# Negative properties of continuously compounded returns

I'm taking an intro to quant finance course online. I'm trying to better understand the properties and uses of continuously compounded returns. Specifically for this question, how can you have negative returns less than -100%? In an example problem given, there is a monthly cc return of -.20. We are told if we get that same RoR each month for 12 months, what is our return on a year? Answer given: 12 x (-.20) = -2.4 (-240%).

Is this of any practical use?

• This question appears to be off-topic because it is primarily theoretical. "Continuously compounded returns" are an academic exercise in the highest possible compounding frequency. Sep 11, 2013 at 9:28
• "Specifically for this question, how can you have negative returns less than -100%?" Because you can go into debt. For example, using leverage, you can easily lose more than you put in, i.e. face a return of less than -100%. Sep 11, 2013 at 11:16
• You could also keep investing, i.e. throwing good money in after bad, thus keeping the balance above zero. Sep 11, 2013 at 12:08
• Losses of 20% are calculated by multiplying by .8, or .8^N for N periods. .8^12 is .068. You are left with just under 7% of your starting amount. Sorry, in my opinion, -240% is incorrect and simplistic. Sep 11, 2013 at 23:42

What you're missing is the continuous compounding computation doesn't work that way. If you compound over n periods of time and a rate of return of r, the formula is e^(r*n), as you have to multiply the returns together with a mulitplicative base of 1. Otherwise consider what 0 does to your formula. If I get a zero return, I have a zero result which doesn't make sense. However, in my formula I'd still get the 1 which is what I'm starting and thus the no effect is the intended result.

Continuous compounding would give e^(-.20*12) = e^(-2.4) = .0907 which is a -91% return so for each \$100 invested, the person ends up with \$9.07 left at the end. It may help to picture that the function e^(-x) does asymptotically approach zero as x tends to infinity, but that is as bad as it can get, so one doesn't cross into the negative unless one wants to do returns in a Complex number system with imaginary numbers in here somehow.

For those wanting the usual compounding, here would be that computation which is more brutal actually: For your case it would be (1-.20)^12=(0.8)^12=0.068719476736 which is to say that someone ends up with 6.87% in the end. For each \$100 had in the beginning they would end with \$6.87 in the end.

Consider someone starting with \$100 and take 20% off time and time again you'd see this as it would go down to \$80 after the first month and then down to \$64 the second month as the amount gets lower the amount taken off gets lower too. This can be continued for all 12 terms. Note that the second case isn't another \$20 loss but only \$16 though it is the same percentage overall.

Some retail stores may do discounts on discounts so this can happen in reality. Take 50% off of something already marked down 50% and it isn't free, it is down 75% in total. Just to give a real world example where while you think a half and a half is a whole, taking half and then half of a half is only three fourths, sorry to say. You could do this with an apple or a pizza if you want a food example to consider.

Alternatively, consider the classic up and down case where an investment goes up 10% and down 10%. On the surface, these should cancel and negate each other, right? No, in fact the total return is down 1% as the computation would be (1.1)(.9)=.99 which is slightly less than 1.

Continuous compounding may be a bit exotic from a Mathematical concept but the idea of handling geometric means and how compounding returns comes together is something that is rather practical for people to consider.

• Good explanation, but its still annual compounding in your example. The formula for continuous compounding is (amount)= (principle) * ( e ^(r*t)). e there is the math constant 'Napier's number' and about 2.78. Continuous compounding is still an academic construct and doesn't exist in any applications most people here will encounter. Sep 11, 2013 at 21:15
• @theao I thought that at first also. But the OP has a monthly rate, and the formula doesn't take a monthly rate. Sep 11, 2013 at 21:32
• I now have both computations in my answer. There is something to be said for the continuous one not being quite as bad as the other case. Sep 11, 2013 at 21:36
• But you have to turn 20% per month into a continuously compounded rate. Sep 11, 2013 at 21:38
• Which is why I previously voted for the thing to be closed. To be precise a continuously compounded rate is the academic exercise of compounding infinitely. Compounding at any other frequency will get you close enough. JB King has made a killer answer out of it though Sep 11, 2013 at 21:40

You cannot use continuous compounding for returns less than or equal to 100% because a natural logarithm can only be taken for a positive amount. This answer includes the accurate way to ascertain `r`, for which many people use an approximation.

