What a loss costs to get back
Fall 20% and you need 25% to get back; fall 50% and you need 100%. Turn that into years, and then add withdrawals, which make it far worse.
- Compute the gain a loss needs, and why it grows faster than the loss
- Turn a recovery into years at a steady return
- Show what withdrawals do to the recovery
- Find the largest fall a recovery horizon allows, and scale it to a mix
The intuition
A client's $100,000 falls 20% to $80,000. Getting back to $100,000 is a 25% gain, not 20%, because it is earned on the smaller balance. Fall 50% and the gain needed is 100%. The deeper the hole, the faster the required gain grows, which is the most useful thing to show a client when setting risk: a drawdown is not a number, it is years of waiting.
At a steady 6% a 20% fall takes about 3.8 years to recover. Add a retiree taking 4% of the balance out every year and the net growth is only 1.76% a year, so the same fall takes nearly thirteen years; take out more than the return and it never recovers at all. Money withdrawn during the recovery never takes part in it, which is why the same drawdown that is an inconvenience for a saver can be permanent for a retiree.
Gain needed after a loss d = d ÷ (1 − d). Years to recover at return g = ln(1 ÷ (1 − d)) ÷ ln(1 + g). With withdrawals of w a year: replace (1 + g) with (1 + g)(1 − w). Largest loss recoverable in Y years = 1 − (1 + g)^−Y. A mixed portfolio's loss with bonds flat = equity weight × equity loss.
Why it works
- The conventions here: a single drawdown followed by a constant annual return. Recovery means the pre-loss dollar value. Withdrawals are a fixed share of the current value at the end of each year, so the value grows by (1 + g)(1 − w) a year. Years are fractional.
- Losses and gains are not symmetric. The gain is earned on what is left, so d ÷ (1 − d) rises faster than d. The arithmetic average of −50% and +50% is zero; the money is down 25%.
- Years turn a percentage into a plan. Growth needed as a factor is 1 ÷ (1 − d); the years are its log over the log of the growth rate.
- Withdrawals shrink the base every year. The net factor (1 + g)(1 − w) is what compounds; at or below one, the old value is never seen again, and the drawdown is a permanent cut to what the plan can pay.
- Run it backwards for a limit. A client who must be whole within Y years can stand a fall of at most 1 − (1 + g)^−Y. Divide by a stress-test equity crash to turn that into an equity weight.
- The geometric average is the one that describes what happened. After −50% and +50% it is √(0.5 × 1.5) − 1 ≈ −13.4% a year; the gap between it and the arithmetic average grows with volatility.
| Gain needed: 20% ÷ 80% | 25.00% |
| Years to recover: ln 1.25 ÷ ln 1.06 | 3.83 |
| Net factor with withdrawals: 1.06 × 0.96 | 1.0176 |
| Years with withdrawals: ln 1.25 ÷ ln 1.0176 | 12.8 |
| Largest fall recoverable in 5 years: 1 − 1.06⁻⁵ | 25.27% |
| 60/40 in a 30% crash: loss 18%; gain needed 21.95% | 3.41 years to recover |
Withdrawals more than tripled the recovery. Fall 50% instead of 20% and the saver needs a 100% gain and about 11.9 years at 6%.
The formulas
Earned on what is left, so it outgrows the loss.
The log of the growth needed over the log of the growth rate.
The net factor is what compounds.
The years formula run backwards.
Scale the crash to the mix.
Worked example
The growth needed as a factor is one over what is left. Take logs to turn the factor into years at the stated return.
See it move
Same client and the same return after the fall ("New company and numbers" changes it). Change the size of the fall, the withdrawal rate, the recovery horizon and the equity weight of the mix.
- Raise the fall. The gain needed rises faster than the fall, and every recovery takes longer.
- Raise the withdrawals. The net growth factor falls toward one, and nothing else moves; the slider stops a point below the return, because at the return the recovery never ends.
- Lengthen the years the client can wait. The largest fall the horizon allows grows; nothing else moves.
- Raise the equity weight of the mix. The mixed portfolio's loss and its recovery time rise toward the all-equity case.
Run it backwards
The client can wait a stated number of years and expects a stated return. What is the largest fall they can stand?
Set (1 − d)(1 + g)^Y = 1 and solve: d = 1 − (1 + g)^−Y. It is the years formula run backwards.
The follow-up turns the limit into an allocation by dividing by a stress-test equity fall: the same conversion the previous chapter used.
Traps
Say it in the interview
“How would you show a client what a big loss really costs?”
Check yourself
4 fresh questions, with new numbers. Answer each one correctly to finish the chapter. Get one wrong and you will see the full working, then you can try it again with new numbers.
Answers within 1% are marked right. Type the number; $, %, x and M are fine. First tries count toward Learned: the topic is Learned once every chapter is done and 75% of first tries were right.
- Gain needed = d ÷ (1 − d); it outgrows the loss.
- Years = ln(1 ÷ (1 − d)) ÷ ln(1 + g); with withdrawals, use (1 + g)(1 − w).
- Withdrawals at or above the return make a drawdown permanent.
- Largest recoverable fall = 1 − (1 + g)^−Y; divide by the crash for a weight.