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Expectation Value and Notation

Updated on January 3, 2012

Symbols and notation.

You will find that the letter E is used in mathematics for Expected value. In English, E is for “Expectation”, for Germans “Erwartungswert”, and for French “Espérance mathématique”.

Function notation:

It is common to denote a function like y=2x+5 and y=4x^2 and any other where the independent variable is x as

f(x)

Pronounced, "f of x"

In this way, you do not need to name an independent variable, and it's convenient to refer to a whole swag of functions as 'functions of x'.

E(something)= ...

therefore fits this kind of notation where, in context, the E is representing an operation that produces an expectation value using the things in the braces.

Probability space

Before we can proceed, you need to know what is a probability space.

If you throw die, pick a card, do an experiment, record an event and so on, there will be certain possible outcomes. When this is based on randomness, we need to define the possible set of results. We need three parameters to do this rigorously. Note: one or more elementary outcomes may be interesting enough to be called an 'event'.

  • Ω is the set of all possible elementary outcomes no matter how interesting or how likely.
  • Σ is the set of all interesting outcomes. (See previous paragraph for definition).
  • P measures the likelihood (probability) assigned to each event.

Collectively, these two sets, the sample space Ω, and Σ which is a sigma algebra, σ-algebra, and the corresponding set of probabilities P are a probability triple . We also call it the probability space.

Clearly, Σ is a subset of Ω because one or more outcomes make up an event.

Upon measurement or observation of an event, we call it a single outcome. It is given the symbol ω.

When we have ω, it will be contained in one or more of Σ. At this point, an example will help. When you pick a card from a pack, there are 52 possible elementary outcomes. An interesting event ω might be that of an ACE. There are four elementary elements which could be an ACE. Alternatively, that ACE might be of hearts, for which there are 13 interesting events to which it could belong since there are 13 hearts in the pack. The probability to pick an ace is 4 out of 52 and the probability of picking a heart is 13 out of 52. This is why we need the triple as we need to know all possible outcomes, those that we are classifying, and the probability that it can occur.

An example.

You can use an expectation value in gambling to work out your average payout, and also why it makes little sense to gamble.

If you gamble, say, 4 times and bet $5, $6, $2 and 7$ to throw a six on a standard die, and if your banker pays 2 times your bet if you win, and nothing otherwise, then these are all the possibilities:


lose $5 or win $(5 x 2) meaning 5/6 chance to lose $5 and 1/6 chance to win $10
lose $6 or win $(6 x 2) meaning 5/6 chance to lose $6 and 1/6 chance to win $12
lose $2 or win $(2 x 2) meaning 5/6 chance to lose $2 and 1/6 chance to win $4
lose $7 or win $(7 x 2) meaning 5/6 chance to lose $7 and 1/6 chance to win $14


This can be written:

"Ways to win" - "Ways to lose"

1/6(1012414 -5/6(5+6+2+7


This is 6.66-16.66 = an expectation of a $ 10 loss. This represents a fair estimate should you decide to abort the game early and work out the most likely outcome. Of course you *might* do better, but you cannot expect it. In fact if the experiment is repeated many times, then the average will be a loss of $10.

A more general example could have assigned to each outcome, different probabilities.


Probability density function.

If you know all the individual outcomes and the probability that they can occur, then all of these probabilities must add up to 1. As a simple example, a fair die will have the following possible outcomes: X={1,2,3,4,5,6} and no others and each will occur with frequency 1/6. The probability that any one of these events happens is written P(x=X) and we can write the expectation value that this particular function of x is:

E(f(x)) ≡ 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6

= 1/6(6) = 1

This is so easy that it seems to be a big who-hah about nothing. However, what if the die was heavily biased. Lets assume that the centre of gravity of the die is nearer the number one. This will make it more likely to fall with a six showing upwards. We might guess that the probability density function is now:

P(x=1) =1/27

P(x=2) =2/27

P(x=3) =2/27

P(x=4) =2/27

P(x=5) =2/27

P(x=6) = 2/3

Again, these all sum to one since the probability of getting a number is certain because you can't consider the impossible case where the die lands on a corner and balances.

1/27+2/27+2/27+2/27+2/27+2/3 = 1

In this case, the probablilty to get any outcome P(X=x) is some function of the physics involved in a biased die. So we write P(f(x)) as before. Having embodied each individual outcome into some computable function, we can compress the notation and generalize it to show the expectation value for any event x, where x is an element of a set of all possible outcomes X like this:

E(f(x)) ≡ Σx P(X=x)f(x),

x is an element of X and the sum is performed over all elements of X.

Don't confuse Σ with 'sigma' used before in the description of probability space. In this case it is the summation notation.

The equation reads,"

The expectation value for a system that is a function of x is the same as adding up for each x in the set X of all individual probabilities of x as a function of x.


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      Luckie 3 years ago

      I was looking evrwyrheee and this popped up like nothing!

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