For this question

If you are long a FRA and short a ED future with the same fixing dates, do you have positive convexity or negative convexity?

The answer is positive convexity, because a Eurodollar future has no convexity and the FRA has positive convexity. What are the implications of this from a trading perspective? Specifically, under what conditions does this lead to an opportunity for arbitrage?


hypothetically if we assume that $R_{fra}=R_{fut}-\frac{1}{2} \cdot \sigma^2\cdot T^2$ holds (convexity adjustment) and you are able to observe $R_{fra}$, $R_{fut}$ and $T$ then you can extract implied volatility of reference interest rate. If your view on volatility is different then you can make a bet: long convexity position (if you expect volatility should be higher); or short convexity position (if you expect volatility should be lower).

  • $\begingroup$ is the following an accurate non-quantitative interpretation of this? As yields rise, Eurodollars will decrease in value linearly, whereas a short FRA will increase in value, but at an increasing rate This would suggest that this long-FRA/short-ED is a good position if yields are expected to decline; Is that not true? Is that actually a bet on increasing convexity? How does this position profit when yields rise? $\endgroup$ Dec 17 '15 at 16:45
  • $\begingroup$ ED future is quoted as 100 less yield (i.e. price 98.5 implies 1.5% yield); which means long position in ED future is effectively a bet on a decrease of underlying rate (and vice versa short ED future - on an increase of rate). Long FRA (pay fixed; receive float) benefits from an increase of underlying rate. So in your example "long-FRA/short-ED" is actually a magnified position on an increase of rates. $\endgroup$
    – Nicholas
    Dec 21 '15 at 10:24
  • $\begingroup$ To construct a bet on increasing convexity, you can decompose sensitivity [to change in rates] of each instrument (ED, FRA) into linear (duration) and non-linear (convexity) components; then you compute weights of FRA, ED in your constructed position so that your position is duration hedged (linear component becomes zero); in the end you are left only with non-linear component aka convexity $\endgroup$
    – Nicholas
    Dec 21 '15 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.