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I am trying to derive the campbell shiller log linear relation, and i got stuck with something (i believe) quite simple. Before we are using the first-order tayler expansion is where i got stuck, because i can't figure out, how that $e$ got in there:

$\ln \left(1+\frac{D_{t+1}}{P_{t+1}}\right)=\ln \left(1+\exp \left\{\ln \left(D_{t+1}\right)-\ln \left(P_{t+1}\right)\right\}\right)$

When i use log rules i get the $\ln \left(D_{t+1}\right)-\ln \left(P_{t+1}\right)$ part, but why does that get raised to $e$?

Does it have something to do with continuously compounding?

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2 Answers 2

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You can simply start with the definition of gross returns \begin{align*} R_{t+1}&=\frac{D_{t+1}+P_{t+1}}{P_t} \\ &=\frac{1+P_{t+1}/D_{t+1}}{P_t/D_t}\frac{D_{t+1}}{D_t}, \end{align*} where the first fraction contains now your price dividend ratio. Going to log-returns, \begin{align*} r_{t+1} &= \ln\left(1+\frac{P_{t+1}}{D_{t+1}}\right) - \ln\left(\frac{P_t}{D_t}\right)+\ln\left(\frac{D_{t+1}}{D_t}\right) \\ &= \ln\left(1+e^{\mathfrak{d}_{t+1}}\right)-\mathfrak{d}_t+\Delta d_{t+1}, \end{align*} where $\mathfrak{d}_t=\ln\left(\frac{P_t}{D_t}\right)$ is the log-price dividend ratio and $\Delta d_{t+1}$ the log dividend growth.

Now, you can use Taylor's theorem $$\ln(1+e^x)\approx\ln(1+e^{x_0})+\frac{e^{x_0}}{1+e^{x_0}}(x-x_0).$$ We normally choose the long-term log-price-dividend ratio, $\bar{\mathfrak{d}}=\ln(\bar{\mathfrak{D}})$ as point $x_0$. Then,

\begin{align*} r_{t+1} &\approx \ln(1+\bar{\mathfrak{D}})+\frac{\bar{\mathfrak{D}}}{1+\bar{\mathfrak{D}}}(\mathfrak{d}_{t+1}-\bar{\mathfrak{d}}) -\mathfrak{d}_t+\Delta d_{t+1} \\ &= k+\rho \mathfrak{d}_{t+1}-\mathfrak{d}_t+\Delta d_{t+1}, \end{align*} where $k,\rho$ are constants.

As you see, your next period return is high if

  • prices today are low (and hence $\mathfrak{d}_t$ is low).
  • prices tomorrow are high (and hence $\mathfrak{d}_{t+1}$ is large).
  • dividend growth ($\Delta d_{t+1}$) is high.

All of this makes intuitive sense!

Of course, you can iteratively apply the above relationship to obtain the decomposition in cashflow component and discount rate component. This also has huge implications for the return predictability literature (the movements in the price-dividend ratio imply that either future dividend growth rates or future returns or both are (partially) predictable).

To guide with the above equations, $\rho\approx0.96$ and $\bar{\mathfrak{D}}\approx25$ (4% dividend price ratio) seem good guesses to me.

Note that all these relationships follow directly from the definition of returns and do not depend on any model assumptions.

Log-linearisation is frequently used to solve all sorts of asset pricing models (e.g. long run risk models). However, be careful. The Taylor approximation is just an approximation of order one. Pohl, Schmedders, and Wilms (2018, JF) show that log-linearisation can yield wrong results because higher order terms are neglected.

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    $\begingroup$ Congratulations on the beautiful use of fonts $\mathfrak{abcdefghijklmnopqrstuvwxyz}$ $\endgroup$
    – nbbo2
    May 11, 2020 at 19:44
  • $\begingroup$ And don't forget the beautiful capital letters $\mathfrak{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$. Mathfrak is clearly not used often enough in finance (and maths)! $\endgroup$
    – Kevin
    May 11, 2020 at 19:50
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The second expression is just another representation of the former and has nothing to do with continuous compounding. Instead note that $\log(a)-\log(b)=\log\left(\frac{a}{b}\right)$ from which the result should become immediately clear.

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  • $\begingroup$ But why is it raised to e? When $\log (a)-\log (b)=\log \left(\frac{a}{b}\right)$, shouldnt we just get: $\ln \left(1+\frac{D_{t+1}}{P_{t+1}}\right)=\ln \left(1+\left\{\ln \left(D_{t+1}\right)-\ln \left(P_{t+1}\right)\right\}\right)$ $\endgroup$
    – mbih
    May 11, 2020 at 17:01
  • $\begingroup$ No, note that your initial expression is not $\log\left(1+\log\left(\frac{D_{t+1}}{P_{t+1}}\right)\right)$ (in which case your second expression would be valid), but $\log\left(1+\frac{D_{t+1}}{P_{t+1}}\right)$. So you have to get rid of the $\log(\cdot)$ part. $\endgroup$
    – sp59b2
    May 11, 2020 at 17:07
  • $\begingroup$ So actually, $\ln \left(1+\exp \left\{\ln \left(D_{t+1}\right)-\ln \left(P_{t+1}\right)\right\}\right)$ is equal to $\ln\left(1+\left(D_{t+1}\right)-\left(P_{t+1}\right)\right)$? and the only reason we use exp is, that we are going to use tayler series? $\endgroup$
    – mbih
    May 11, 2020 at 17:10
  • $\begingroup$ No we have $\log(1+\exp(\log(D_{t+1})-\log(P_{t+1})))=\log\left(1+D_{t+1}\times\frac{1}{P_{t+1}}\right)$ because $\exp(a-b)=\frac{\exp(a)}{\exp(b)}$ $\endgroup$
    – sp59b2
    May 11, 2020 at 17:14
  • $\begingroup$ And yes we put it into this representation to then do a Taylor expansion. $\endgroup$
    – sp59b2
    May 11, 2020 at 17:22

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