Appendix: Explanation of the Math behind the Individual Shock Model of the Interbank Rate

This appendix explains the mathematical foundations for the "individual shock model of liquidity management.” All that's required to understand this a basic familiarity with probability and calculus.

Here's the basic mathematical representation of the model:

The integral in the equation is an expected value calculation. We can understand it by first thinking of an expected value calculation without calculus.

Assume a function, f(x), of a random variable, x, equals 0 or 1 with 50% probability each. If I asked you to calculate the expected value of f(x), you’d probably be able to figure out intuitively that the answer is: 0*.5 + 1*.5 = .5. We simply multiple each possible value of f(x) by its respective probability of occurring. We can represent this in mathematical terms as E[f(x)] = ∑f_ip_i, where E[] represents the expected value of whatever we’re interested in, f_i represents each possible value of f(x), and p_i represents the probability of a given value of f(x), and ∑ represents the summation of each f_ip_i pair.

In the example above, the random variable is discrete, as there were a finite number of values f(x) could take on. Thus the ∑f_ip_i calculation works. However, if we’re thinking about an infinite number of values, the random variable is continuous, and we need to use calculus. The expected value calculation analog for a continuous random variable is

E[f(x)] = ∫f(x)*Φ(x)dx, integrated from negative infinity to positive infinity

Φ(x) is the probability density function of the continuous random variable x. You can see this is very similar to the discrete calculation. We’ve pretty much just substituted an integral for the ∑ sign, and a probability density function for p_i.

To address the equation in the post specifically, let’s match up the notation of the individual shock model to the general example E[f(x)] = ∫f(x)*Φ(x)dx.

C(q) corresponds to E[f(x)]. Our f(x) is the following step function:

f(x) = i_B*(q+x) , if x < -q
f(x) = 0 , if x > -q

So now we just need to explain the limits on the integral. Before I said that the expected value integral integrates from negative to positive infinity. Why is it from negative infinity to –q in this equation? The answer is that you don’t have to integrate to positive infinity, because once the random liquidity shock x ­­> -q, then C(q) = 0 by definition. In other words, at all values of x where x ­­> -q, the liquidity shock is not negative enough to drive the bank’s reserves below zero, in which case the bank will face 0 borrowing costs. C(q) will thus be 0, and taking the integral of 0 across any range of values is 0. Thus, it would be unnecessary to continuing integrating, but you could if you wanted.

And that’s it!