Calculating standard deviation on compounded investment returns
March 25, 2008 1:20 AM Subscribe
Is it possible to estimate the standard deviation of an investment's return after "y" years, if you know the investment's mean annual return and standard deviation?
In other words, I understand how to estimate the probability that next year's return might be n standard deviations above the mean, or between +n and -n standard devations around the mean.
But what if you bought and held that investment for y years. Is there a way you could estimate the standard deviation of the investment's value then?
E.g., let's say that i put $10,000 into an index fund. That fund has an average return of 8%, and a standard deviation of 16%, and I plan to hold it for 20 years. According to Excel's FV function, the portfolio's expected value will be $46,609.57 by then.
But I don't know how to estimate the standard deviation of the portfolio's possible value after 20 years of compounded growth. Is it possible to calculate/estimate it from just these measures?
I figure that simulation will get me an answer, but I'm hoping for something more generalizable...
Many thanks in advance for any advice or insight (or better yet, a formula!)
In other words, I understand how to estimate the probability that next year's return might be n standard deviations above the mean, or between +n and -n standard devations around the mean.
But what if you bought and held that investment for y years. Is there a way you could estimate the standard deviation of the investment's value then?
E.g., let's say that i put $10,000 into an index fund. That fund has an average return of 8%, and a standard deviation of 16%, and I plan to hold it for 20 years. According to Excel's FV function, the portfolio's expected value will be $46,609.57 by then.
But I don't know how to estimate the standard deviation of the portfolio's possible value after 20 years of compounded growth. Is it possible to calculate/estimate it from just these measures?
I figure that simulation will get me an answer, but I'm hoping for something more generalizable...
Many thanks in advance for any advice or insight (or better yet, a formula!)
If the standard deviation for one period is s, then the standard deviation for n periods is s * sqrt(n).
In your example, s = 16% and n = 20, so the 20-year standard deviation would be 16% * sqrt(20) = 72%.
posted by blue mustard at 3:31 AM on March 25, 2008
In your example, s = 16% and n = 20, so the 20-year standard deviation would be 16% * sqrt(20) = 72%.
posted by blue mustard at 3:31 AM on March 25, 2008
that's true when the quantities are added blue mustard, but in the case of returns they are all multiplied together. unfortunately we're stuck using the ugly equation i wrote above.
posted by njgo at 4:12 AM on March 25, 2008
posted by njgo at 4:12 AM on March 25, 2008
What I'm picturing is a graph with time on the bottom, and $ on the left side. There's a line starting at $10,000 and going up to $46000. Then a shaded area surrounding that line representing the deviation. However, I'm at a loss as to where to go from there. I don't think your assumptions are correct- the standard deviation of financial markets is different depending on the term. What happens in the 20th year will depend on what has happened in years 1-19. I'm sure there is a way to calculate this (calculus, I think), but I'm not sure it's going to be correct for this purpose. Maybe I'm over complicating it...
posted by gjc at 5:06 AM on March 25, 2008
posted by gjc at 5:06 AM on March 25, 2008
1. Be careful to note that, while the yearly return may be normally distributed with mean 1.08 and standard deviation 0.16, the compounded return after 20 years will not be normally distributed. Instead, it will have a long tail. For example, the mean (which you discovered using Excel's FV is 4.66096) is quite far from the mode (most probable outcome), which is near 2.0.
2. That said, the standard deviation on the 20-year return in your example is approximately 3.44 +/- 0.02. I calculated this via simulation.
3. In general, you are asking for the probability distribution of the product of n=20 independent normally distributed variables. This is not simple even in the n=2 case.
4. I assume this is a fairly fundamental and frequent calculation in quantitative finance. Thus, there may be useful approximations or specialised numerical methods for solving this difficult-in-general problem.
posted by hAndrew at 6:01 AM on March 25, 2008
2. That said, the standard deviation on the 20-year return in your example is approximately 3.44 +/- 0.02. I calculated this via simulation.
3. In general, you are asking for the probability distribution of the product of n=20 independent normally distributed variables. This is not simple even in the n=2 case.
4. I assume this is a fairly fundamental and frequent calculation in quantitative finance. Thus, there may be useful approximations or specialised numerical methods for solving this difficult-in-general problem.
posted by hAndrew at 6:01 AM on March 25, 2008
gjc, i'm not denying that there's causality in the markets when I say that the increments of the share prices are uncorrelated. this statement is the mathematical equivalent of saying that I cannot predict future performance of the market based solely on past performance. I must take new information that the market is unaware of into account if I want to gain an advantage. if this wasn't true, the markets would be "rigged" in some sense, and it would be very easy to capitalize off the imbalances. take at weak form efficiency here: http://en.wikipedia.org/wiki/Efficient_market. anyway, calculus is not required to do this calculation, the equation I wrote down above gives you the answer...
