Charta is on Google Play: practise on the go. Get the app
Charta

Explain how effective duration and convexity are used to hedge a position.

FRM® Part I · Valuation and Risk Models · Applying Duration, Convexity, and DV01 · FRM1.VR.12.h

The idea

A duration hedge matches dollar duration, which is duration × market value and is equivalent to matching DV01. For a position with market value V and duration D, hedged with an instrument of duration Dh, the market value to hold in the opposite direction is Vh = V × D ÷ Dh. Market value converts to face by dividing by the price per unit of face. That removes first-order exposure to a small parallel shift only. Because the position and the hedge usually have different convexities, the hedged book is still exposed to large moves. If the position is more convex, it gains from big moves either way, and if it is less convex, it loses. To neutralize both, use two hedge instruments and solve two equations, one for dollar duration and one for dollar convexity. Hedges also drift as rates change, so they need rebalancing.

Formulas

Market value of a duration hedge

Vh = V × D ÷ Dh

V
the market value of the position
D
the position's effective duration
Dh
the hedge instrument's effective duration

Equates dollar durations, so it covers only small parallel shifts and is held in the opposite direction. The trap is inverting the ratio, or matching market values and ignoring the durations.

Profit on a duration-neutral book from convexity

ΔV = 0.5 × NDC × (Δy)²

NDC
net dollar convexity: convexity × market value, longs less shorts
Δy
the parallel yield change as a decimal

Only the second-order term is left once dollar durations offset, so the sign is the same whichever way yields move. The usual slip is dropping the 0.5.

A practice question

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

A risk manager hedges a bond position using effective duration. What is the hedge designed to achieve?

  1. To make the market value of the hedge equal to the market value of the position.Why not: Equal market values do not offset sensitivity when the instruments have different durations.
  2. To eliminate all losses from any size and shape of yield move.Why not: A duration hedge covers only small parallel moves; large and non-parallel moves remain.
  3. To make the dollar duration of the hedge equal and opposite to that of the position, so small parallel yield changes have no first-order effect. - correct
  4. To make the hedge's convexity equal to that of the position.Why not: Matching convexity is a second-order refinement; a duration hedge matches the first-order sensitivity.

A duration hedge sets duration x market value of the hedge equal and opposite to the position. This gives zero first-order sensitivity to a small parallel shift.

A harder one

A USD 80 million portfolio with an effective duration of 5.0 is hedged with a short position in an instrument of effective duration 1.8. The market value of the hedge required is closest to:

  1. USD 28.8 millionWhy not: Inverts the duration ratio, using 80 x 1.8 / 5.0.
  2. USD 80 millionWhy not: Matches the market values of the position and hedge, ignoring durations.
  3. USD 142 millionWhy not: Hedges only the excess of the position duration over the hedge duration, using 80 x (5.0 - 1.8) / 1.8.
  4. USD 222 million - correct

Match dollar durations: 80 x 5.0 = Vh x 1.8, so Vh = 400 / 1.8 = USD 222.2 million.

The hardest tier

A USD 100 million position has duration 6 and convexity 60. A manager sells two instruments to hedge both duration and convexity. Instrument A has duration 3 and convexity 15, and instrument B has duration 12 and convexity 200. The amount of instrument B, in market value, to sell is closest to:

  1. USD 50.0 millionWhy not: Hedges duration alone using B only.
  2. USD 21.4 million - correct
  3. USD 114.3 millionWhy not: Takes the amount of instrument A from the two-equation solution instead of B.
  4. USD 30.0 millionWhy not: Hedges convexity alone using B only.

Equations: 3a + 12b = 600 and 15a + 200b = 6,000. From the first, a = 200 - 4b. Then 15(200 - 4b) + 200b = 6,000, so 140b = 3,000 and b = 21.4 million (a = 114.3 million).

In the app: 6 practice questions on this outcome, spaced review that brings it back before you forget it.

Start free →

More in Applying Duration, Convexity, and DV01