Michael Woodford's Individual Liquidity Shock Model of the Interbank Rate


This post gives a more complete explanation of the individual shock model introduced here. Bindseil gives a somewhat incomplete treatment of it in his book, given that he prefers the aggregate liquidity management model for its relative simplicity and so devotes most of his attention to it. In my own search for a more detailed explanation of the individual shock model, I found this 2007 paper from Bindseil, which reviews a more detailed individual shock model from Michael Woodford’s 2001 Jackson Hole paper, Monetary Policy in the Information Economy. This post explains that model and the math behind it in more detail than can be found in those papers.

A summary of the model is as follows. A bank reserve manager, j, seeks to minimize his expected costs of refinancing, C(s), which is represented by the following equation:
  • C(sj) = isj – iBEj[min(sj+ εj,0)] – iDEj[max(sj+ εj,0)], where sj is the quantity of reserves bank chooses to hold following the interbank market session, ε is a random variable for the end-of-day liquidity shock, i is the interbank rate, iB is the central bank's borrowing facility rate, and iis the central bank's deposit facility rate
The quantity of reserves a bank manager thus seeks to trade in the interbank market, sj, is given by the first order condition:
  • sj = -F-1[(i-iD)/(iB-iD)], where F-1 is the inverse cumulative distribution function of ε   
The market clearing price brings quantity of reserves demanded equal to quantity supplied:
  • ∑sj = R, where R is the quantity of reserves supplied by the central bank via its OMO
Substituting in the expression for sj and solving for i yields:
  • i = iD + F(-R/(∑ j)*(iB – iD), where F is the cumulative distribution function of ε
Thus, the central bank can control i by altering iD, iB, and/or R.

Model Assumptions

We assume banks must end the day with a non-negative amount of reserves – i.e., reserve requirements are set at 0. If a bank ends up with negative reserves (say they incur a costless intraday overdraft), it must borrow the balance at the central bank’s lending facility interest rate iB. If a bank ends up with excess reserves, it earns interest on its reserves at the central bank’s deposit facility interest rate iD.

Woodford’s model assumes a similar same one-day reserve maintenance period that was discussed in the aggregate liquidity management model post. One slight difference is that instead of considering the liquidity shock to be an aggregate autonomous factor shock, we view it as any shock that can affect an individual bank’s reserve position. As such, the liquidity shock may only lead to a redistribution of reserves across banks rather than aggregate change in quantity of reserves.

Specifically, the model assumes that the central bank sets the quantity of reserves to be traded in the interbank market session via its OMO. Following that, a perfectly competitive interbank market session takes place. A stochastic liquidity shock occurs following the market session, which alters banks’ liquidity positions. Banks then seek recourse to the central bank’s lending and deposit facilities as necessary. The stochastic uncertainty created by the end-of-day liquidity shock is central to the model, as we’ve noted previously, and as Woodford notes here:

“The existence of residual uncertainty at the time of trading in the interbank market is crucial; it means that even after banks trade in the interbank market, they will expect to be short of funds at the end of the day with a certain probability, and also to have excess balances with a certain probability. Trading in the interbank market then occurs to the point where the risks of these two types are just balanced for each bank.

Denote sj as the quantity of reserves banks choose to hold by dealing in the interbank market. Denote εj as the random variable representing the stochastic liquidity shock to an individual banks’ reserve position at the end of the day. Assume the shocks are independently distributed across banks with an expected value of 0 and standard deviation σj.

The Bank Reserve Manager’s Optimization Problem

A risk-neutral bank reserve manager thus seeks to minimize the expected cost of refinancing, C(sj), according to the following equation:

C(sj) = is– iBEj[min(sjj,0)] – iDEj[max(sjj,0)]                                                                             (1)

It is critical to understand that this equation represents the total expected costs of refinancing at the time of trading in the interbank market – not actual costs at the end of day. Banks can’t perfectly tell the future. They have to take this uncertainty into account when trading reserves.

The first term on the right side of the equation represents the costs associated with borrowing sj reserves in the interbank market. These costs are clearly known for certain at the time of trading in the interbank market. The rest are not. The second term represents the expected costs of borrowing (denoted by the notation E[.]) at the central bank’s lending facility at the end of the day. The min function is used since costs are only incurred if a bank’s total reserve position (sj+ εj) is less than 0. A minus sign is attached to this term since it is only non-zero when (sj+ εj) is negative, and we want this part of the equation to represent a net addition to expected costs. Finally, the third term represents net revenue from interest on reserves at the central bank’s deposit facility. The max function is used since it only comes into play when the total quantity of bank reserves is non-negative. A negative sign is used since interest on reserves decreases bank costs. Note that the goal is not necessarily to maximize this portion of the equation, since in order to acquire reserves, they must be borrowed in the interbank market at a cost i.

