The basic principle of BOP accounting is double entry bookkeeping. In short, every transaction enters twice. As a consequence of this, by definition

$$\text{Current account } (CA) + \text{Capital account } (KA) = \text{Financial account } (FA)$$

Note that this may differ depending on how you classify BOP accounts. Some sources classify it as CA = KA (and do not have an FA), but the principle is the same.

In my definition, the FA includes the official reserve account (OA).

What I don't understand is - if every transaction automatically generates a credit and debit entry, how can it be possible to have:

$$CA + KA \neq (FA \; \backslash \; OA)$$

I know I am wrong to think this, but it seems to me like the 'need' for central bank intervention to absorb a surplus/deficit in the other components violates the definition of double-entry bookkeeping, which implies BOP = 0 at any point in time. So:

1) Why am I wrong wrt the above?

2) If the OA does not absorb the deficit/surplus from the rest of the BOP, what happens?

  • $\begingroup$ Ive responded to your edit now. $\endgroup$ Commented Aug 3, 2019 at 14:48
  • $\begingroup$ What does the "$\backslash$" symbol mean (in "$FA \; \backslash \; OA$")? $\endgroup$
    – user18
    Commented Jan 22, 2021 at 9:25

1 Answer 1


If OA is a part of FA. The second equation follows from the first: $CA + KA = FA \wedge OA \neq 0 \Rightarrow FA - OA = CA +KA -OA \neq CA +KA$.

Response to your EDIT:

$CA + KA \neq \frac{FA}{OA}$ follows from the first equation as well:

$CA + KA = FA \wedge OA \neq 1 \Rightarrow \frac{FA}{OA}= \frac{CA+KA}{OA} \neq CA +KA$ in general. (I've checked your LateX expression it really says: "(FA \backslash OA)"). I don't understand why you would think that BOP has anything to do with division. For example Feenstra and Taylor say in their textbook:

$OSB=-Reserve FA=CA+KA+Nonreserve FA$, where OSB is your OA and their version of the BOP identity is $CA + KA + FA = 0$, with your FA as their -FA.

  • $\begingroup$ I meant that it was not equal to the FA, excluding the OA. I've amended notation to make this clearer. $\endgroup$
    – Thev
    Commented Aug 3, 2019 at 10:58

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