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STATLECT > Fundamentals of probability

Indicator functions

Let Omega be a sample space, let $Esubseteq Omega $ be an event and denote by [eq1] the probability assigned to the event E. The indicator function of the event E (or indicator random variable of the event E), denoted by $1_{E}$, is a random variable defined as follows:[eq2]In other words, the indicator function of the event E is a random variable that takes value 1 when the event E happens and value 0 when the event E does not happen.

Example_ We toss a die and one of the six numbers from 1 to $6$ can appear face up. The sample space is:[eq3]Define the event [eq4]i.e. E is the event "An odd number appears face up". A random variable that takes value 1 when an odd number appears face up and value 0 otherwise is an indicator of the event E.

From the above definition, it can easily be seen that $1_{E}$ is a discrete random variable with support [eq5]and probability mass function:[eq6]

Indicator functions are often used in probability theory to simplify notation and to prove theorems.

Properties

Indicator functions enjoy the following properties.

Powers

The n-th power of $1_{E}$ is equal to $1_{E}$:[eq7]because $1_{E}$ can be either 0 or 1 and:[eq8]

Expected value

The expected value of $1_{E}$ is equal to [eq9]:[eq10]

Variance

The variance of $1_{E}$ is equal to [eq11]. Thanks to the usual variance formula and the powers property above, we obtain:[eq12]

Intersections

If E and F are two events, then:[eq13]In fact:[eq14]

Indicators of zero-probability events

Let E be a zero-probability event and X an integrable random variable. Then:[eq15]While a rigorous proof of this fact is beyond the scope of this introductory exposition, this property should be intuitive. The random variable [eq16] is equal to zero for all sample points omega except possibly for the points $omega in E$. The expected value is a weighted average of the values $X1_{E}$ can take on, where each value is weighted by its respective probability. The non-zero values $X1_{E}$ can take on are weighted by zero probabilities, so [eq17] must be zero.

Solved exercises

Below you can find some exercises with explained solutions:

  1. Exercise set 1 (use of the indicator function)

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