ETF / Leverage / Leveraged ETF / Mathematical Finance / Quantitative Analysis / Quants

Barclays quants error on leveraged ETFs

Eric14 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)

and

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…

14 thoughts on “Barclays quants error on leveraged ETFs”

  1. interesting to see where this goes. i hadn’t spotted the mistake but their conclusion no longer holds. maybe this will tell us why.

    Reply

    • Do you have Mathematica?

      The next step is to just perform a series expansion. I did it on a napkin, but wanted to check my work before posting. Then life intervened 🙂

      The basic idea is to note that

      E(G_x) = 1 – x + x E(G).

      Setting

      E(G_x) = exp(x mu’ dt)

      we have

      mu’ = log[1 – x + x exp(mu dt)]/(x dt).

      With Mathematica, you could expand this in terms of mu dt symbolically.

      If I recall correctly, you get something like

      mu’ = x mu – mu (x^2 – x) + O[(mu dt)^2].

      So you do see the negative drift there, but it does not depend on sigma.

      I still hope to post more on this, but am recovering from a string of 17+ hour workdays and am suffering from TLA syndrome, i.e. trying to add some more three-letter acronyms to my business card).

      Reply

  4. Thanks for an excellent post. You should really continue work on this and publish either here or elsewhere.

    Anyway, another indication that the Cheng and Madhavan [not Madhaven] final result must be wrong is that it depends on the final value of the underlying asset (the non-leveraged ETF) only and not on the trajectory. This is obviously not the case, as the following classic example shows:
    Let’s take two days (t = 2) in which the underlying asset does not move in total,
    S_2 = S_0.
    This might arise if it didn’t move at all also in the middle, S_1 = S_0, in which case there is nothing to leverage and obviously A_2 = A_0 also.
    On the other hand, S could have fallen by 20% and then rose by 25%, S_1 = 0.8 S_0; in this case, taking for example a two-fold leverage, x = 2, would give A_1 = 0.6 S_0 and finally A_2 = 1.5 A_1 = 0.9 S_0.

    Reply

    • Hi Ehud,

      Thanks for your comments. If you have any thoughts on where to go next with this, I’d be happy to explore it. It has been a while and I’ve been distracted with other things lately.

      Reply

  6. Talking practically: If we assume the difference between tn and tn-1 is one day then the subsequent maths does hold. However, shouldn’t the volatility term x*sigma actually be |x|*sigma as volatility can’t be negative?

    Joe

    Reply

    • Hi Joe. Thanks for the comment, but no, the subsequent maths is not correct. The problem is that the expression

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

      is false. From this point, everything else is incorrect, including any relationship between the underlying vol and the leveraged vol.

      Reply

  7. Hi, can you also check 3.2 Order flow and prices? it is substituting and rearranging (12) and (13) with no reason and I can’t understand where the error is.

    Reply

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