and hAndrew, the distribution does not need to be normal to have a standard deviation. SD is just defined as the square root of the variance, which is defined in the way I showed above. normal curves are neat in that they are completely defined by their SD and mean, but SD is a completely general concept.
posted by njgo at 7:48 AM on March 25, 2008
and hAndrew, the distribution does not need to be normal to have a standard deviation. SD is just defined as the square root of the variance, which is defined in the way I showed above. normal curves are neat in that they are completely defined by their SD and mean, but SD is a completely general concept.
posted by njgo at 7:48 AM on March 25, 2008
Response by poster: All,
Many thanks for these responses! A lot of insight!
njgo, I'm having a little trouble following the formula you posted - I was hoping you could provide a little clarification.
The formula you provided was:
v(V) = (v(N1) + {N1}^2)(v(N2) + {Nn}^2)...(v(Ni) + {Ni}^2) - {N1}^2*{N2}^2*...*{Nn}^2
I'm assuming you're using "n" and "i" interchangeably, and I think there's a typo in the second term. So, is this correct?
v(V) = (v(N1) + {N1}^2)(v(N2) + {N2}^2)...(v(Nn) + {Nn}^2) - {N1}^2*{N2}^2*...*{Nn}^2
Assuming it is, I glean from your post that Nn is the return of the investment in year n, and v(Nn) is the variance of returns in year n.
Since I'm trying to estimate what this investment might be worth in n years, all I have is the mean annual return and standard deviation.
So, in your formula, is N1, N2, ... Nn all the same? Likewise, is v(N1) == v(Nn)? And is {Nn} == Nn? Or, have I completely misunderstood these terms?
I guess, basically, I don't really understand how to apply the formula, given my three inputs (years, mean, and s.d.).
Again, many thanks in advance.
posted by stuehler at 8:44 AM on March 25, 2008
Many thanks for these responses! A lot of insight!
njgo, I'm having a little trouble following the formula you posted - I was hoping you could provide a little clarification.
The formula you provided was:
v(V) = (v(N1) + {N1}^2)(v(N2) + {Nn}^2)...(v(Ni) + {Ni}^2) - {N1}^2*{N2}^2*...*{Nn}^2
I'm assuming you're using "n" and "i" interchangeably, and I think there's a typo in the second term. So, is this correct?
v(V) = (v(N1) + {N1}^2)(v(N2) + {N2}^2)...(v(Nn) + {Nn}^2) - {N1}^2*{N2}^2*...*{Nn}^2
Assuming it is, I glean from your post that Nn is the return of the investment in year n, and v(Nn) is the variance of returns in year n.
Since I'm trying to estimate what this investment might be worth in n years, all I have is the mean annual return and standard deviation.
So, in your formula, is N1, N2, ... Nn all the same? Likewise, is v(N1) == v(Nn)? And is {Nn} == Nn? Or, have I completely misunderstood these terms?
I guess, basically, I don't really understand how to apply the formula, given my three inputs (years, mean, and s.d.).
Again, many thanks in advance.
posted by stuehler at 8:44 AM on March 25, 2008
njgo is right that you can quickly and easily compute the standard deviation of the 20-year return via njgo's formula, which is Sqrt[(0.16^2 + 1.08^2)^20 - (1.08^2)^20].
I would again warn, however, that the standard deviation of the 20-year return is probably not of any use, because the probability distribution of the 20-year return is not at all well described by its first two moments (mean and standard deviation). Its skewness is approximately 2.5.
posted by hAndrew at 8:53 AM on March 25, 2008
I would again warn, however, that the standard deviation of the 20-year return is probably not of any use, because the probability distribution of the 20-year return is not at all well described by its first two moments (mean and standard deviation). Its skewness is approximately 2.5.
posted by hAndrew at 8:53 AM on March 25, 2008
stuehler: exactly, in your case all the Ni have the same distribution, so v(Ni)s and {Ni}s are all the same. sorry about the type in the second term. hAndrew's expression is correct, and gives you an uncertainty of 344%.
posted by njgo at 9:01 AM on March 25, 2008
posted by njgo at 9:01 AM on March 25, 2008
Response by poster: All,
Again, many, many thanks for these excellent responses.
Judging from hAndrew and njgo's posts, which seem to settle the question, I guess there isn't a simple, yet reasonably accurate, formula for this problem.
The skewness and kurtosis of the n-year distribution make the formulas above too subject to error (both measures get much worse when the stdev gets higher).
I suppose there are probably very sophisticated techniques for handling those complications, but they're probably miles beyond my understanding.
I'm happy to use simulation - I just didn't want to go through the trouble if there was a simple formula that I was overlooking. I guess I don't have to worry about that.
Cheers
posted by stuehler at 9:20 AM on March 25, 2008
Again, many, many thanks for these excellent responses.
