3
$\begingroup$

I have been playing with $V=Pe^{rT}$ and have been thinking about how to apply it to situations involving different assets growing at different rates.

Suppose I had data for the historic sale prices of a row of houses in a street, with two data points per house. My assumption here is going to be that each house's price grows according to $V=Pe^{rT}$, with $r$ constant.

House no. Year A Sold for Year B Sold for
1 2000 \$100,000 2015 \$500,000
2 2005 \$60,000 2019 \$111,000
3 1999 \$30,000 2018 $80,000
4 2002 \$125,000 2005 \$126,000

Using $V=Pe^{rT}$, I could calculate each house's instantaneous rate of interest $r=\frac{\ln{V}-\ln{P}}{T}$, and its APR $\hat{r}=e^{r}-1$, where $T=B-A$ is the number of years elapsed.

House no. Year A Sold for Year B Sold for $r$ $\hat{r}$
1 2000 \$100,000 2015 \$500,000 10.73% 11.33%
2 2005 \$60,000 2019 \$111,000 4.39% 4.49%
3 1999 \$30,000 2018 $80,000 5.16% 5.30%
4 2002 \$125,000 2005 \$126,000 0.27% 0.27%

My question is, how should I meaningfully combine the growth rates of the different houses? Say I want a figure for the street's overall growth rate in order to evaluate whether a particular house has been growing more quickly or more slowly than this rate. I have various ideas:

  • Take the average of the $r_{i}$, which is 5.138%
  • Use $-1+e^{5.138\%}=5.272\%$
  • Take the average of the $\hat{r}_{i}$, which is 5.346%
  • Work out the houses' prices in two base years (say, 2000 and 2021), using $P=Ve^{-rT}$, and work out what my $r$ or $\hat{r}$ would have been if I'd bought the entire street in 2000 and kept the houses until 2021.
House no. $r$ 2000 2021
1 10.73% \$100,000 \$951,827
2 4.39% \$48,165 \$121,197
3 5.16% \$31,589 \$93,400
4 0.27% \$124,338 \$131,470
Total - \$304,092 \$1,297,894

Using the totals above as my new $P$ and $V$, I get that $r=6.91\%$ and $\hat{r}=7.15\%$.

Is this a legitimate method? If so, what is it called? I am concerned that it might not be legitimate, because changing the two base years gives different results:

House no. $r$ 2005 2008
1 10.73% \$170,998 \$235,930
2 4.39% \$60,000 \$68,455
3 5.16% \$40,892 \$47,741
4 0.27% \$126,000 \$127,008
Total - \$397,890 \$479,134

Here, $r=6.19\%$ and $\hat{r}=6.39\%$, which seems strange, since each individual house's growth rate was constant. Does this mean that the "What if I'd bought the whole street in year $X$ and waited until year $Y$" method is wrong, or does it still have an interpretation?

What is the best way of defining the "overall" rate of growth for the whole street of houses?

Thanks

$\endgroup$

1 Answer 1

1
$\begingroup$

I don't think there is a unique best way to do this.

A flexible approach might be to compute the average growth rate by regression. Let $P_i, V_i$ and $T_i$ be the variables for house $i$. Then we have the formula: $$ \ln(V_i) - \ln(P_i) = r_i T_i $$ Putting this into a regression framework gives: $$ \ln(V_i) - \ln(P_i) = r T_i + \varepsilon_i $$ where now $r$ is the "average" growth rate and, by definition $\varepsilon_i = (r_i - r) T_i$.

If we assume that $\mathbb{E}(\varepsilon_i T_i) = 0$ (e.g. this holds if deviations of $r_i$ from $r$ are zero conditional on $T_i$). Then we can multiply this equation by $T_i$ and take expectations to obtain: $$ \mathbb{E}[(\ln(V_i)- \ln(P_i)) T_i] = r \mathbb{E}[T_i^2] $$ So: $$ r = \frac{\mathbb{E}[(\ln(V_i) - \ln(P_i))T_i]}{\mathbb{E}[T_i^2]} = \frac{\mathbb{E}(r_i T_i^2)}{\mathbb{E}(T_i^2)}. $$

An estimate can be constructed by replacing the expectation by the sample mean. $$ \hat r = \frac{\sum_i r_i T_i^2}{\sum_i T_i^2} $$ For your data set, this gives an estimate of 0.065

An alternative might be to also include an intercept. $$ \ln(V_i) - \ln(P_i) = \alpha + r T_i + \varepsilon_i. $$ Doing this gives an estimate of 0.0742

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.