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)


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
  • $\begingroup$ 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. $\endgroup$
    – Alex C
    Commented Apr 28, 2019 at 14:40
  • $\begingroup$ @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! $\endgroup$ Commented Apr 28, 2019 at 14:56

1 Answer 1


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?

  • $\begingroup$ 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? $\endgroup$ Commented Apr 28, 2019 at 16:15
  • $\begingroup$ 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. $\endgroup$
    – dm63
    Commented Apr 28, 2019 at 16:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – dm63
    Commented Apr 28, 2019 at 16:50

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