Judging from hAndrew and njgo's posts, which seem to settle the question, I guess there isn't a simple, yet reasonably accurate, formula for this problem.
The skewness and kurtosis of the n-year distribution make the formulas above too subject to error (both measures get much worse when the stdev gets higher).
I suppose there are probably very sophisticated techniques for handling those complications, but they're probably miles beyond my understanding.
I'm happy to use simulation - I just didn't want to go through the trouble if there was a simple formula that I was overlooking. I guess I don't have to worry about that.
Cheers
posted by stuehler at 9:20 AM on March 25, 2008
This is a bit late, but asset prices are frequently modeled as being lognormally distributed. This model actually behaves quite nicely, since products of lognormally distributed variables (such as day-to-day price changes) are also log normal.
I think blue mustard was right--the standard deviation increases with the square root of time.
http://www.bionicturtle.com/learn/article/lognormal_distribution_introduction/
posted by xbalto at 10:49 PM on March 25, 2008
I think blue mustard was right--the standard deviation increases with the square root of time.
http://www.bionicturtle.com/learn/article/lognormal_distribution_introduction/
posted by xbalto at 10:49 PM on March 25, 2008
As others have said, you can do what you want using standard deviation and the square root of forward periods with a lognormal price model, but it's debatable how useful the result will be for a couple of reasons. First, you need to assume the price process follows a lognormal distribution, but as others have pointed out long-tail behavior is real and gets more significant as time goes on.*
Second, your measure assumes that your holding behavior will not change with the fund's performance. IRL, you'll likely dump a poorly-performing fund after X years of tanking -- or it may even wind down (less likely for an index fund, but it happens. It's probably even more likely with the possibility of major changes in the global economy on the horizon.) Mathematically, this contributes to the long-tailiness by making it even less likely you'll see lower returns. When you pull the trigger, your losses on that investment stop. Your risk tolerance, after having taken a hit, may well have changed by that point, and then are you going to put the leftovers into an investment with a similar risk-reward profile or into something safer? Well, IRL you probably won't know that until you are in that position.
I guess what I'm saying is this is an interesting mathematical exercise, and the kind of thing I might throw at a finance student to see if they really get the relationship between volatility and time, but I think you'll find such a wide uncertainty band that you'll confirm that investments can be dangerous if not monitored. I'd guess anything less than an order-of-magnitude difference in the size of the band between two investments is probably too narrow to carry much significance. And I don't know how well it will characterize your expectation of results assuming real-life investor behavior.
None of the above should be taken as a criticism of the initial question, by the way -- it's a great one.
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*Okay, I don't like the lognormal model all that well even for short periods for certain assets, but this is not the place to debate that.
posted by Opposite George at 6:18 AM on March 26, 2008
Second, your measure assumes that your holding behavior will not change with the fund's performance. IRL, you'll likely dump a poorly-performing fund after X years of tanking -- or it may even wind down (less likely for an index fund, but it happens. It's probably even more likely with the possibility of major changes in the global economy on the horizon.) Mathematically, this contributes to the long-tailiness by making it even less likely you'll see lower returns. When you pull the trigger, your losses on that investment stop. Your risk tolerance, after having taken a hit, may well have changed by that point, and then are you going to put the leftovers into an investment with a similar risk-reward profile or into something safer? Well, IRL you probably won't know that until you are in that position.
I guess what I'm saying is this is an interesting mathematical exercise, and the kind of thing I might throw at a finance student to see if they really get the relationship between volatility and time, but I think you'll find such a wide uncertainty band that you'll confirm that investments can be dangerous if not monitored. I'd guess anything less than an order-of-magnitude difference in the size of the band between two investments is probably too narrow to carry much significance. And I don't know how well it will characterize your expectation of results assuming real-life investor behavior.
None of the above should be taken as a criticism of the initial question, by the way -- it's a great one.
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*Okay, I don't like the lognormal model all that well even for short periods for certain assets, but this is not the place to debate that.
posted by Opposite George at 6:18 AM on March 26, 2008
This thread is closed to new comments.
V = N1*N2*...*Nn
The variance (the square of the standard deviation) is defined by
v(V) = {V^2} - {V}^2 where I'm using curly braces to represent average values. For the product we care about,
v(V) = {N1^2}*{N2^2}*...*{Nn^2} - {N1}^2*{N2}^2*...*{Nn}^2
Now we can again use the definition {Ni^2} = v(Ni) + {Ni}^2 to get
v(V) = (v(N1) + {N1}^2)(v(N2) + {Nn}^2)...(v(Ni) + {Ni}^2) - {N1}^2*{N2}^2*...*{Nn}^2
which I believe is the formula you seek. Sorry that html math isn't easier on the eyes.
posted by njgo at 3:13 AM on March 25, 2008