If an nth Moment Exists, Moments of Lower Orders than n Also Exist
Theorem
If there exists $E( X^n )$ for a random variable $X$ and a natural number $n$, then $E( X^m ), m=1,2,3,\cdots, n$ also exists.
Description
Regardless of the degree, if a certain moment exists, moments of lower degrees always exist, although naturally the converse is not true. Of course, in practice, it is rare for higher-order moments to be given first, but this theorem does save a significant amount of space when listing the conditions for some theorems.
Proof
Strategy: The proof circumvents by using the absolute value of the random variable to show the inequality relationship and then returns to the original inequality intended to be shown. Although this proof is for continuous probability distributions, the same method can be applied to discrete probability distributions as well.
Let’s say the probability density function of the random variable $X$ is $f$.
Part 1. $E \left( X^{n} \right) < \infty \implies E \left( |X|^{n} \right) < \infty$
Assuming $E \left( |X|^{n} \right) = \int_{-\infty}^{\infty} |x|^{n} f(x) dx= \infty$, we have $$ \int_{0}^{\infty} |x|^{n} f(x) dx= \infty $$ or $$ \int_{-\infty}^{0} |x|^{n} f(x) dx= \infty $$ Meanwhile, since $\begin{cases} \displaystyle \int_{0}^{\infty} |x|^{n} f(x) dx= \int_{0}^{\infty} x^{n} f(x) dx \\ \displaystyle \int_{-\infty}^{0} |x|^{n} f(x) dx = (-1)^{n} \int_{-\infty}^{0} x^{n} f(x) dx \end{cases}$, we have $$ E \left( X^{n} \right) = \int_{-\infty}^{\infty} x^{n} f(x) dx = \int_{0}^{\infty} x^{n} f(x) dx - \int_{-\infty}^{0} x^{n} f(x) dx = \infty $$
Part 2. $E \left( |X|^{n} \right) < \infty \implies E \left( |X|^{m} \right) < \infty$
$$ \begin{align*} E \left( |X|^{m} \right) =& \int_{-\infty}^{\infty} |x|^{m} f(x) dx \\ =& \int_{|x|<1} |x|^{m} f(x) dx + \int_{|x|>1} |x|^{m} f(x) dx \\ \le & \int_{|x|<1} f(x) dx + \int_{|x|>1} |x|^{n} f(x) dx \\ \le & \int_{-\infty}^{\infty} f(x) dx + \int_{-\infty}^{\infty} |x|^{n} f(x) dx \\ \le & 1 + E \left( |X|^{n} \right) \\ <& \infty \end{align*} $$
Part 3. $E \left( |X|^{m} \right) < \infty \implies E \left( X^{m} \right) < \infty$
$$ \begin{align*} E \left( X^{m} \right) =& \int_{-\infty}^{\infty} x^{m} f(x) dx \\ \le & \left| \int_{-\infty}^{\infty} x^{m} f(x) dx \right| \\ \le & \int_{-\infty}^{\infty} | x | ^{m} f(x) dx \\ =& E \left( |X|^{m} \right) \\ <& \infty \end{align*} $$ Collecting Part 1. ~ Part 3., we obtain $E \left( X^{n} \right) < \infty \implies E \left( X^{m} \right) < \infty$.
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