Surely obvious if you know probability theory well, but it took me a while to realize:

The expected value of a random variable (or of a function of a random variable) is the same as the Lebesgue integral of the random variable (or of the function thereof) over the probability measure.

This is why the following are equivalent statements:

  • \(X\) is integrable
  • The expected value of \(X\) exists and is well defined

Conventional definition

For a discrete random variable \(X\) the expected value is defined as

\[\begin{equation} E[X] = \sum_{i\ \ \ \ } x_i P_{x_i}, \end{equation}\]

where the sum is over all indexes \(i\), and \(P_{x_i}\) is the probability of \(x_i\).

Similarly, for a continuous random variable \(X\) that takes on real values, the expected value is defined as

\[E[X] = \int\limits_{-\infty}^{\infty} x P(x) dx.\]

Lebesgue integral interpretation

But this is just the Lebesgue integral of \(X\) over the probability measure:

\[E[X] = \int X dP.\]

To see this, it’s easier to consider the prior form:

\[E[X] = \int\limits_{-\infty}^{\infty} x P(x) dx.\]

Partition the range of \(X\) into intervals; for each interval we multiply (1) the measure of the event mapped by the random variable \(X\) into that interval (this corresponds to \(P(x)dx\)) with (2) the expression to be integrated (here it is \(X\) because we are integrating \(X\)); sum over all intervals; and take the limit as the partition is made more finely.


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