# Proof that IRR(A) < IRR(A+B) < IRR(B) ? Ie that the IRR of two cashflows together must be within the range of the IRR of the two cashflows?

## 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

• The statement that IRR(A) <= IRR(A+B) <= IRR(B) is not true in general, so you cannot use it to prove what you are trying to prove. – Alex C Apr 28 '19 at 14:40
• @alex-c , can you show me a counter-example where IRR(A) <= IRR(A+B) <= IRR(B) doesn't hold? I must say I couldn't think of one. Also, do you maybe have any suggestions on how I could prove that IRR(A)=IRR([0,0,0,A])? Thanks! – Pythonista anonymous Apr 28 '19 at 14:56

Another way to write:

IRR(A) = x and IRR(0,0,A) = x is:

PV(A;x)=0 and PV(0,0,A;x)=0

where PV=present value, and x is the discount rate. Since we are using the same discount rate x, we can just add these up:

PV(A,0,A;x)=0 which means that IRR(A,0,A) = x

Is it clear?

• Yes, thanks! In fact, I now realise my question was actually rather banal :( As for the point raised by Alex-C, can you think of a counter-example where IRR(A) <= IRR(A+B) <= IRR(B) does not hold? – Pythonista anonymous Apr 28 '19 at 16:15
• A=(-100,110) and B=(101,-101) is a counterexample. If you restrict A, B to have the sign always negative on first cashflow and positive on subsequent cashflows, the statement is true. – dm63 Apr 28 '19 at 16:23
• For a proof: let IRR(A)=a, IRR(B)=b, and assume b>a. PV(A;x) and PV(B;x) are both decreasing functions of x , with PV(A;a)=0 and PV(B;b)=0. We have PV(A+B;a)=PV(A;a)+PV(B;a)=PV(B;a)>0. Also PV(A+B;b)=PV(A;b)+PV(B;b)=PV(A;b)<0. Then By the intermediate value theorem the function PV(A+B;x) has a root between a and b. – dm63 Apr 28 '19 at 16:50