有界随机变量是否存在无限期望？含连续性情形的证明问询

Let's cut straight to the chase: No, a bounded random variable (whether discrete or continuous) cannot have an infinite expectation. Both of your questions boil down to the same core idea, so let's break down why this is true, with straightforward proofs for both discrete and continuous cases.

First, what does "bounded" mean here? #

A random variable (X) is bounded if there exists some positive real number (M) such that the probability that (X) falls outside the interval ([-M, M]) is zero. In formal terms: (P(|X| \leq M) = 1) for some (M > 0).

Proof for discrete bounded random variables #

Suppose (X) is discrete, with possible values (x_1, x_2, x_3, ...) and corresponding probabilities (P(X=x_i) = p_i) (where (\sum p_i = 1)). By the boundedness condition, every (|x_i| \leq M).

The expectation of (X) is defined as:
[E[X] = \sum_{i} x_i p_i]

Taking the absolute value of the expectation, we can use the triangle inequality for sums:
[|E[X]| = \left| \sum_{i} x_i p_i \right| \leq \sum_{i} |x_i| p_i]

Since each (|x_i| \leq M), substitute that in:
[|E[X]| \leq \sum_{i} M p_i = M \sum_{i} p_i = M \times 1 = M]

This means (|E[X]| \leq M), so the expectation is bounded (and therefore finite, not infinite).

Proof for continuous bounded random variables #

Now take a continuous random variable (X) with probability density function (pdf) (f(x)). Again, by boundedness, (f(x) = 0) for all (x) outside ([-M, M]).

The expectation of (X) is:
[E[X] = \int_{-\infty}^{\infty} x f(x) dx]

Since (f(x)) is zero outside ([-M, M]), this integral simplifies to:
[E[X] = \int_{-M}^{M} x f(x) dx]

Taking the absolute value and using the triangle inequality for integrals:
[|E[X]| = \left| \int_{-M}^{M} x f(x) dx \right| \leq \int_{-M}^{M} |x| f(x) dx]

Again, since (|x| \leq M) over the entire interval of integration:
[|E[X]| \leq \int_{-M}^{M} M f(x) dx = M \int_{-M}^{M} f(x) dx = M \times 1 = M]

Just like the discrete case, the absolute value of the expectation is bounded by (M), so it can't be infinite.

Wrapping up #

The key takeaway here is that boundedness imposes a hard limit on how large (or small) the random variable can be. Since expectation is a weighted average of the variable's possible values, that average can never "blow up" to infinity when all values are constrained within a finite range.

内容的提问来源于stack exchange，提问作者Jan Vainer