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?
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$\begingroup$ Your formula can't be right. $1 - \int_0^\infty (1 - F(t+y))\; dy$ could well be negative, so it can't be interpreted as probability. $\endgroup$– Robert IsraelCommented Aug 12, 2014 at 15:28
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$\begingroup$ @RobertIsrael Fixed, forget to divide by E[X]. Thanks $\endgroup$– dffCommented Aug 12, 2014 at 17:46
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1 Answer
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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, $$E(G(\hat X))=\frac{E(\hat G(X))}{E(X)},\qquad \hat G(x)=\int_0^xG.$$
