Why Delayed Repetitions of a Cash Flow Preserve Its IRR

# 45350


# 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
```

## Answer by dm63 (score 2, accepted)

https://quant.stackexchange.com/a/45352

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?