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I'm following the famous paper Carry of Pedersen et al. I have a particular question about the section Global Fixed Income Carry.

My main questions are around equation 15. They define Carry as

$$C_t:=\frac{S_t-F_t}{F_t} (1)$$

for a fully funded position. My first question, why is this equivalent to for fixed income securities?

$$\frac{P_{t+1}^{T-1}+D\cdot 1_{[t+1\in \text{coupon dates}]}-P_t^T}{P_t^T} (2)$$

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  • $\begingroup$ How does he move from first to second line? Add and substract $P_{t+1}^{T-1}(y_t^T)$ and use the definition of yield (discussed before in the same section). Agree that some of their notation is a bit confusing. $\endgroup$
    – fes
    Commented Oct 17, 2020 at 13:43
  • $\begingroup$ @fesman could you please give some more details in an answer so that I can also accept it? thanks! $\endgroup$
    – swissy
    Commented Oct 22, 2020 at 9:05
  • $\begingroup$ @swissy it is a previous version of the paper. So check also the new one. However, I would be also be very interested in seeing why these two definition for carry are the same $\endgroup$
    – math
    Commented Oct 26, 2020 at 11:26
  • $\begingroup$ I think they only claim that they are approximately the same. $\endgroup$
    – fes
    Commented Oct 26, 2020 at 11:57
  • $\begingroup$ @fesman in a newer version the use $C_t=\frac{S_{t+1}^{T-1}-F^T_t}{F^T_t}=\frac{S^{T-1}_{t+1}-(1+r^f)S^T}{(1+r^f)S^T}$, looks like the $(1+r^f)$ is somehow too "much" on the equation $\endgroup$
    – swissy
    Commented Oct 26, 2020 at 13:35

1 Answer 1

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As per the definition:

$C_t:=\frac{S_t-F_t}{F_t}$

Per the article, it should actually be: (tomorrow minus today ) divided by today:

$C_t=\frac{F_{t+1}-F_t}{F_t}$

but they assume price does not change so $S_{t+1}=S_t$. An equivalent assumption for the bond would be that the YTM does not change. But there are two predictable (model free) characteristics of bonds: it pays coupon and its maturity shrinks as time progresses. So tomorrow (t+1) the same bond will have one fewer day to maturity, and if tomorrow happened to be a coupon date, then the bond holder will get coupon as well, so the equivalent of $F_ {t+1}$ is $P^{T-1}_{t+1}+D (\,\mathrm{if}\; t+1 \;\mathrm{is \,coupon \,date}\,)$

Re-comment, the price of the asset or exchange rate is a random process so it will vary over time, and then there is the arbitrage/parity type relationship between the current price and the forward/future price. In the calculation of their carry, they assumed that the price remains constant over time, but the parity relationships are deterministic so they hold. If you have a non-dividend paying stock, then $F_t=S_t (1+r)$. If you substitute into the carry equation:

$C_t:=\frac{S_t-F_t}{F_t}=\frac{S_t-S_t (1+r^f)}{F_t}=-r^f \frac{S_t}{F_t}$

So the carry is minus funding rate. And if you have a dividend paying stock or FX, then $F_t=S_t (1+r)-E[D]$,so carry will be dividend minus the funding rate:

$C_t:=\frac{S_t-F_t}{F_t}=\frac{S_t-S_t (1+r^f)+E[D]}{F_t}=\left(\frac{E[D]}{S_t}-r^f \right)\frac{S_t}{F_t}$

So essentially they assume the price is not random over time, but parity type relationships hold.

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  • $\begingroup$ thanks for your answer. But the questions is, why $F_t = S_t$. I do agree with your form of $F_{t+1}$. The problem is for $F_t$ we use $(1+r^f)S_T$. If $F_{t+1}=P^{T-1}_{t+1}+D(\text{if }t+1\text{ is coupon date})$, then $F_t =S_t$, which seems wrong $\endgroup$
    – swissy
    Commented Oct 29, 2020 at 8:16
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    $\begingroup$ Is it clear what I mean :) otherwise let me know as I'm very thankful for your help $\endgroup$
    – swissy
    Commented Oct 30, 2020 at 18:44
  • $\begingroup$ Thanks! Just to clarify, you mean $f_t=S_t(1+r^f)$? This would be the case for a stock that pays no dividends? $\endgroup$
    – Magic is in the chain
    Commented Oct 30, 2020 at 23:10
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    $\begingroup$ If you check the new version, link the comments above one can see that they write $F^\tau_t = (1+r^f_t)S_t^\tau$ $\endgroup$
    – swissy
    Commented Oct 31, 2020 at 10:11
  • $\begingroup$ thanks, let me add details based on the current version, and we can then correct me if i get it wrong based on what you have seen in the new version! $\endgroup$
    – Magic is in the chain
    Commented Oct 31, 2020 at 17:41

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