Random variable with infinite $(k+1)$th moment I was asked to give an example of a random variable which has finite $i$'th moments for $i=1,2,\ldots,k$ and has an unbounded $(k+1)$th moment.
Obviously, a Student-distributed $t_{k+1}$ random variable will work, but I suppose giving a proof link to Wikipedia is not an option, and dealing with density which involves Gamma function is far beyond my current level of maths.


*

*Is there a simpler way to show the desired property for Student distribution? If not,

*Are there any simpler examples of random variables with this desired property?


Thank you very much in advance! 
 A: It is actually not that hard to show that a random variable that follows Student distribution with $k+1$ degrees of freedom satisfies this property. Since the probability density function is proportional to $\left( \frac{k+1}{k+1+x^2} \right)^{1+k/2}$ and since, for $k \geqslant 1$,
$$
 \left| x^r \left( \frac{k+1}{k+1+x^2} \right)^{1+k/2} \right| \geqslant \frac{ \vert x \vert^r }{\left( 1+ x^2 \right)^{1+k/2}}
$$
The integral defining $r$-th moment is only convergent for $1 \leqslant r \leqslant k$:
$$
    \int_{-\infty}^\infty \left\vert x^r \mathcal{N}_k \left( \frac{k+1}{k+1+x^2} \right)^{1+k/2} \right\vert \, \mathrm{d} x \geqslant \mathcal{N}_k \int_{-\infty}^\infty 
   \frac{ \vert x \vert^r }{\left( 1+ x^2 \right)^{1+k/2}} \mathrm{d} x = 2 \mathcal{N}_k \int_{0}^\infty \frac{ x^r }{\left( 1+ x^2 \right)^{1+k/2}} \mathrm{d} x
$$
The latter integral only converges at $+\infty$ if $r \leqslant k$.
A: Consider the discrete random variable with the probability as follows $$P_{\epsilon}(X = n) = \frac{\frac1{n^{k+1+\epsilon}}}{\zeta(k+1+\epsilon)}$$ for any $\epsilon \in (0,1]$.
$$\mathbb{E}(X^m) = \left \{ \begin{array}{cl} \frac{\zeta(k+1+\epsilon - m)}{\zeta(k+1+\epsilon)}  < \infty, &~~ \forall m \in \{1,2,\ldots,k\}\\
\infty, &~~ \forall m>k \end{array}\right.$$
A: Hint:  Let $X$ have density function $f(x)=\dfrac{c}{x^{k+2}}$ for $x\ge 1$, and $f(x)=0$ elsewhere, where $c$ is chosen to make the integral equal to $1$. 
You can also produce a discrete analogue of the above construction. For $n=1, 2, 3, \dots$, let $P(X=n)=\dfrac{c}{n^{k+2}}$ for a suitable constant $c$.  
A: Take $X_n$ defined over $[1,\infty)$ such that 
$$
\mathbb P(X \le x) = \frac{\int_1^x \frac 1{x^{n+2}} \, dx}{C}, 
$$
where $C$ is
$$
C = \int_1^{\infty} \frac 1{x^{n+2}} \, dx.
$$
Therefore, 
$$
\mathbb E[X_n^k] = \frac 1C \int_1^{\infty} \frac 1{x^{n+2-k}} \, dx < \infty \qquad \text{ if } k = 1, \dots, n
$$
but diverges when $k = n+1$.
Hope that helps,
