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I have found a statement here equation 23 without explanation that $E\frac{X^k}{(\omega X^2+\sigma)^r}<\infty$

for $\omega,\sigma>0, k\in\{0,1,...,2r\}$, $r\in \mathbb{N}$, where we don't know if $EX^r$ exists. I have tried to understand it, obviously $E\frac{X^k}{(\omega X^2+\sigma)^r}<E\frac{X^k}{\omega^r X^{2r}}$, but maybe $\omega$ would be too small, so that the numerator is greater than the denominator, so we can not use 1 to bound it. Also I don't know if $E\frac{1}{X^{2r-k}}$ exists.

Thanks for any hints!

toki
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  • put the whole problem statement. Your link is behind paywall. What kind of random variable is $X$ ? – dezdichado Nov 28 '23 at 18:11
  • @dezdichado X is a random variable for random coefficient AR(1) model, that is $X_t=(\phi+\psi_t)X_{t-1}+\epsilon_t$ where $\epsilon_t$ and $\psi_t$ has 0 mean and finite variance and $\phi$ a constant. $EX^2_t$ might be infinite. I guese the only useful thing is that $EX^\delta_t<\infty$ for a $\delta>0$ – toki Nov 28 '23 at 18:44
  • what I meant is you need to put all the necessary background in the problem description. Otherwise, your question won't get much interests. I think that roughly speaking the expectation should be finite since it's $\sim X^{k-2r}$ of order and $k<2r.$ – dezdichado Nov 28 '23 at 21:21
  • @dezdichado Sorry because I also don't know what X is. That was already all what I know. I think $X$ can not be constant because of term $\epsilon_t$ – toki Nov 28 '23 at 21:32

1 Answers1

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Let $k\in\{0,\dots,2r\}$. Replacing $X$ by $\lvert X\rvert$ if necessary, we assume that $X$ is non-negative.

Define $Y=\frac{X^k}{(\omega X^2+\sigma)^r}$. One has $$ Y\mathbf{1}_{\{X\leqslant 1\}}\leqslant \frac 1{\sigma^r} $$ hence $Y\mathbf{1}_{\{X\leqslant 1\}}$ is bounded.

Moreover, $$ Y\mathbf{1}_{\{X> 1\}}\leqslant \frac 1{\min\{w^r,\sigma^r\}}\frac{X^k}{(1+X^2)^r}\mathbf{1}_{\{X> 1\}} \leqslant \frac 1{\min\{w^r,\sigma^r\}}\left(\frac{X}{1+X^2}\right)^r\mathbf{1}_{\{X> 1\}}\leqslant \frac 1{\min\{2^rw^r,2^r\sigma^r\}} $$ hence $Y$ is bounded.

Davide Giraudo
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