Infinite expected value of a random variable

Let $X$ be a random variable that is equal to $2^n$ with probability $2^{-n}$ (for positive integer $n$). Then $${\mathbb E} X = \sum_{n=1}^\infty 2^{-n} \cdot 2^n = \sum_{n=1}^\infty 1 = \infty.$$

Cauchy Distribution is an example of a continuous distribution that doesn't have an expectation.

Once you consider probabilistic experiments with infinite outcomes, it is easy to find random variables with an infinite expected value. Consider the following example (which is just a game that yields an example similar to the one Yuri provided):

• You throw a coin until it lands tails.
• You then get paid $2^{n}$ dollars, where $n$ is the amount of heads you got.

It is easy to construct the expected value funcion of your payment (let's name it $X$):

$$E[X] = \frac{1}{2} \times 2^0 + \frac{1}{4} \times 2^1 + \dots = \sum_{n=1}^{\infty} 2^{-n}\times 2^{n-1} = \sum_{n=1}^{\infty} \frac{1}{2} = \infty $$

This game is also known as St. Petersburg paradox. Why does this occur and how can we interpret it?

From a construction point of view, it is easier to understand. In this particular case, the probability of each outcome decreases exponentially. Since the number of outcomes is infinite, the payout scheme only has to grow at the same rate as the probability of the outcome decreases in order for the series to diverge.

What this means in practice is that, although the payout is always finite, if you average the payouts from $k$ consecutive games, this average will (with high probability) be higher the greater $k$ is. As $k$ approaches infinity, so does the average of the $k$ payouts. Behind this boundless growth is the fact that everytime an unlikely outcome happens, the payout is so large that, when averaged with the payout of more likely outcomes, the average is skewed up.