The question
The IRR of two sets of cashflow is not (necessarily) the weighted average of each set of cashflows. E.g. if
A = (-100,110)
B = (-80,100)
C = (-180,210)
then
IRR(A) = 10%
IRR(B) = 25%
IRR(C) = 16.666%
unweighted average IRR = 17.5%
weighted average IRR = 20%
However, is there a mathematical proof that the IRR of the sum must be within the range of the two IRRs, i.e. that
IRR(A) <= IRR(A+B) <= IRR(B) ?
Intuitively, I get the concept, but is there a generic mathematical proof, that holds regardless of the items in the cashflow, i.e. regardless of the degree of the polynomials?
There was a discussion here, but I am not sure it fully answers the question (or, if it does, I'm not sure I fully understood it), especially for a general case regardless of the degree of the polynomial.
The background
Note: the rest below is just for colour.
Why do I need this? Because I need to prove that the IRR of one project + the same project starting a few periods laters is the same as the IRR of the single project, e.g.:
IRR(-100,0,121) = IRR(-100,0,121,-100,0,121)
We see that
IRR(-100,0,121) = 10%
If the cashflows start some periods later, it can be proven that the IRR is still the same:
IRR(0,0,0,-100,0,121) = 10%
The IRR of the sum is still the same in this example,
IRR(-100,0,121,-100,0,121)= 10%
but is there a mathematical proof for this? Proving that
IRR(A) <= IRR(A+B) <= IRR(B)
would prove it, because delaying cashflows doesn't affect the IRRs. Proving this is quite simple. Say the cashflow is over 3 periods, and the IRR is the i which solves:
$a + \frac{b}{(1+i)} + \frac{c}{(1+i)^2} = 0$
Delaying it by one period simply means dividing each item by $(1+i)$:
$0 + \frac{a}{(1+i)} + \frac{b}{(1+i)^2} + \frac{c}{(1+i)^3} = 0$
which can of course be simplified away.
So, to recap, we know that
IRR(A) = x
IRR(0,0,A) = x
if we can prove that IRR(A) <= IRR(A+B) <= IRR(B) then it follows that
IRR(A,0,A) = x , too