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## Fractional reserve banking

Fractional reserve banking is at the heart of the way money is created. It means that banks do not keep cash reserves equal to the balances of their depositors, instead they just hold a fraction of their depositors’ balances in reserve. This allows them to use the rest of the depositors’ money to make loans to others. It also relies on the gamble that depositors are not going to want to withdraw all their money at once but will only demand access to a small amount of their deposits at any one time.

For instance if you had a bank that had £1000 of deposits from customers, that £1000 is a liability to the bank because it owes £1000 to its customers should they wish to withdraw it. If the bank keeps £1000 in reserves then it is fully covered, but has no money free to make loans to others – this would be 100% reserve banking. If instead the bank said that it would operate a reserve ratio of 10% (or 0.1) then it would just keep £100 in reserves and then make £900 of loans to other customers. This £900 would then be an asset to the bank as it is owed by the customers to the bank – of course the bank would charge interest on this loan as well.

Now what would happen if that bank then received a new deposit of £100? It now has £1100 of deposits from customers and £200 in reserves. But if it is just keeping a reserve ratio of 0.1, then for £1100 of deposits it only needs to keep £110 in reserves, so it can reduce its reserves by lending out another £90. So from £100 deposit the bank creates £90 of new loans. The reserve ratio of 0.1 means that the bank lends out 0.9 of its new deposits.

If that £90 gets lent to a customer that goes and deposits it in their bank (or spends it and it ends up deposited in another customer’s bank). Their bank will keep 0.1 x £90 = £9 extra in reserves and lend out 0.9 x £90 = £81 to another customer.

And so the pattern repeats itself. That customer deposits the £81 which goes into another bank. When their bank gets it they keep back 0.1 x £81 = £8.10 in reserves and lend out 0.9 x £81 = £72.90. The original deposit of £100 is creating new loans through every round of lending and depositing but each round gets smaller.

Where will it end? We can think of it in terms of a geometric series. If we call the initial deposit $d$ and the reserve ratio $\theta$ then the rounds of spending look like this:

$d + d(1-\theta) + d(1-\theta)^2 + d(1-\theta)^3 +....d(1-\theta)^n = d[1 + (1-\theta) + (1-\theta)^2 + (1-\theta)^3 +....(1-\theta)^n ]$. This is a geometric series with geometric ratio $1-\theta$.

As it’s a geometric series it will sum to $d[\frac{1-(1-\theta)^{n+1}}{1-(1-\theta)}] = d[\frac{1-(1-\theta)^{n+1}}{\theta}]$ (see here for why). As $0<(1-\theta)<1$ then it means when $n \rightarrow \infty$, $(1-\theta)^{n+1} \rightarrow 0$ so $d[\frac{1-(1-\theta)^{n+1}}{\theta}] \rightarrow d[\frac{1}{\theta}]$.

This is the bank multiplier, $\frac{1}{\theta}$. The total amount of money created from the initial deposit is found by multiplying the initial deposit by $\frac{1}{\theta}$.

We can extend this model by taking account of the fact that customers will hold some of their wealth in cash, they won’t just deposit everything straight back into the bank.

Before making the model lets lay out all the definitions we need.
$M$ is the total amount of money people have.
$D$ is the total amount customers deposit with banks.
$CU$ is the total amount of currency people hold in cash.
$c$ is the currency ratio, the proportion of their total money that they hold in cash, so $cM = CU$ and $(1-c)M = D$
$R$ are the cash reserves banks keep in order to keep enough liquidity to cover the expected withdrawals of consumers.
$\theta$ is the reserve ratio, ie $R = \theta D$

Now if we extend the original model by saying there is a reserve ratio of 0.1 and a currency ratio of 0.4, this time the initial £100 meant again that the bank could lend out 0.9 x £100 = £90 to a customer.

That now has £90, but will hold 0.4 x £90 = £36 in cash and deposit 0.6 x £90 = £54 into a bank. The bank will then lend out 0.9 x £54 = £48.60 to another customer who will deposit 0.6 x £48.60 = £29.16 into theirs.

The pattern continues as before but this time the multiplier is smaller. The geometric series is:
$d[1 + [(1-\theta)(1-c)] + [(1-\theta)(1-c)]^2 + [(1-\theta)(1-c)]^3 +....[(1-\theta)(1-c)]^n ]$, the geometric ratio this time is $(1-\theta)(1-c)$.

So this geometric series will sum to $d[\frac{1-[(1-\theta)(1-c)]^{n+1}}{1-[(1-\theta)(1-c)]}] = d[\frac{1-[(1-\theta)(1-c)]^{n+1}}{1-[1-\theta-c+\theta c]}]=d[\frac{1-[(1-\theta)(1-c)]^{n+1}}{\theta+c-\theta c}]$.

Again, as $0<\theta<1$ and $0 then when $n \rightarrow \infty$, $[(1-\theta)(1-c)]^{n+1} \rightarrow 0$. So $d[\frac{1-[(1-\theta)(1-c)]^{n+1}}{\theta+c-\theta c}] \rightarrow d[\frac{1}{\theta+c-\theta c}]=d[\frac{1}{c+\theta(1-c)}]$

This is the money multiplier, $\frac{1}{c+\theta(1-c)}$. It is smaller than the bank multiplier.