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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.

  1. Is there a simpler way to show the desired property for Student distribution? If not,
  2. Are there any simpler examples of random variables with this desired property?

Thank you very much in advance!

Michael Hardy
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wh1t3cat1k
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4 Answers4

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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 $r \leqslant k$.

Sasha
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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 $\epsilon \in (0,1]$.

$$\mathbb{E}(X^m) = \left \{ \begin{array}{cl} \frac{\zeta(k+1+\epsilon - m)}{\zeta(k+1+\epsilon)} < \infty, &~~ \forall m \in \{1,2,\ldots,k\}\\ \infty, &~~ \forall m>k \end{array}\right.$$

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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 $c$.

André Nicolas
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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,

Patrick Da Silva
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