<p>If OA is a part of FA. The second equation follows from the first:
<span class="math-container">$CA + KA = FA \wedge OA \neq 0 \Rightarrow FA - OA = CA +KA -OA \neq CA +KA$</span>.</p>

<p>Response to your <strong>EDIT</strong>:</p>

<p><span class="math-container">$CA + KA \neq \frac{FA}{OA}$</span> follows from the first equation as well:</p>

<p><span class="math-container">$CA + KA = FA \wedge OA \neq 1 \Rightarrow  \frac{FA}{OA}= \frac{CA+KA}{OA} \neq CA +KA$</span> 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:</p>

<p><span class="math-container">$OSB=-Reserve FA=CA+KA+Nonreserve FA$</span>, where OSB is your OA and their version of the BOP identity is <span class="math-container">$CA + KA + FA = 0$</span>, with your FA as their -FA.</p>

CC BY-SA 4.0
2019-08-03T10:58:23.567Z
I meant that it was not equal to the FA, excluding the OA. I've amended notation to make this clearer.
Thev
CC BY-SA 4.0
2019-08-03T08:42:44.523Z
2019-08-06T05:22:50.423Z
2019-08-06T05:22:50.423Z
Grada Gukovic
<p>The basic principle of BOP accounting is <strong>double entry bookkeeping</strong>. In short, every transaction enters twice. As a consequence of this, by definition</p>

<p><span class="math-container">$$\text{Current account } (CA) + \text{Capital account } (KA) = \text{Financial account } (FA)$$</span></p>

<p>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.</p>

<p>In my definition, the FA includes the official reserve account (OA).</p>

<hr>

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

<p><span class="math-container">$$CA + KA \neq (FA \; \backslash \; OA)$$</span></p>

<p>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:</p>

<p>1) Why am I wrong wrt the above?</p>

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

CC BY-SA 4.0
2019-08-03T14:48:03.567Z
Ive responded to your edit now.
Grada Gukovic
<international-economics><balance-of-payments>
What happens if the Central Bank does not absorb a BOP deficit/surplus?
Thev
# What happens if the Central Bank does not absorb a BOP deficit/surplus?

(Asked by: username_0 on 2019-08-03)

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).

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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?

**Tags:** international-economics, balance-of-payments

> **Comments:**
> **username_1**: Ive responded to your edit now.  

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## 1 answers

### Answer by username_1 on 2019-08-03. Score: 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.

> **Comments:**
> **username_0**: I meant that it was not equal to the FA, excluding the OA. I've amended notation to make this clearer.  

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