# What are your expectations for 2019?

[Originally posted on December 31th, 2018. Revised a couple sentences on January 2nd, 2019 — sketching under pressure and without proofreading, while the wife is showering, before going to a New Year’s Eve celebration is not really the way to go 😀 ]

As promised in Active inactivity, here is a new blog post, before the end of the year!

Since New Year’s Eve is upon us, I think it is only fair to begin this introduction to De Finetti’s definition of probability with a preparatory introduction to the concept of expectation.

In Statistics, the word expectation has somehow a peculiar meaning, that to me represents an improvement on the everyday meaning of the word; the layman’s definition of the word expectation, according to the Oxford Dictionary is “A strong belief that something will happen or be the case.“. Is this enough for the statistician?

Well, yes and no.

Yes, in the sense that the act of making a statement about the future is somehow maintained, at least for a suitable realization of the abstract definition of probability. No, in the sense that we are not interested in making a generic statement about what we believe that will happen in the future; we want to make a statement that reasonably encompasses everything that could happen, resulting in a statement about the average outcome that I can expect.

Let’s make a simple example: say that you draw a card from a deck, and that you gamble such that you win 10 euros is you get a King, and you lose 10 euros if you get a black card. What do you expect to happen on average?

The relevant useful concept here is that of expectation: each outcome (King, anything but a King) has two numbers associated to it: the probability of obtaining that outcome, and the value or pay-off that you get if that particular outcome happens. For example, there are 4 Kings out of the 52 cards that make the deck, so the probability of obtaining a King is $p(K) = \frac{4}{52} \simeq 0.08$, whereas the pay-off you get if the outcome happens is $V(K)=10$, euros. By converse, the probability of obtaining not-a-King is $p(nonK) = \frac{48}{52} = 1- \frac{4}{52} \simeq 0.92$, and the value you stand to win is $V(nonK)=-10$, euros (negative, since you would incur in a loss).

To know what you can reasonably expect to happen on average in this situation, it is necessary to think a bit. If the situation was simpler, for example if you stand to win 10 euros regardless of the outcome of the card draw, you can expect that you will win 10 euros. By converse, if you stand to loose 10 euros regardless of the outcome, you can expect that you will loose 10 euros. But in our fictitious situation different outcomes are rewarded with different values; it is then crucial to have a way of estimating a global, average value for what you can expect. Since each pay-off will happen only its corresponding outcome happens, a natural choice is to weight each possible pay-off with the probability that its corresponding outcome will actually happen. It turns out then that, following this line of reasoning, your expectation is given by an average of the pay-offs, weighted by the probability that each outcome happen. For our concrete case, $E[gamble] = p(K)\times V(K) + p(nonK)\times V(nonK) = 0.08\times 10 + 0.92\times(-10) =-8.4$; in words, the expected value you get out of your gamble is lose 8.4 euros.

You can already use such considerations to find the expected value of any kind of situation in which you can gamble on some well-known outcomes; this works for example for any gamble on a deck of cards, in which you can easily calculate the probability of any outcome by simply counting the cards—or combinations thereof—in the deck).

In order to interpret such statement, and to really get to the bottom of the meaning of expected value, we need to make a small step back and look into how we can define and compute an essential element of the formula for the expectation: probabilities.

However, you will have to wait for the New Year, because my wife has finished showering, and we need to get ready for this night’s party!

See what I did here? Not only have I described briefly the concept of expectation, but I have also given you a way of computing what is the value you expect to get from this blog for the first week of 2019: what is the probability you assign to me writing and publishing the next blog post before January 7th? What is the value you assign to the blog post coming out, and what is the value you assign to the blog post not coming out?

Try to get your probabilities and pay-offs figured out before midnight!