# The Balducci Assumption

### The Balducci Assumption

Numerical example

Suppose we are given that $q_{70}=0.15$ and we wish to calculate $_{0.2}q_{70}$ using the Balducci Assumption

The Balducci assumption states that $_{1-t}q_{x+t} = (1-t) q_x$

We are interested in the period aged $70$ to $70.2$ so we need to set $t=0.2$ so that we are considering mortality to the age $70.2$ not the age $70.8$

So we have that $_{1-0.2}q_{70.2}={}_{0.8}q_{70.2}=(1-0.2)q_{70} = 0.8 \times 0.15 = 0.12$

We have from basic probability that: ${}_1p_{70} = {}_{0.2}p_{70}\times {}_{0.8} p_{70.2}$

So ${}_{0.2}q_{70}=1-{}_{0.2}p_{70}= 1- \frac{{}_1p_{70}}{{}_{0.8} p_{70.2}}=1- \frac{1-{}_1q_{70}}{1-{}_{0.8} q_{70.2}}=1- \frac{1-0.15}{1-0.12}=0.3409$

Equivalence to Linear Interpolation of Asymptotes

We start from the Balducci Assumption: ${}_{1-t}q_{x+t} = (1-t) q_x$

$\therefore 1-{}_{1-t}p_{x+t} = (1-t) (1-p_x)$

$\therefore 1-\frac{S(x+1)}{S(x+t)} = (1-t) (1-\frac{S(x+1)}{S(x)})$

$\therefore 1-\frac{S(x+1)}{S(x+t)} = -\frac{S(x+1)}{S(x)}\times(1-t) +1-t$

$\therefore \frac{S(x+1)}{S(x+t)} = \frac{S(x+1)}{S(x)}\times(1-t) + t$

$\therefore \frac{1}{S(x+t)} = \frac{1-t}{S(x)} + \frac{t}{S(x+1)}$

And so we can see that the Balducci Assumption is equivalent to the linear interpolation of he reciprocal of the survival function