain $$ \frac{1}{\lambda} f'(t) = - f(t) $$ So $ f(t) = C e^{-\lambda t} $. The constant C is determined by normalization.