Is it true that if the $k$-th moment exists, then the $j$-th moment exists for all $0 \leq j \leq k$?

Question

Assume $X$ is a random variable. Is it true that if $E(X^k)$ exists, then the $E(X^j)$ exists for all $0 \leq j \leq k$?

Answer 1

Yes this is true, and follows easily from one of two choices of classical inequality.

Jensen's Inequality

Note that for $s \geq r$, the function $\varphi(x) = x^{s/r}$ is convex. Therefore using $X^r$ inside Jensen's inequality with $\varphi$ we have

$$\mathbf E[ X^s] = \mathbf E[ \varphi(X^r)] \geq \varphi \left( \mathbf E[ X^r] \right)= \mathbf E[X^r]^{s/r},$$ and hence if $\mathbf E[X^s] < \infty$ then we have $\mathbf E[X^r] < \infty$.

Hölder's Inequality

Consider Holder's inequality in the case of two random variables $X^r$ and the other $\mathbf 1$ the identity variable which is equal to $1$ almost surely. Let $p = \frac{s}{r}$ then

$$\mathbf E[|X|^r] = \mathbf E[ |X|^r \mathbf 1] \leq \mathbf E[ |X|^{r(s/r)}]^{(r/s)} \mathbf E[ \mathbf 1 ]^{s/(s-r)} = \mathbf E[|X|^s]^{s/r}, $$ hence if $s > r$ and $\mathbf E[|X|^s] < \infty$ it follows that $\mathbf E[|X|^r] < \infty$.

Note that $\mathbf E[|X|^r] < \infty$ then implies $\mathbf E[X^r] < \infty$.