## Interpolation May 27, 2007

Posted by Peter in Exam 3/MLC.

We use interpolation whenever we want to construct a continuous model from discrete data. Interpolation methods range from the rudimentary (linear interpolation) to the sophisticated (polynomial splines). Since splines are no longer on the 4/C syllabus, we’ll instead talk about forms of interpolation based on the relation

$\varphi(s(x+t)) = (1-t)\varphi(s(x)) + t\varphi(s(x+1)), \quad t \in [0,1].$

Here, s(x) represents the function to be interpolated, $\varphi$ is an interpolation assumption, and t is a parameter on [0,1]. The simplest instance of the above is when $\varphi$ is the identity function:

$s(x+t) = (1-t)s(x) + t s(x+1).$

This is called linear interpolation, and it is the basis of the uniform distribution of deaths (UDD) assumption in life contingencies, where s is the survival distribution and x is age. But we also use this relationship (albeit slightly modified) when interpolating values in, say, a normal distribution table:

$\Phi\left(x+\frac{t}{100}\right) = (1-t)\Phi(x) + t\Phi\left(x+\frac{1}{100}\right).$

This assumes that adjacent entries in the table are listed in increments of 0.01. For instance, suppose we want to find Φ(1.263). 1.263 is between 1.26 and 1.27, so we have

$\Phi(1.263) \approx (1-0.3)\Phi(1.26) + 0.3 \Phi(1.27),$

and looking up the values in the table, we get Φ(1.263) = (0.7)(0.8962) + (0.3)(0.8980) = 0.89674.

In life contingencies, we are also sometimes interested in the constant force of mortality interpolation assumption; that is to say, deaths are not uniformly distributed at fractional ages, but rather, survival is exponentially distributed between integer ages. In this case, $\varphi(s) = \log s$ and the interpolation relation becomes

$\log s(x+t) = (1-t) \log s(x) + t \log s(x+1)$

or equivalently,

$s(x+t) = s(x)^{1-t} s(x+1)^t.$

To see that this indeed results in a constant force of mortality between integer ages, we differentiate the above with respect to t to obtain

$\displaystyle \mu_x(t) = -\frac{d}{dt}\left[\log s(x+t)\right] = \log s(x) - \log s(x+1) = \log \frac{s(x)}{s(x+1)} \ge 0,$

since s(x) ≥ s(x+1). Finally, there is the Balducci, or hyperbolic, interpolation assumption, where we set $\varphi(s) = 1/s$:

$\displaystyle \frac{1}{s(x+t)} = \frac{1-t}{s(x)} + \frac{t}{s(x+1)}.$

This model is so called because the survival function at fractional ages forms an arc of a hyperbola. In all cases, we can use the resulting relation on the survival function to derive the other life table functions $l_{x+t}, \,_t p_x, \,_t q_x, \mu_x(t)$, etc. But as we have seen, these three interpolation assumptions are not the only ones we can use, even in the very simple case of two-point interpolation.