Given a random variable $X$ with survival function $S_X(x)$, I know that I can find the survival function of the residual lifetime rv $T_t=X-t|X>t$ by $S_{T_t}(x)=\frac{S_X(t+x)}{S_X(t)}$. But can I go the other way around, and find the survival function of $X$ starting with knowing the survival function of $T_t$? Here's the problem:
Given a random variable $X>0$, the associated residual lifetime random variable is defined as $T_t=X-t|X>t$ for a constant $t\geq 0$. Given that $S_{T_t}(y)=e^{-y^2-(t+5)y}$, $y>0$, determine the hazard rate of $X$, i.e. $h_X(s)$.
Here's what I tried: I attempted to get the survival function of $X$ by setting $t=0$ and using that to calculate the hazard rate. But when I set $t=0$ I get $S_X(y)=e^{-y^2-5y}$. When plugging that back into the formula for the residual lifetime rv survival function I get $$S_{T_t(y)}=\frac{S_X(t+y)}{S_X(y)}=\frac{e^{-(t+y)^2-5(t+y)}}{e^{-t^2-5t}}=\frac{e^{-t^2-2ty-y^2-5t-5y}}{e^{-t^2-5t}}=e^{-y^2-2yt-5y}$$ This no longer matches the original problem where $S_{T_t}(y)=e^{-y^2-(t+5)y}$. Where am I going wrong?
