Phorgy Phynance

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

14 Responses

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  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.


    May 6, 2009 at 6:29 am

  2. Good to see you filthy.

    I’ll post what I found when I get a few extra minutes.


    May 6, 2009 at 8:03 am

  3. come on eric. i’m hoping you can shed some light on a possibly profitable pattern.


    May 20, 2009 at 11:22 am

    • 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).


      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).


      May 20, 2009 at 11:43 am

      • thanks. when i get time i’m going to rework the paper myself.


        May 21, 2009 at 12:03 pm

    • Hey! We all have a bit of Mathematica now via Wolfram Alpha.

      Here is the correct expansion:

      mu’ = x mu – .5 (x^2 – x) mu^2 + O(mu^3)

      where I made a slight change to

      E(G_x) = exp(mu’)

      and set dt = 1.


      May 21, 2009 at 12:08 pm

  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.

    Ehud Schreiber

    December 31, 2009 at 4:37 am

    • Sorry, I typo’ed S instead of A twice in the final line, but the idea should be clear.

      Ehud Schreiber

      December 31, 2009 at 4:41 am

    • 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.


      December 31, 2009 at 9:14 pm

  5. […] might find the three articles by here, the previous post and here, respectively. Possibly related posts: (automatically generated)Mixed Bag – getting […]

  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?



    September 11, 2011 at 9:43 pm

    • 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.


      September 13, 2011 at 4:20 pm

  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.


    October 13, 2011 at 12:29 am

    • Hi Pasquale,

      It looks like they have a typo in (12). It should be

      S_{t_{n+1}}- S_{t_n} = \lambda \left[ q_{t_n, t_{n+1}} + \phi \Delta_{t_{n+1}} + (1 - \phi) \Delta_{t_n} \right] + w_{t_n, t_{n+1}}.


      October 15, 2011 at 5:49 pm

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