So how does one find the quantity sj that minimizes this equation? If you recall from calculus, we take the partial derivative (since there are multiple variables) of this equation, set it equal to 0, and solve for sj. This is called the “first order condition.” Before doing this, let’s write out (1) with full calculus notation (the subscript j is omitted in the following for ease of typing). 

See here for a basic explanation of why the expected value operator E[.] turns into an integral (here, f(ε) is the probability density function for ε). Once that is understood, consider how it applies to the min and max functions. Take the expected value of the min function for example. Note that when ε is less than –s, the min function equals (s+ε), since it will be less than 0. However, when ε is greater than –s, the min function equals 0. As such, we can split the integration up according to these two regions, as is displayed in (2). We can eliminate the integral with 0, since the integral of 0 is always 0. The same reasoning applies to the max function, except the relevant region of integration is reversed. Equation (2) simplified is:


Okay, now we can begin to take ∂C(s)/∂s and set it equal to 0. To do this, I’ll break the equation up into three parts.

  1. ∂C(s)/∂s of the term [is] is clearly just i
  2. ∂C(s)/∂s of the  term requires use of the fundamental theorem of calculus, since we’re taking a derivative of an integral with respect to a variable that is in the limits of integration. If you distribute (s+ε)*f(ε), split it into two separate integrals, and then properly use the product and chain rules, you should be left with , which is equivalent to –iB*F(-s), where F(.) is the cumulative distribution function (CDF) of ε. The CDF gives the cumulative probability that a random variable is less than or equal to a certain value; the integral above is evaluating the probability density function (pdf) from negative infinity to –s, which is the same as evaluating the CDF at –s (the probability ε takes on any value less than –s). 
  3. ∂C(s)/∂s of the term attached to iD is a bit more complicated, but uses the same principles as above. You should end up with –iD*(1-F(-s)).

Setting the sum of these three partials equal to zero and doing some algebraic manipulation results in the first order condition displayed in the Bindseil and Woodford papers*:

(iD - 1)(1 - F(-s)) + (iB-i)*F(-s) = 0                                                                                                         (4)

Solving for s, which is what the bank reserve manager is looking to do, we move the i, iD, and i­B variables to the right side and then take the inverse of the function, to get:

s = -F-1[(i-iD)/(iB-iD)]                                                                                                                           (5)

This is an important equation. It shows that the quantity of reserves that a bank will want to trade is dependent on a) the probability distribution of the liquidity shock and b) the relative distance of i between iD and iB. The absolute levels of the interest rates do not matter, which should make sense, since the bank is always trying to minimize its cost of refinancing regardless of the level of interest rates. Equation (5) is made clear in the following graph.

Essentially, [(i-iD)/(iB-iD)] will always be some fraction that corresponds to a probability on the y-axis of the CDF. This probability then corresponds to a certain quantity of reserves according to the CDF curve, and this is the quantity (multiplied by -1) that a particular bank will want to trade. For example, an i closer to iD than iB means the above expression will be less than 50%. A probability less than 50% corresponds to a particular negative value of ε, which (5) converts to a positive value of sj – meaning a bank will want to borrow sj reserves. In other words, with an interbank rate less than halfway in between the standing facility rates, a bank can acquire a certain quantity of reserves at a cost now that nonetheless lowers its net expected costs going forward. Vice versa for the case where i is closer to iB than iD.

The Market Clearing Price of Reserves

We’re not done yet, though, because we still need to figure out what i should be. In a perfectly competitive market, the market equilibrium price of reserves, i, should bring quantity demanded equal to quantity supplied. The quantity supplied, R, is set by the central bank following the OMO. The quantity demanded is simply the sum of equation (5) across all banks.

∑sj = R                                                                                                                                               (6)

Solving for i:

∑-F-1[(i-iD)/(iB-iD)] = R

-1*n* (F-1[(i-iD)/(iB-iD)]) = R (assuming all n banks are identical)

F-1[(i-iD)/(iB-iD)] = -R/n

(i-iD)/(iB-iD) = F(-R/n) 

i = iD + F(-R/n)*(iB - iD)                                                                                                                  (7)

Equation (7) shows that, given a probability distribution for ε, the central bank can control i through a combination of controlling the standing facility rates and the quantity of aggregate reserves. Furthermore, i can only be within the bounds set by iand iD, as F(-R/n) is a fraction between 0 and 1. This is the same result we achieved using the aggregate liquidity management model.


* Bindseil and Woodford normalized the distribution of ε to ε/σ, which is why you see the terms F(-s/σ) in their equations. 

No comments:

Post a Comment