# Infinite expected value of a random variable

How can a positive random variable $X$ which never takes on the value $+\infty$, have expected value $\mathbb{E}[X] = +\infty$?

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The realization of a random variable has to be a finite number. – Patrick Li Nov 17 '12 at 15:35
The term I've come across for such variables is that their distributions have a "too fat tail", signifying that high-value outcomes have such a big probability that even though it all sums up to $1$, once you calculate the expected value, the big outcomes kick in and inflate the result. – Arthur Nov 17 '12 at 16:05
The result is quite counter-intuitive. How can summing up products of finite numbers (the values of the random variable) with finite numbers (the probability of the random variable taking on that value) be infinite? – lodhb Nov 25 '12 at 11:08
See here for a statistical perspective: stats.stackexchange.com/questions/94402/… – kjetil b halvorsen Nov 15 '15 at 15:25

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.

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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.

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