Phorgy Phynance

Archive for May 2009

Barclays quants error on leveraged ETFs

with 14 comments

In a recent article, Cheng and Madhaven from Barclays Global Investors published a good article on leveraged ETFs

The Dynamics of Leveraged and Inverse Exchange-Traded Funds
April 8, 2009

Check it out.

The Error

They begin from a fairly standard starting point

dS_t = \mu S_t dt + \sigma S_t dW_t

However, they proceed to state that since

\frac{A_{t_i}-A_{t_{i-1}}}{A_{t_{i-1}}} = x\frac{S_{t_i}-S_{t_{i-1}}}{S_{t_{i-1}}}

“holds for any period”, then it follows that

\frac{dA_t}{A_t} = x\frac{dS_t}{S_t}

where A_t is the ETF NAV and x is the leverage factor.

Unfortunately, that is not correct. The problem is that

\frac{A_{t_i}-A_{t_{i-1}}}{A_{t_{i-1}}} = x\frac{S_{t_i}-S_{t_{i-1}}}{S_{t_{i-1}}}.

only holds when t_i - t_{i-1} is 1 day. Otherwise, we could let t_i - t_{i-1} be 1 year and this would say that the 1-year return of the ETF is x times the 1-year return of the index, which we already know is not true.

This should have also been obvious by plugging t=1 into their final expression

\frac{A_t}{A_0} = \left(\frac{S_t}{S_0}\right)^x \exp\left[\frac{\left(x-x^2\right)\sigma^2 t}{2}\right],

which violates the relation defining leveraged ETFs they started with. As a result of this error, their discussion of return dynamics in Section 4 must be re-examined

The Solution

The correct way to look at this is to let

G_{i-1,i} =\frac{S_{t_i}}{S_{t_{i-1}}} and G_{x,i-1,i} = \frac{A_{t_i}}{A_{t_{i-1}}}.

If \Delta t is 1 day, then

\begin{aligned} G_{x,i-1,i} &= 1 + x \left(G_{i-1,i} - 1\right) \\ &= (1-x)\left[1+\left(\frac{x}{1-x}\right) G_{i-1,i}\right]\end{aligned}

so that

\begin{aligned} G_{x,0,n} &= \prod_{i=1}^n G_{x,i-1,i} \\ &= (1-x)^n\prod_{i=1}^n \left[1+\left(\frac{x}{1-x}\right) G_{i-1,i}\right].\end{aligned}

If we assume S_t is a geometric Brownian motion (as they do), then

G_{i-1,i} = \exp\left(\bar\mu \Delta t + \sigma \sqrt{\Delta t} W_{\Delta t}\right),

where \bar\mu = \mu - \frac{\sigma^2}{2}. With a slight abuse of notation, we can drop the indices and let

G =\exp\left(\bar\mu \Delta t + \sigma\sqrt{\Delta t} W_{\Delta t}\right)

so that

G^i =\exp\left(\bar\mu i \Delta t + \sigma\sqrt{i\Delta t}W_{i \Delta t}\right).

This allows us to rewrite (using the definition of the binomial coefficient)

\begin{aligned} G_{x,0,n} &= (1-x)^n \left[1+\left(\frac{x}{1-x}\right) G \right]^n \\ &=(1-x)^n \sum_{i=0}^n \binom{n}{i}\left(\frac{x}{1-x}\right)^i G^i. \end{aligned}

Noting that

E(G) = \exp\left[\left(\bar\mu + \frac{\sigma^2}{2}\right)\Delta t\right] = \exp\left(\mu\Delta t\right)


E(G^i) = \exp\left(\mu i\Delta t\right) = E(G)^i.

we arrive at a disappointingly simple, yet important, expression

\begin{aligned} E(G_{x,0,n}) &=(1-x)^n \sum_{i=0}^n \binom{n}{i}\left(\frac{x}{1-x}\right)^i E(G)^i \\ &= (1-x)^n \left[1+\left(\frac{x}{1-x}\right) E(G) \right]^n \\ &= \left[1-x+x E(G)\right]^n. \end{aligned}

The expression above governing leveraged ETFs is the starting point for further analysis. We will come back to this in a subsequent post.

To be continued…

Written by Eric

May 4, 2009 at 7:38 pm