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Describe how a portfolio can be built so as to hedge exposure to several factors.

FRM® Part I · Foundations of Risk Management · The Arbitrage Pricing Theory and Multifactor Models of Risk and Return · FRM1.FR.6.d

The idea

To hedge exposure to k factors, build a portfolio whose net beta to every factor is zero. With pure factor portfolios, the position in each is minus the portfolio's beta to that factor times its value. When the hedging instruments are impure, each loads on several factors. Then you must solve k simultaneous equations, so that total exposure to each factor, from the portfolio plus every hedge position, equals zero. That needs at least k instruments with linearly independent exposures. In a single-factor case, the hedge is sized as follows. Notional = portfolio beta × value ÷ instrument beta. If the hedged portfolio then still earns more than the risk-free rate, that excess is an alpha, which under APT signals arbitrage. The classic mistakes are hedging factors one at a time, ignoring cross-exposures, and getting the sign wrong on negative betas.

Formulas

Single-factor hedge notional

hedge notional = βp × V ÷ βh

βp
the portfolio's beta to the factor
V
the portfolio value
βh
the hedge instrument's beta to the factor

The position is a short when βp is positive. With several factors and impure instruments, solve all the zero-exposure equations together instead.

A practice question

From the question bank, with the reason each wrong option is wrong.

A portfolio manager wants to neutralise a portfolio's exposure to three distinct systematic factors. The minimum number of hedging instruments with linearly independent factor exposures required is:

  1. three, one for each factor. - correct
  2. one, the market portfolio, because all factors are reflected in the market.Why not: Confuses the multifactor model with the single-factor CAPM; one instrument can set only one exposure.
  3. two, because the risk-free asset takes care of the third factor.Why not: The risk-free asset has no factor exposure to offset; each factor needs its own independent instrument.
  4. four, one per factor plus one for the risk-free asset.Why not: The risk-free asset is not a hedging instrument; k independent factors need k instruments, not k + 1.

Setting k factor betas to zero is a system of k equations, which needs k hedging positions with linearly independent factor exposures. Fewer instruments cannot neutralise all exposures; extra ones are unnecessary.

A harder one

A USD 100 million portfolio has betas of 1.0 to factor 1 and 0.6 to factor 2. The manager sells short USD 80 million of an instrument whose betas are 1.1 to factor 1 and 0.3 to factor 2. The hedged portfolio's betas (net exposure divided by USD 100 million) are closest to:

  1. factor 1: 0.12; factor 2: 0.60Why not: Assumes the hedge instrument has no effect on factor 2, though it carries a beta of 0.3 to it.
  2. factor 1: 0.20; factor 2: 0.36Why not: Offsets factor 1 by 0.8 and ignores the instrument's beta of 1.1; the effect is 0.8 x 1.1 = 0.88.
  3. factor 1: 0.12; factor 2: 0.84Why not: Adds the instrument's factor 2 exposure instead of subtracting it; a short position reduces exposure.
  4. factor 1: 0.12; factor 2: 0.36 - correct

Net beta = portfolio beta - 0.8 x instrument beta. Factor 1: 1.0 - 0.8 x 1.1 = 0.12. Factor 2: 0.6 - 0.8 x 0.3 = 0.36. An impure hedge instrument changes every factor exposure it loads on.

The hardest tier

A USD 100 million portfolio has betas of 1.0 to factor 1 and 0.6 to factor 2. Two hedging instruments are available. Instrument X has betas of 1.2 to factor 1 and 0.3 to factor 2; instrument Y has betas of 0.5 to factor 1 and 1.0 to factor 2. Which positions (USD million) set both net factor exposures to zero?

  1. Short 66.7 of X and short 40.0 of Y. - correct
  2. Short 83.3 of X and short 60.0 of Y.Why not: Hedges each factor separately with one instrument, ignoring the cross-exposures (X on factor 2, Y on factor 1).
  3. Short 78.1 of X and short 21.0 of Y.Why not: Solves the system with the beta matrix transposed, mixing up instruments and factors.
  4. Short 66.7 of X and long 40.0 of Y.Why not: Solves correctly for X but flips the sign on Y; both positions are shorts.

Solve 1.2x + 0.5y = -100 and 0.3x + 1.0y = -60 (x, y in USD million). From the second equation y = -60 - 0.3x; substituting gives 1.05x - 30 = -100, so x = -66.7, y = -40.0. Both are short positions, and each hedge instrument changes both exposures.

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