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Let $X$ be a non-negative random variable with all finite moments. Does it hold that $$\lim_{k \to \infty} \frac{\mathbb{E}[X^k 1_{ \{X \ge k \}} ]}{\mathbb{E}[X^k]} =0$$

What I Tried This clearly holds for bounded random variables, and variables whose distributions have fast decaying tails. However, I cannot find a general argument.

Thành Nguyễn
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Boby
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2 Answers2

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We prove that OP's claim does not hold in general. Let $X$ have the density function of the form

$$ f(x) = \frac{1}{\sqrt{\pi/2} \, x} e^{-\frac{1}{2}\log^2 x} \mathbf{1}[x \geq 1]. $$

Then for any $k$,

\begin{align*} \mathbb{E}[X^k] &= \frac{1}{\sqrt{\pi/2}} \int_{1}^{\infty} x^{k-1} e^{-\frac{1}{2}\log^2 x} \, \mathrm{d}x \\ &= \frac{1}{\sqrt{\pi/2}} \int_{0}^{\infty} e^{-\frac{u^2}{2} + ku} \, \mathrm{d}u \\ &= \frac{e^{k^2/2}}{\sqrt{\pi/2}} \int_{-k}^{\infty} e^{-u^2/2} \, \mathrm{d}u. \end{align*}

This shows that $f$ is indeed a valid density function such that $\mathbb{E}[X^k]$ is finite for all $k$. Also, by a similar computation, for $k \geq 1$,

\begin{align*} \mathbb{E}[X^k \mathbf{1}_{\{X \geq k\}}] &= \frac{1}{\sqrt{\pi/2}} \int_{k}^{\infty} x^{k-1} e^{-\frac{1}{2}\log^2 x} \, \mathrm{d}x \\ &= \frac{1}{\sqrt{\pi/2}} \int_{\log k}^{\infty} e^{-\frac{u^2}{2} + ku} \, \mathrm{d}u \\ &= \frac{e^{k^2/2}}{\sqrt{\pi/2}} \int_{\log k - k}^{\infty} e^{-u^2/2} \, \mathrm{d}u. \end{align*}

Finally, if we define $G(x) = \int_{-x}^{\infty} e^{-u^2/2} \, \mathrm{d}u$, then $G(\infty) = \sqrt{2\pi}$, and hence

$$ \lim_{k\to\infty} \frac{\mathbb{E}[X^k \mathbf{1}_{\{X \geq k\}}]}{\mathbb{E}[X^k]} = \lim_{k\to\infty} \frac{G(k - \log k)}{G(k)} = 1. $$

Here is a more natural example:

Exercise. When $X \sim \operatorname{Exponential}(1)$, show that

$$ \lim_{k\to\infty} \frac{\mathbb{E}[X^k \mathbf{1}_{\{X \geq k\}}]}{\mathbb{E}[X^k]} = \frac{1}{2}. $$

Hint. Show that the ratio can be written as $\mathbb{P}(N_k \leq k)$, where $N_t$ is the Poisson process with unit rate, and then invoke the central limit theorem.

Sangchul Lee
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This is currently not an answer but tracks

We can write it :

$$ E[X^k]=E[X^k1_{X\geq k}] + E[X^k1_{X\leq k}]$$

Then :

$$ 1=\dfrac{E[X^k1_{X\geq k}]}{E[X^k]} + \dfrac{E[X^k1_{X\leq k}]}{E[X^k]}$$

Now $X$ being non-negative and by increasing caracter of expectancy:

$$ E[X^k1_{X\leq k}]=\int X^k1_{X\leq k}dP$$

By transfer theorem on $\mathbb{R}^+$ :

$$ E[X^k1_{X\leq k}]=\int_{0}^{k}x^kf_X(x)dx$$

The question is now :

Does the following equivalence hold ? :

$$\int_{0}^{k}x^kf_X(x)dx \underset{k \to \infty}{\sim} E[X^k] $$

Under the conditions :

$$ \int_{\mathbb{R}} f_X(x)dx =1$$ $$ \forall k\geq 1, \int_{\mathbb{R}}x^kf_X(x)\triangleq M_k<+ \infty$$

What we have :

  • By the convergence of all moments :

$$ \forall k\geq 1, f_X(x)=\underset{x \to \infty}{o}(\dfrac{1}{x^k}) $$

(which is by the way powerful)

  • We suppose $X$ non-bounded (otherwise as you said it is easy)

  • We have because of moments to be finite, for all $k$ :

$$ x^kf_X(x) \sim_x x^k1_{x\leq k}f_X(x)$$

$$ \int_y^{\infty}x^kf_X(x) \sim_y\int_y^{\infty} x^k1_{x\leq k}f_X(x)dx$$

Hope it gives ideas.

EDX
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