Fdtxjr

If you're comfortable with survival analysis, please feel free to skip to the Question section. Otherwise, I set up notation for my question here.

Setup

Let $L$ be a nonnegative continuous random variable such that

$$
mathbbPL geq t = e^-H(t) = e^- int_0^t h(s) ds
$$

where $H(t)$ is the hazard function and $h(t)$ is the hazard rate function. I want to show that

$$
h(t) u approx mathbbPL < t + u mid L geq t
$$

In words, the probability that you do not survive longer than $u$ given that you have already survived for $t$ is proportional to an infinitesimal slice of area under $h(t)$ or roughly the height of $h(t)$.

Consider this derivation:

$$
beginalign
mathbbPL geq t + u mid L geq t
&= fracmathbbPL geq t + u, L geq tmathbbPL geq t
\\
&stackreltextind= fracmathbbPL geq t + umathbbPL geq t
\\
&= frace^-H(t+u)e^- H(t)
\\
&= frace^-int_0^t+u h(s)dse^-int_0^t h(s)ds
\\
&= e^-int_0^t+u h(s)ds + int_0^t h(s)ds
\\
&= e^- int_t^t+u h(s)ds
endalign
$$

So the probability of not surviving in $u$ years if you have lived $t$ years is given by

$$
mathbbPL < t + u mid L geq t = 1 - mathbbPL geq t + u mid L geq t = 1 - e^- int_t^t+u h(s)ds
$$

Now let $F_t(u) = 1 - e^- int_t^t+u h(s) ds$. Therefore,

$$
beginalign
mathbbPL < t + u mid L geq t &= F_t(u)
\\
lim_u rightarrow 0 mathbbPL < t + u mid Lgeq t &= lim_u rightarrow 0 F_t(u)
\\
lim_u rightarrow 0 fracmathbbPL < t + u mid L geq tu &= lim_u rightarrow 0 fracF_t(u)u
\\
lim_u rightarrow 0 fracmathbbPL < t + u mid L geq tu &= lim_u rightarrow 0 fracF_t(u) - F_t(0)u
endalign
$$

where the last step holds because $F_t(0) = 1 - e^0 = 0$.

Question

Here is what I don't understand. The next step in the course notes is

$$
lim_u rightarrow 0 fracF_t(u) - F_t(0)u = fracpartialpartial u F_t(u) Bigg|_u=0 = h(t) tag$star$
$$

But I am used to the definition of a derivative as being

$$
lim_h rightarrow 0 fracf(x+h) - f(x)h = fracpartialpartial x f(x)
$$

and I am unable to see why $star$ holds.

If you have a function of two variables $F(t,u)$ (written $F_t(u)$ for your example), then the partial derivative with respect to $u$ is defined by $$fracpartialpartial uF(t,u) =limlimits_sto 0fracF(t, u+s)-F(t,u)s.$$
To evaluate this at $u=0$, we put $u=0$ on the right-hand side, so
$$fracpartialpartial uF(t,u)Bigg|_u=0=limlimits_sto 0fracF(t, s)-F(t,0)s.$$
Writing it with the subscript $t$ notation used in your example (so $F(t,s)=F_t(s)$ and $F(t,0) = F_t(0)$), we see that we have
$$fracpartialpartial uF_t(u)Bigg|_u=0 =limlimits_s to 0fracF_t(s)-F_t(0)s.$$
You can now change the $s$ to $u$ if you want, it is just a dummy variable for the limit. This matches the equation your notes had. (If you want to see why this equals $h(t)$, you will need to differentiate $F_t(u) = 1-e^-int_t^t+uh(s), ds$ with respect to $u$ (using the chain rule and Fundamental Theorem of Calculus), and then substitute in $u=0$.)