College Math Teaching

August 28, 2017

Integration by parts: why the choice of “v” from “dv” might matter…

We all know the integration by parts formula: \int u dv = uv - \int v du though, of course, there is some choice in what v is; any anti-derivative will do. Well, sort of.

I thought about this as I’ve been roped into teaching an actuarial mathematics class (and no, I have zero training in this area…grrr…)

So here is the set up: let F_x(t) = P(0 \leq T_x \leq t) where T_x is the random variable that denotes the number of years longer a person aged x will live. Of course, F_x is a probability distribution function with density function f and if we assume that F is smooth and T_x has a finite expected value we can do the following: E(T_x) = \int^{\infty}_0 t f_x(t) dt and, in principle this integral can be done by parts….but…if we use u = t, dv = f_x(t), du = dt, v = F_x we have:

\

t(F_x(t))|^{\infty}_0 -\int^{\infty}_0 F_x(t) dt which is a big problem on many levels. For one, lim_{t \rightarrow \infty}F_x(t) = 1 and so the new integral does not converge..and the first term doesn’t either.

But if, for v = -(1-F_x(t)) we note that (1-F_x(t)) = S_x(t) is the survival function whose limit does go to zero, and there is usually the assumption that tS_x(t) \rightarrow 0 as t \rightarrow \infty

So we now have: -(S_x(t) t)|^{\infty}_0 + \int^{\infty}_0 S_x(t) dt = \int^{\infty}_0 S_x(t) dt = E(T_x) which is one of the more important formulas.

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