# The law of how late people choose to be

An organizer of a small research workshop, observing that all the locals were late, once remarked that “the closer people are to the venue the later they come.” If I remember my school days correctly, the same could have been said of my classmates. Of course, this is not a precise law, as people differ in how much they dislike being late (or early) to things, but on the whole, there could well be some empirical truth in this statement. In this post, I will sketch a little decision theoretic model that delivers this law.

We want to model a person deciding when to leave from home to go to work (or school or a conference venue or whatever their destination is). Let me phrase this as the problem of choosing an intended delay in getting there. Let me call this chosen intended delay $\displaystyle d$, which can be any real number. A negative delay means you are early. But $\displaystyle d$ is only the intended delay, the actual delay depends on traffic and such things. Call the actual delay $\displaystyle d + X$, where $\displaystyle X$ is a random variable with zero mean and some non-negative variance $\displaystyle \sigma^2$. The variance is my attempt to capture the random traffic conditions. A person who lives far from their destination quite possibly faces a higher variance in their travel time than a person who lives closer to their destination. The law I am looking for would then state that the higher the variance that a person faces for their travel time the earlier this person will be on average.

The decision maker chooses the intended delay, but what is their goal? I think it makes sense to assume that they prefer not to be late, and that in such a way that the later they are the worse it is. But it is probably also not great to be too early, as they could use the waiting time better at home before they leave. One option would be to consider preferences as captured by the utility function (often used as the loss function in statistics) $\displaystyle v(d) = -(d+x)^2.$

This utility function captures both ideas: ideally people would show up exactly on time, that is the realized actual delay $\displaystyle d+x$ is zero, giving a utility of zero; they dislike being early and also being late, as in both cases the utility is negative. Also, the further away the realized actual delay is from zero (in both directions) the lower is this person’s utility.

However, a person with this utility function would choose to have zero expected delay regardless of the variance. To do so they would choose an intended delay $\displaystyle d$ of zero. This means that everybody will arrive on time on average, and those with a higher variance will be more random in their delay, sometimes being quite early, sometimes quite late. There is no law that links the variance of travel times to a person’s average delay.

This is because under the given utility function the decision maker cares equally about being early and being late, the two concerns exactly offsetting each other. This may not be such a reasonable assumption for most cases. In many cases people probably worry more about being late than about being early. A simple way to accommodate such preferences is to consider the utility function $\displaystyle u(d) = -(d+x)^2 e^{\frac{d}{\sigma}}.$

The multiplicative term, the function $\displaystyle e^{\frac{d}{\sigma}},$ is always positive, but is smaller for small (such as negative) delays and larger for large (such as positive) delays. Dividing by $\displaystyle \sigma$ in the exponent is a kind of normalization, that makes my life easier below. Different such positive and increasing functions would lead to different, yet qualitatively similar, laws at the end.

So, suppose our decision maker behaves as if they had such preferences as captured by this utility function. What would be their optimal choice? Their optimal intended delay would be the one that maximizes their expected utility. This problem can be written as $\displaystyle \max_{d} - \mathbb{E} \left[(d+X)^2 e^{\frac{d}{\sigma}}\right],$

where, $\displaystyle \mathbb{E}$ indicates the expectation with respect to the random additional delay $\displaystyle X.$ Equivalently, the problem can be written as $\displaystyle \min_{d} \mathbb{E} \left[(d+X)^2 e^{\frac{d}{\sigma}}\right].$

We can solve this by differentiating (finding the first order conditions, using the product rule) with respect to $\displaystyle d.$ We obtain $\displaystyle \mathbb{E} \left[2(d+X) e^{\frac{d}{\sigma}} + (d+X)^2 \frac{1}{\sigma} e^{\frac{d}{\sigma}} \right] = 0.$

Dividing by $\displaystyle e^{\frac{d}{\sigma}},$ multiplying by $\displaystyle \sigma,$ noting that the expectation of a sum is the sum of expectations, and recalling that $\displaystyle \mathbb{E} X = 0,$ we get $\displaystyle 2 \sigma d + \mathbb{E} \left[(d+X)^2\right] = 0.$

By the fact that $\mathbb{E} \left[(d+X)^2\right] = \mathbb{E} \left[d^2+2dX+X^2\right] = \mathbb{E} \left[d^2+X^2\right],$ and noting that $\mathbb{E} \left[X^2\right] =\sigma^2,$ the variance of $X,$ we get $\displaystyle 2 \sigma d + d^2 + \sigma^2 = 0.$

Note that $\displaystyle 2 \sigma d + d^2 + \sigma^2 = (d+\sigma)^2,$ and, thus, the optimal intended delay is simply $d= - \sigma.$

In words the decision maker would optimally aim to arrive one standard deviation early and we have our law of how late people are as a function of how far they start from their destination, or more correctly as a function of how much variance they face or at least perceive for their travel time.

Our very specific law, finally, states that people aim to be one standard deviation (of their random travel time) early! 😉

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