Downside Capture Ratios: Compounded Returns and Ratios of Means

# Downside Market Capture Ratio: compute with sum or product?


# Downside Market Capture Ratio: compute with sum or product?












I'm computing downside market capture ratio in R. The PerformanceAnalytics R package has a built-in UpDownRatios function which does this, but it computes the ratio using sums of returns, not products.

Since the ratio is the "compound return when the benchmark was down divided by the benchmark's compound return when the benchmark was down", shouldn't that be product?

## Answer by David Addison (score 2, accepted)

https://quant.stackexchange.com/a/33667

I am not familiar with that R package, but I've written a few performance tracking libraries in my past life, so I might be able to add some insight.

While it is indeed true that logarithmic returns may be added and subtracted, all non-quant investors and hedge funds present their performances in percent returns. The reason one can't simply add and subtract percent returns is because the denominators changes instantaneously.

For a Numeraire, $\mathbb{N}_t$ representing one's bank account, the wealth process may evolves as such:

$\frac{\mathbb{N}_{t}}{\mathbb{N}_{t-1}} = (1 + m_{t-1}) =e^{ \mu_{t-1}} \approx e^{m_{t-1} - \frac{\sigma^2}{2}} $

where: $m_t$ is the geometric rate of return; $\mu$ is logarithmic rate of return; and, $\sigma^2$ is the variance of $\mathbb{E}[\mu]$.

If you're programming is taking the sum of percent returns, then clearly it will result in the following inequality:

$$(\sum_{t}^{T} m_t) +1 \ne \frac{\mathbb{N}_{T}}{\mathbb{N}_{t}}$$

You are correct, however, about the following equality:

$$\prod_{t}^{T}( 1 + m_t) = \frac{\mathbb{N}_{T}}{\mathbb{N}_{t}} = e^{\Sigma_t^T \mu_t}$$

If changing something about the source code is too onerous, there are two options which will get you the same result:

Option A: convert the percent return to natural logs through the following:

$\mu_t = ln(1+m_t) \, , \forall t \in T $

And then convert them back to percents in the final step:

$m_t = e^{\mu_t} -1 \, , \forall t \in T $

Option B: take the ratios of the compounded periodic returns as such

$\frac{\prod_{t}^{T}( 1 + m_t)}{\prod_{0}^{t-1}( 1 + m_t)}$

## Answer by Brian G. Peterson (score 2)

https://quant.stackexchange.com/a/33679

David Addison's discussion of log versus simple(arithmetic) returns in his answer is correct, but this particular calculation has nothing to do with arithmetic versus log returns.

Up Capture is defined by Bacon(2004), p. 47 as:

$$Up Capture = \frac{\bar{r+}}{\bar{b+}} $$

(mean of the asset returns over mean of the benchmark return)

So simple versus log returns makes no difference in this calculation.

The code is implemented using

$$ \frac{sum(UpRa)}{sum(UpRb)}$$

which is what is causing the confusion.

The ratio of sums and the ratio of means is the same. So the calculation in the code is correct.

Sum is a more efficient vectorized calculation than mean. mean involves an additional division by the total number of observations. This is why the calculation is implemented the way it is.

Another way to think about why the calculations are equivalent is that both means would have the same denominator. So you could consider the ratio of sums as having factored out the common denominators (the number of observations).

Ref: Bacon, Carl. Practical Portfolio Performance Measurement and Attribution, Second Edition. Wiley. 2004. p. 47