Hi Ross
It is possible to write the integral with either the range 0 to infinity or 5 to infinity, depending on how you define the variable of integration. I'll show you what I meant. Let's start with the definition of the complete expectation of life, as the integral of tpx, and split it out into the first 5 years and the subsequent 5 years:
\[ e^o_0 = \int_0^{\infty} {}_tp_0 \,dt = \int_0^5 {}_tp_0 \,dt + \int_5^{\infty} {}_tp_0 \,dt\]
The expression inside the first integral is just exp(-mu*t). The expression inside the second integral can be rewritten as follows, using the principle of consistency (for when \( t > 5 \) ).
\[ {}_tp_0 = {}_5p_0 * {}_{t-5}p_5 \]
Thus we have:
\[ e^o_0 = \int_0^{\infty} {}_tp_0 \,dt = \int_0^5 e^{-\mu * t} \,dt + \int_5^{\infty} {}_5p_0 * {}_{t-5}p_5 \,dt\]
Now, substituting in expressions with exponentials for the probabilities in the second integral we get:
\[ e^o_0 = \int_0^{\infty} {}_tp_0 \,dt = \int_0^5 e^{-\mu * t} \,dt + \int_5^{\infty} e^{-\mu * 5} * e^{-\lambda * (t-5)} \,dt\]
Being careful to include the (t-5) in the exponent for the second probability, as this is the duration, not t.
You could evaluate the integral from there and you should get the same answers as the specimen paper solutions.
However, we can represent the second integral slightly differently by doing a substitution (in the second integral only):
\[ s = t - 5 \]
this gives:
\[ e^o_0 = \int_0^{\infty} {}_tp_0 \,dt = \int_0^5 e^{-\mu * t} \,dt + \int_0^{\infty} e^{-\mu * 5} * e^{-\lambda * (s)} \,ds\]
This is the form in the specimen paper and is equally valid.
Hope this helps
Andy
