There are plenty of things written about leveraged ETFs and their apparent decay based on empirical observation that both bull and bear products often underperform the benchmark. The usual explanation (and the one that makes most sense to me) is that ETFs don't really decay - just appear to be doing so because higher leverage lowers median - the most likely - price, but expected value remains unchanged. Jev Kuznetsov covered the topic few months ago on his blog.
Recently I started working on leveraged ETF models, not for the underlying but for options on leveraged ETFs, and spent more time thinking about the basics. Someone asked me if distribution of "regular" ETF returns is not gaussian, would leveraged ETF price still be a martingale. In retrospect the answer should have been trivial: if E[S*(1+r)]=S, then E[r]=0, and E[L*r]=0, and therefore E[S*(1+L*r)]=S, where S is price, r is return, L is leverage, and E is the expectation operator. However I did not figure this out right away, and was not sure about the answer. Most of the "basic" illustrations are based on binomial tree that converges to gaussian, and I did not see the same illustration being done for distributions with excess skew or kurtosis.
Here is the usual binomial tree for an ETF. The instrument moves up 10%, or down 10 %. Implicit in the plot are probabilities of up and down move - in this case both equal to 50%. It is easy to verify that the ETF is a martingale, i.e. the expected value does not change. Leveraged version of the ETF is also a martingale.
Now, let's consider a different type of binomial tree, where probabilities are not the same.
Here the ETF moves up x and moves down 1/(1+x). In the first case x = 0.1, and stock moves up to 100 * (1+x) = 100 * 1.1 = 110, and moves down to 100 * 1/(1+x) = 100 * 1/1.1 = 100 / 1.1 = 90.90. Probabilities are of course cannot be the same. To calculate probabilities I solve equation [1 = p*u + (1-p)/u] for p; p = (1/u - 1)/(1/u - u). In our case (1/1.1-1)/(1/1.1-1.1) = 0.47619 , and 100 = 110 * 0.47619 + 90.90909 * (1-0.47619). Similarly for the second step, terminal probabilities are 0.47619^2 = 0.226757, 0.47619 * (1-0.47619) = 0.249433, and (1-0.47619)^2 = 0.274376; expected value is 0.226757 * 121 + 0.249433 * 100 + 0.249433 * 100 + 0.274376 * 82.64463 = 100
Double-leveraged version works as expected. All these examples are trivial - ETF is a martingale, binomial is gaussian in limit. However we can also construct a trinomial tree that can simulate probability distributions other than gaussian.
Up and down moves are chose arbitrarily, while the middle move is set to sqrt(up*down), so the tree recombines. For clarity's sake I drew all nodes. Probabilities are constrained to sum up to 1, and to produce a martingale - p1 * up + p2 * mid + (1-p1-p2)*down = 1. In the illustration my price multipliers and probabilities are:
Leveraged version does not produce a trinomial tree that recombines, but price still remains a martingale.
In the following posts I will address the topic of constructing consistent volatility skews for regular, inverse, and leveraged ETFs.
Recently I started working on leveraged ETF models, not for the underlying but for options on leveraged ETFs, and spent more time thinking about the basics. Someone asked me if distribution of "regular" ETF returns is not gaussian, would leveraged ETF price still be a martingale. In retrospect the answer should have been trivial: if E[S*(1+r)]=S, then E[r]=0, and E[L*r]=0, and therefore E[S*(1+L*r)]=S, where S is price, r is return, L is leverage, and E is the expectation operator. However I did not figure this out right away, and was not sure about the answer. Most of the "basic" illustrations are based on binomial tree that converges to gaussian, and I did not see the same illustration being done for distributions with excess skew or kurtosis.
Here is the usual binomial tree for an ETF. The instrument moves up 10%, or down 10 %. Implicit in the plot are probabilities of up and down move - in this case both equal to 50%. It is easy to verify that the ETF is a martingale, i.e. the expected value does not change. Leveraged version of the ETF is also a martingale.
Now, let's consider a different type of binomial tree, where probabilities are not the same.
Here the ETF moves up x and moves down 1/(1+x). In the first case x = 0.1, and stock moves up to 100 * (1+x) = 100 * 1.1 = 110, and moves down to 100 * 1/(1+x) = 100 * 1/1.1 = 100 / 1.1 = 90.90. Probabilities are of course cannot be the same. To calculate probabilities I solve equation [1 = p*u + (1-p)/u] for p; p = (1/u - 1)/(1/u - u). In our case (1/1.1-1)/(1/1.1-1.1) = 0.47619 , and 100 = 110 * 0.47619 + 90.90909 * (1-0.47619). Similarly for the second step, terminal probabilities are 0.47619^2 = 0.226757, 0.47619 * (1-0.47619) = 0.249433, and (1-0.47619)^2 = 0.274376; expected value is 0.226757 * 121 + 0.249433 * 100 + 0.249433 * 100 + 0.274376 * 82.64463 = 100
Double-leveraged version works as expected. All these examples are trivial - ETF is a martingale, binomial is gaussian in limit. However we can also construct a trinomial tree that can simulate probability distributions other than gaussian.
Up and down moves are chose arbitrarily, while the middle move is set to sqrt(up*down), so the tree recombines. For clarity's sake I drew all nodes. Probabilities are constrained to sum up to 1, and to produce a martingale - p1 * up + p2 * mid + (1-p1-p2)*down = 1. In the illustration my price multipliers and probabilities are:
Leveraged version does not produce a trinomial tree that recombines, but price still remains a martingale.
In the following posts I will address the topic of constructing consistent volatility skews for regular, inverse, and leveraged ETFs.
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