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When MIRR becomes IRR

A short algebraic proof that the IRR is a special case of the MIRR.

Setup

Consider a fund with a serie of cash flows where capital calls are negative and distributions are positive. Let Ct denote calls (negative values) and Dt denote distributions (positive values).

The MIRR is defined as:

MIRR DEFINITION
$$\text{MIRR} = \left(\frac{FV}{PV}\right)^{\frac{1}{N}} - 1$$
Where FV is the future value:
$$\;FV = \displaystyle\sum_{t} D_t \cdot (1+r_{reinv})^{N-t}\;$$
And PV the present value:
$$\;PV = \displaystyle\sum_{t} \frac{|C_t|}{(1+r_{uncalled})^{t}}$$

Here runcalled​ is the uncalled capital return (for calls) and rreinv is the reinvestment return (for distributions).

The claim

If we set runcalled​ = rreinv = r and the MIRR also equals r, then the result is equivalent to the standard IRR definition:

IRR DEFINITION
$$\sum_{t} \frac{CF_t}{(1 + \text{r})^{t}} = 0$$

Proof

Step 1: Set all rates equal to r
$$FV = \sum_{t} D_t \cdot (1+r)^{N-t} \qquad PV = \sum_{t} \frac{|C_t|}{(1+r)^{t}}$$
Step 2: MIRR = r means
$$\left(\frac{FV}{PV}\right)^{\frac{1}{N}} - 1 = r \quad \Longrightarrow \quad \frac{FV}{PV} = (1+r)^N$$
Step 3: Rearrange
$$FV = PV \cdot (1+r)^N$$
$$\sum_{t} D_t \cdot (1+r)^{N-t} \;=\; \sum_{t} \frac{|C_t|}{(1+r)^{t}} \cdot (1+r)^N$$
Step 4: Simplify the right side
$$\sum_{t} D_t \cdot (1+r)^{N-t} \;=\; \sum_{t} |C_t| \cdot (1+r)^{N-t}$$
Step 5: Subtract and factor
$$\sum_{t} \bigl(D_t - |C_t|\bigr) \cdot (1+r)^{N-t} = 0$$
Dividing both sides by (1+r)N:
$$\sum_{t} \bigl(D_t - |C_t|\bigr) \cdot \frac{1}{(1+r)^{t}} = 0$$
Step 6: Recognize the net cash flow
Since CFt = Dt − |Ct| (distributions minus calls = net cash flow):
$$\sum_{t=0}^{N} \frac{CF_t}{(1+r)^t} = 0$$

Q.E.D.
This is exactly the definition of the IRR: the rate rr that makes the net present value of all cash flows equal to zero.
The standard IRR is therefore a special case of the MIRR where the finance rate, the reinvestment rate, and the resulting return are all assumed to be the same.
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