Taylor Approximation for Leveraged ETF Growth and Volatility Drag

# Deriving the approximation for leveraged ETF returns?


# Deriving the approximation for leveraged ETF returns?












From this article (and I've seen the same approximation in multiple places):

> The formula for the long term compound annual growth rate of a leveraged ETF cannot be written in terms of just the benchmark return and volatility. It also involves terms containing the skewness and kurtosis of the benchmark. Its derivation and form is published in the paper that accompanies this article. It does not assume that benchmark returns are Gaussian or that returns are continuous as do formulae derived using Ito’s lemma. But it turns out that for the world’s stock markets and for low levels of leverage (up to about 3) the formula can be approximated by this formula: $$R = k\mu - \frac{k^2\sigma^2}{2(1 + k\mu)} $$ where $R$ is the compound daily growth rate of the ETF, $k$ is the ETF leverage, $\mu$ is the mean daily return of the benchmark, and $\sigma$ is the daily volatility (i.e. standard deviation) of the daily return of the benchmark.

How is this derived? Is there a reference for this approximation?

The linked article (Alpha Generation and Risk Smoothing Using Managed Volatility, by Tony Cooper, 2010) says:

> It is derived using a Taylor series expansion (details available from the author). It does not assume that benchmark returns are Gaussian or that returns are continuous as do formulae derived using Itoís lemma.

## Answer by T123 (score 1, accepted)

https://quant.stackexchange.com/a/79843

Here is my thought: As mentioned in the paper, this is a Taylorseries of the function $$ f(\mu)=ln(1+k\mu) $$

The first derivative of this is $f'(\mu)=\frac{k}{(1+k\mu)}$, the second derivative is $f''(\mu)=-\frac{k^2}{(1+k\mu)^2}$, the Taylorseries around $\mu_0=0$ is:

$$ f(\mu)=f(\mu_0)+f'(\mu_0)(\mu-\mu_0)+\frac{f''(\mu_0)}{2!}(\mu-\mu_0)^2+\dots$$

Using the derivatives from above $$ ln(1+k\mu)=ln(1+k\mu_0)+\frac{k}{(1+k\mu_0)}(\mu-\mu_0)-\frac{k^2}{2(1+k\mu_0)^2}(\mu-\mu_0)^2 $$

The first term is zero, the second one is $k\mu$, the third one uses the notation for the variance $\sigma^2=\mu^2-2E(\mu_0)\mu-E(\mu_0)^2$ Finally we get the expression $ k\mu - \frac{k^2\sigma^2}{2(1 + k\mu)} $

## Answer by Mahavir Bhattacharya (score 0)

https://quant.stackexchange.com/a/79839

Thank for sharing this; was an interesting read.

You can find a more detailed discussion in this paper by the same authors. They have also cited references for the actual mentions and derivations in the same.

Hope it helps!