Appendix C: Velocity Calculations | RDP 2018-12: Where's the Money? An Investigation into the Whereabouts and Uses of Australian Banknotes

RDP 2018-12: Where's the Money? An Investigation into the Whereabouts and Uses of Australian Banknotes Appendix C: Velocity Calculations

Richard Finlay, Andrew Staib and Max Wakefield

December 2018

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In this appendix, we justify multiplying the turnover rate by (1 + 2r). We deal with the case of estimating the time cash spends in wallets, but note that the same logic applies to tills and ATMs. Further, since we are only concerned about the flow of a generic dollar, not individual banknotes per se, we assume that cash flows out of a wallet in the same order in which it flows in (first-in first-out) and that cash is spent at a constant rate halfway through each day.

Case 1: No buffer stock. Suppose that the number of days between cash top-ups is n. Then 1/n^th of cash is spent at 0.5 days; the next 1/n^th of cash is spent at 1.5 days; and so on, until the last 1/n^th of cash is spend at n − 0.5 days. The average time that any dollar spends in the wallet is:

\frac{1}{n} (\frac{1}{2} + \frac{3}{2} + \dots + n - \frac{1}{2}) = \frac{1}{n} (1 + 2 + \dots + (n - 1)) + \frac{1}{2} = \frac{n - 1}{2} + \frac{1}{2} = \frac{n}{2}

We would estimate that cash spends n/2 days on average in a wallet.

Case 2: Buffer stock. Suppose again that the number of days between cash top-ups is n. If the buffer share is r (where r is a number between 0 and 1), then we need to recognise that in any period a person first spends their buffer stock from the previous period. The buffer stock is exhausted after (rn − 0.5) days. Any cash spent from the buffer spends (n + b + 0.5) days in the wallet, where b is the number of days since the most recent top-up and is an integer in [0, rn − 1]. Cash that is not part of the previous buffer is then spent from the rn + 0.5 day until the n − 0.5 day. The average time that a dollar spends in the wallet is then:

\begin{array}{l} \frac{(n + 0.5) + (n + 1.5) + \dots + (n + r n - 0.5) + (r n + 0.5) + \dots + (n - 0.5)}{n} = \frac{r n^{2} + (0.5 + 1.5 + \dots + n - 0.5)}{n} \\ = r n + \frac{n}{2} = \frac{n}{2} (1 + 2 r) \end{array}

This is the formula we use.