Cashing a blank check

January 26, 2021. Suppose you find a blank check on the ground, and unscrupulously decide to cash it in. If overdrawing gets you nothing, how much should you cash it in for? Assuming wealth follows the 80-20 rule, the answer is: almost nothing!

Introduction

In the film “Blank Check” (1994), 11-year old Preston Waters is handed a blank check, and cashes it in for a million dollars. Luckily, this is precisely the amount of money that the check’s signer, a convict attempting to launder his ill-gotten gains, has left with the bank’s president. But what if Preston overdrew, asking for, say, $10$ billion? This would probably have raised the suspicions of the complicit bank president and the check would have bounced altogether. When I was a kid, I thought it was incredibly lucky for Preston to find the check in the first place. I now think drawing the precise amount of money held in trust is infinitely luckier. But this raises the question: if you find a blank check, and you don’t want it to bounce, how much should cash it in for?

Expected return

I’ll assume we know nothing about the identity of the signee, and that if they have a balance of $b$, and we make out the value of the check to be $v$, then the check will bounce if $v > b$. Our strategy will be to calculate the expected return for $v$ and then maximise it. If $f(b)$ is the probability distribution for bank balances, then the expected return for $v$ is simply $v$ multiplied by the probability $b> v$:

\[E(v) = v \int_v^\infty f(b) \, db = v[1 - F(v)] = v \bar{F}(v),\]

where $F$ is the cumulative distribution function, and the $\bar{F} = 1 -F$ the tail. To maximise this, we assume the curve is smooth, differentiate and set to $0$, using $\bar{F}’ = -f$:

\[E'(v) = \bar{F} - vf(v) = 0 \quad \Longrightarrow \quad v = \frac{\bar{F}(v)}{f(v)}.\]

Any $v$ which satisfies this equation is an extremum.

Long and short tails

Now the question is how to model the distribution of bank balances. This is the sort of thing expected to follow a power-law curve like the Pareto distribution, the proverbial “80-20” curve. This is simply defined by its power-law tails:

\[\bar{F}(v) = \left(\frac{L}{v}\right)^\alpha,\]

where $L$ is the minimum amount to keep a bank balance open (say a monthly fee), and $\alpha > 0$ is a shape parameter we will “leave blank” for the moment. This is well-defined since it heads to zero. The probability density for $v \geq L$ is

\[f(v) = -\bar{F}'(v) = \frac{\alpha L^\alpha}{v^{\alpha + 1}}.\]

The optimal draw then obeys

\[v = \frac{\bar{F}(v)}{f(v)} = \left(\frac{L}{v}\right)^\alpha \cdot \frac{v^{\alpha + 1}}{\alpha L} = \alpha v.\]

For $\alpha \neq 1$, the only solutions are $v = 0$ and $v = \infty$! For $\alpha > 1$, we can plot the expected return $E(v)\propto v^{1-\alpha}$, and see that it monotonically decreases, with the maximum at $v = L$. Preston should only have asked for a few bucks! But perhaps this is an artefact of the infinite power-law tail. A more realistic choice is the truncated Pareto distribution, where the power law is confined to $L \leq v \leq H$ for an upper limit $H$, say the personal wealth of Jeff Bezos or Elon Musk. The density for the truncated Pareto distribution is simply a conditional probability, conditioned on being in the interval $[L, H]$:

\[f(v) = \frac{\alpha L^{\alpha}v^{-(\alpha+1)}}{1 - (L/H)^\alpha},\]

and the tail is

\[\bar{F}(v) = \int_v^H \frac{\alpha L^{\alpha}v^{-(\alpha+1)}}{1 - (L/H)^\alpha} dv = \frac{(L/v)^\alpha - (L/H)^\alpha}{1 - (L/H)^\alpha}.\]

Thus, we now have to solve

\[v = \frac{\bar{F}(v)}{f(v)} = \frac{(L/v)^\alpha - (L/H)^\alpha}{\alpha L^{\alpha}v^{-(\alpha+1)}} \quad \Longrightarrow \quad v = (1-\alpha)^{1/\alpha} H.\]

If $\alpha < 1$, then we do get a finite answer, proportional to the upper bound, so for instance if $\alpha = 0.5$, and we take the upper limit to be around 100 billion dollars, then Preston should ask for

\[v \sim \sqrt{1-0.5} \times 10^{11} \approx 70 \text{ billion dollars},\]

or $0.7$ of some other reasonable guess for $H$. But if $\alpha \geq 1$, the prefactor is not real, and as for the full Pareto distribution, the maximum expected return occurs at $L$. And indeed, wealth typically does obey an approximate Pareto distribution with $\alpha > 1$. For instance, the proverbial “80-20” rule corresponds to $\alpha \approx 1.16$, and this analysis of the Forbes 400 richest people in the world finds a shape parameter of $\alpha = 1.49$. So once again, a perfectly rational Preston Waters would ask only for the monthly fee! But this would make for a far less entertaining movie.

Written on January 26, 2021
Mathematics   Statistics   Everyday