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32_1_The_money_multiplier - The total increase in new...

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The money multiplier The money multiplier tells us the maximum amount of new demand-deposit money that can be created by a single initial dollar of excess reserves. This multiplier, m , is the inverse of the reserve requirement, R : m = 1/ R . This note will demonstrate that result. Suppose some initial amount, d 1 , is deposited into the banking system. With a reserve requirement of R , this deposit creates initial excess reserves equal to E 1 = (1 – R ) x d 1 . Assuming that all of this amount is lent out and redeposited within the system, these excess reserves become new money: M 1 = E 1 = (1 – R ) x d 1 . This second deposit creates its own excess reserves equal to E 2 = (1 – R ) x M 1 = (1 – R ) x E 1 . As before, E 2 is new money, so that M 2 = (1 – R ) x E 1 . Continuing on like this indefinitely, we see a pattern develop: M 1 = E 1 M 2 = E 2 = (1 – R ) x E 1 M 3 = E 3 = (1 – R ) x E 2 = (1 – R ) 2 x E 1 M 4 = E 4 = (1 – R ) x E 3 = (1 – R ) 3 x E 1 and so on ad infinitum .
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Unformatted text preview: The total increase in new money (call this “ D ”) can be found by adding up all the successive changes in new money, D = ∆ M 1 + ∆ M 2 + ∆ M 3 + …. Then substituting for ∆ M i , D = E 1 x [1 + (1 – R ) + (1 – R ) 2 + (1 – R ) 3 + …]. Suppose we multiply both sides by the term (1 – R ) and subtract the resulting product from D . D – (1 – R ) x D = E 1 x [1 + (1 – R ) + (1 – R ) 2 + (1 – R ) 3 + …] – E 1 x [ (1 – R ) + (1 – R ) 2 + (1 – R ) 3 + …]. All terms on the right-hand side with the exception of the initial E 1 x 1 would cancel out: D x [1 – (1 – R )] = D x R = E 1 . Finally, divide both sides by R to obtain the desired result: D = E 1 x R 1 . That is, an initial amount of excess reserves equal to E 1 creates new money equal to this amount multiplied by the inverse of the reserve requirement, or m = R 1 ....
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