 Solution

The size-biased distribution associated with ${\xi }_1$ is defined byfor every bounded measurable $h$. Thus, if ${\xi }_1$ has density $f$, then ${\hat{\xi }}_1$ has density $\lambda xf(x)$.

Now suppose that ${\xi }_1$ is exponential with rate $\lambda$. The renewal epochs form a Poisson process. Writeso that $L(t)=A_t+R_t$. Independent stationary increments and the exponential waiting-time law show that $R_t$ is exponential with rate $\lambda$, independent of the history up to $t$ and hence of $A_t$. For fixed $a\ge 0$ and $t>a$,Consequently $A_t$ converges in distribution to another rate-$\lambda$ exponential variable, independent of $R_t$. Therefore $L(t)$ converges to the sum of two independent rate-$\lambda$ exponentials, whose density isThe size-biased density of ${\xi }_1$ is likewiseHence the exponential renewal interval limit is