If the n-th Moment Exists, Then Moments of Order Less Than n Also Exist
Theorem
For a random variable $X$ and a natural number $n$, if $E( X^n )$ exists, then $E( X^m ), m=1,2,3,\cdots, n$ also exist.
Explanation
As long as a moment of some order exists, moments of lower order always exist, but naturally the converse does not hold. Of course, when tackling actual problems, it is rare for a higher-order moment to be given first, yet this is a theorem that can save quite a bit of space when listing the conditions of some theorem.
Proof
Strategy: We detour through the absolute value of the random variable to establish the order relation, and then return to the inequality we originally wanted to show. This proof is for continuous probability distributions, but the case for discrete probability distributions can be proved by the same method.
Let the probability density function of the random variable $X$ be $f$.
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. $E \left( |X|^{m} \right) < \infty \implies E \left( X^{m} \right) < \infty$
$$ \begin{align*} E \left( X^{m} \right) =& \int_{-\infty}^{\infty} x^{m} f(x) dx \\ \le & \left| \int_{-\infty}^{\infty} x^{m} f(x) dx \right| \\ \le & \int_{-\infty}^{\infty} | x | ^{m} f(x) dx \\ =& E \left( |X|^{m} \right) \\ <& \infty \end{align*} $$ Combining Part 1. through Part 3., we obtain $E \left( X^{n} \right) < \infty \implies E \left( X^{m} \right) < \infty$.
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