On GMO's 7-Year Asset Class Return Model

Every month GMO gives forecasts for several different asset classes (sign up for free to access these). What I'm going to do a quick look at is the model they use to derive these projections.

James Montier in What Goes Up Must Come Down (a great piece on profit margins) provides some insight into how they model the returns. You can see this in Exhibit 2 of the paper:

The idea is pretty simple. Returns will be dividends plus price appreciation. What they're doing is breaking down price appreciation into components.

For starters we'll ignore dividends (assume we're dealing with a non-dividend paying stock). Then returns will be:
$$\text{Returns} = \left(\frac{\text{Final Price}}{\text{Initial Price}}\right)^{(1/t)}-1$$
where $t$ is the number of years.

In general, the price can be broken down into three components:
$$\text{Price} = \frac{\text{Price}}{\text{Earnings}} \times \frac{\text{Earnings}}{\text{Sales}} \times \text{Sales}$$
$$\text{Price} =  \text{PE} \times \text{Profit Margin} \times \text{Sales}$$
So that's the basic model. So in order to project the future price (and subsequently, the future returns) we need to know what the future PE will be, the future Profit Margin and future Sales.

GMO projects future sales by assuming a constant growth rate $g$. In other words:
$$\text{Sales}_f = \text{Sales}_i \times (1+g)^t$$
where $t$ is number of years in the future (GMO looks at $t =7 $ years).

That will give us expressions for the initial and final price:
$$\text{Price}_i = \text{PE}_i \times \text{Profit Margin}_i \times \text{Sales}_i$$
$$\text{Price}_f = \text{PE}_f \times \text{Profit Margin}_f \times \text{Sales}_i \times (1+g)^t$$
Therefore returns will be:
$$\text{Returns} = \left(\frac{\text{PE}_f \times \text{Profit Margin}_f \times \text{Sales}_i \times (1+g)^t}{\text{PE}_i \times \text{Profit Margin}_i \times \text{Sales}_i}\right)^{(1/t)} - 1$$
This simplifies to:
$$\text{Returns} = \left(\frac{\text{PE}_f}{\text{PE}_i}\right)^{(1/t)} \left(\frac{\text{Profit Margin}_f}{\text{Profit Margin}_i}\right)^{(1/t)}(1+g)-1$$
Now you'll notice that he's actually adding the components, not multiplying. While technically incorrect, it provides a decent approximation. Using his assumptions (without the dividends), my formula gives:
 $$\text{Returns} = \left(\frac{15.0}{16.1}\right)^{(1/7)} \left(\frac{6.0\%}{7.8\%}\right)^{(1/7)}(1+2.9\%)-1=-1.88\%$$
Compare that to what he's doing (without dividends):
 $$\text{Returns} = \left(\left(\frac{15.0}{16.1}\right)^{(1/7)}-1\right) + \left(\left(\frac{6.0\%}{7.8\%}\right)^{(1/7)}-1\right) +\left(2.9\%\right)$$
$$=-1.00\% + -3.68\% +2.9\% = -1.78\%$$
Then to factor in dividends, Montier adds in the terminal dividend rate. This can be contrasted with John Hussman who takes the average of the initial dividend and final dividend rates.

Of course you have to come up with decent estimates for the final price to earnings ratio, profit margins and sales growth for this model to provide a decent estimate.

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Note regarding why Montier's approximation is OK. One way to show this is by considering the linear approximation to the exponential function: $$e^x \approx 1+x$$ Suppose that we have two rates, $x$ and $y$. We can use the above to make some substitutions into the following expression: $$(1+x)(1+y) - 1 \approx e^x \times e^y - 1 $$ $$ = e^{(x+y)} - 1 $$ $$ \approx (1 + (x+y)) - 1$$ $$ = x+y$$ This will only work well if $x$ and $y$ are close to 0 (which happens to be the case since we're dealing with relatively small percentages.) That's why Montier can get away with using this approximation.

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