For example, using -20% monthly return for 12 months:-

``````percent = 0.01;
RoR = -20 percent
``````

-0.2

``````r = Log[RoR + 1]
``````

-0.223144

``````P = 1;
t = 12;
A = P E^(r t)
``````

0.0687195

Checking:

``````P = 1;
m1 = P*(RoR + 1);
m2 = m1*(RoR + 1);
m3 = m2*(RoR + 1);
m4 = m3*(RoR + 1);
m5 = m4*(RoR + 1);
m6 = m5*(RoR + 1);
m7 = m6*(RoR + 1);
m8 = m7*(RoR + 1);
m9 = m8*(RoR + 1);
m10 = m9*(RoR + 1);
m11 = m10*(RoR + 1);
m12 = m11*(RoR + 1)
``````

0.0687195

``````A == m12
``````

True

Now trying -100% monthly return:-

``````RoR = -100 percent
``````

-1.

``````r = Log[RoR + 1]
``````

Indeterminate

Why? Because a natural logarithm can only be taken for a positive amount.

``````Plot[Log[x], {x, -10, 10}]
`````` So the latter calculation can not be done using (logarithmic) continuous compounding.

Of course, the calculation can still be done using regular compounding. For -100% the results go to zero in the first month, but -150% produces a more interesting result:

``````RoR = -150 percent
``````

-1.5

``````P = 1;
m1 = P*(RoR + 1);
m2 = m1 - (m1*RoR);
m3 = m2 - (m2*RoR);
m4 = m3 - (m3*RoR);
m5 = m4 - (m4*RoR);
m6 = m5 - (m5*RoR);
m7 = m6 - (m6*RoR);
m8 = m7 - (m7*RoR);
m9 = m8 - (m8*RoR);
m10 = m9 - (m9*RoR);
m11 = m10 - (m10*RoR);
m12 = m11 - (m11*RoR)
``````

-11920.9

Well, one can easily have rates below -100%.

Suppose I start with \$100, and end up with \$9 after a year.

What was my rate of return? It could be -91%, -181%, -218%, or -241%, or something else, depending on the compounding method.

We always have that the final amount equals the initial amount times a growth factor G, and we can express this using a rate r and a day count fraction T.

1. Simple compounding: B(T) = B(0) * ( 1 + r*T )
2. Annual compounding: B(T) = B(0) * ( 1 + r )^T
3. Quarterly compounding: B(T) = B(0) * ( 1 + r/4 )^(4*T)
4. Monthly compounding: B(T) = B(0) * ( 1 + r/12 )^(12*T)
5. Continuous compounding: B(T) = B(0) * exp( r*T )

In this case, we have T = 1, and B(T) = B(0) * 0.09, so:

1. Simple: 0.09 = 1 + 1 * r, thus r = (G-1)/T = (0.09-1)/1, so r = -91%
2. Annual: 0.09 = (1 + r)^1, thus r = G^(1/T)-1 = 0.09^1-1, so r = -91%
3. Quarterly: 0.09 = (1 + r/4)^(4*1), thus r = 4(G^(1/4T)-1) = 4(0.09^(1/4)-1), so r = -181%
4. Monthly: 0.09 = (1 + r/12)^(12*1), thus r = 12(G^(1/12T)-1) = 12(0.09^(1/12)-1), so r = -218%
5. Continuous: 0.09 = exp(r*1), thus r = ln(G)/T, so r = -241%

So, depending on how we compound, we have a rate of return of -91%, -181%, -218%, or -241%.

This nicely illustrates that:

• it doesn't make sense (or, at any rate, it's imprecise) to talk of a rate or yield without specifying the compounding method (and furthermore the day count fraction method, though here we just use T=1)
• while all these compounding methods line up well for rates close to zero, they do not for very large (or very small) rates,
• continuous compounding has the nice property that the geometric mean of the growth factors is replaced by the arithmetic mean of the (continuously compounded) rates.
• for simple and annual compounding, it doesn't make sense to talk of rates less than -100%. For more frequent (semi-annual, quarterly etc.) or continuous compounding, we can have rates less than -100%.
• The problem comes when continuous methods are used to calculate the effective annual rate, `r = e^i - 1`, where `i` is the nominal rate. For example, when `i = -2.41` (as you calculated), `r = -91%`. When the effective annual rate, `r <= -100%` the logarithmic calculations cannot be done. (Note:- "The effective annual rate is the total accumulated interest that would be payable up to the end of one year, divided by the principal." - ref) Apr 15, 2014 at 12:02
• Good point. If your growth factor goes negative (as an example, your investment goes from \$100 to \$-50), you would not be able to express this using a continuously compounded rate, as you said, while you would be able to express this using a simple or annually compounded rate (-150%). However, I think the original question did not refer to that case.
– Fab
Apr 21, 2014 at 17:57