1
$\begingroup$

I am hoping someone can help with this - perhaps trivial question of interpreting the formula below. It is the mean residual lifetime (restricted to the window of data) for the case of discrete time.

Restricted Mean Residual Life

Remaining time until event or upper bound--whichever comes first

$$\mu_{[\gamma]}(t|\mathbf{x})=E\left(\min(T, t+\gamma+1)-t|T\ge t, \mathbf{x}\right)=\sum_{j=t}^{t+\gamma}\prod_{k=t}^{j}\left(1-h(k|\mathbf{x})\right)$$

My question is how do you interpret the sigma and product? $t$ is the current time (e.g. "5") and $\gamma$ is the upper bound from my understanding. How do you compute this?

$\endgroup$

1 Answer 1

2
$\begingroup$

$\left(1-h(k|\mathbf{x})\right)$ is the probability of surviving past date $k$ given you survived to the start of date $k$

$\prod_{k=t}^{j}\left(1-h(k|\mathbf{x})\right)$ is the probability of surviving past date $j$ given you survived to the start of date $t$.

Add up these latter probabilities and you get your remaining expectation restricted to the $\gamma+1$ dates starting from the start of date $t$. I would draw a graph where the sum is represented by vertical rectangles, while the expectation can be seen as horizontal rectangles (sum of probabilities of surviving exactly that long multiplied by time survived), something like this

enter image description here

$\endgroup$
2
  • $\begingroup$ I think I understand @Henry. So, gamma here is max time minus t. So, if I have discrete time by month (up to 24) and we are at t=5, then $\gamma$ is 19. Any way you could post an example using the notation given in the question? I am not completely clear on it (with t and j being the same...) $\endgroup$
    – B_Miner
    Oct 18, 2011 at 23:11
  • $\begingroup$ It seems $t+\gamma+1$ is the maximum time [see $\min(T, t+\gamma+1)$ in your formula] while $t$ is the minimum time. $t$ and $j$ are not the same: in the diagram $j$ is the horizontal axis and cannot take a value less than $t$. $\endgroup$
    – Henry
    Oct 18, 2011 at 23:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.