The residual life time distribution of a random variable $X$ with distribution function $F$ is given by the formula
\begin{equation}R(t)=P[X\_\text{res}\leq t] = 1-\frac{1}{\mathbb{E}[X]}\int\_{y=0}^\infty(1-F(t+y))\text{d}y. \end{equation}
However this formule only works if $\mathbb{E}[X]<\infty$. What happens to the residual life time distribution if $\mathbb{E}[X]=\infty$, for example 
\begin{equation}
F(t)=1-\left(\frac{1}{t+1}\right)^{\frac{1}{2}},\text{ for }t>0,
\end{equation}
does it simply cease to exist or is something else going on?



Residual lifetime of heavy-tailed random variable
A partial answer is that, when $X$ is not integrable, $\hat X\_x$ the residual lifetime of $\min\{X,x\}$ converges to infinity in distribution when $x\to\infty$. Since the residual lifetime $\hat X$ of $X$ should logically be bounded below by $\hat X\_x$ for every $x$, this remark points to a "simply cease to exist" verdict.


Note: An equivalent formulation of the transform $X\to \hat X$ when $X$ is integrable is that, for every bounded measurable function $G$ defined on the nonnegative real halfline, $$E(G(\hat X))=\frac{E(\hat G(X))}{E(X)},\qquad \hat G(x)=\int\_0^xG.$$


