Deriving the force of interest

[Fooddude]

Hi all

Apologies for asking what is probably quite a simple question. My excuse is that I'm home studying so I don't have any other access to any actuaries, maths gurus or tutors. It's also been over 10 years since maths a-level!

I want to check whether I have properly understood how delta t is arrived at - and get some help on the last manipulation.

I'll run through it in bits:

1. F(t + h) - F(t) / h

This is the rate of change of the value of the fund - F (the y axis) with respect to time (x axis). Putting this in the limit of h (the small increase in time) tending towards 0 is the same as taking the derivative or writing F'(t).

2. The above is divided by F(t) since we want to know the rate of change as a proportion of where it was a moment ago. Like a percentage - dividing the increase by the starting value.

So that gets me to F'(t)/F(t)

As I understand it, all we've done is rewrite our original formula - we haven't actually differentiated anything? Have I understood this correctly - or have I got to the right place but using faulty logic?

Now how do we get to d/dt ln F(t)?

I understand that d/dt ln(t) = 1/t, but other than that I'm having trouble seeing it. I suspect that I could just be being a bit dense!

Any help appreciated

Cheers

Rob

 

[didster]

Differentiating ln(F(x)) wrt x gives F'(x)/F(x) using the chain rule.

Search for "chain rule" in wikipedia or an A'level maths book.

 

[Fooddude]

Ok, found it - though if I'm honest I'm not sure how to derive that result - I'm sure it doesn't matter.

Thanks for the pointer

 

[didster]

u = F(x)
y = ln(F(x)) = ln(u)

dy/dx = dy/du * du/dx
= 1/u * F'(x)
= F'(x)/F(x) (replacing u with F(x))

No need to know the proof of chain rule if that's what you meant.