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$x \in R$ is a continuous random variable.

Is the statement : IF $\int_{-\infty}^{\infty} x^3 f(x) dx < \infty$ then: $Cov(X,X)<\infty$ .TRUE?

My thought was that Var(x)=Cov(x,x) , so $Var(x)= E(x^2) - E^2(x)$. Hence

both $\int_{-\infty}^{\infty} x^2 f(x) dx < \infty$ and $\int_{-\infty}^{\infty} x f(x) dx < \infty$

so I think the Question can be written as:

Is it true that IF $\int_{-\infty}^{\infty} x^3 f(x) dx < \infty$ then both $\int_{-\infty}^{\infty} x^2 f(x) dx < \infty$ and $\int_{-\infty}^{\infty} x f(x) dx < \infty$?.

It looks False to me. But I am not sure how to prove it.

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    $\begingroup$ IF $\int_{-\infty}^{\infty} x^2 f(x) dx = infty$ does that mean that $\int_{-\infty}^{\infty} x f(x) dx = \infty$?. $\endgroup$
    – Mia
    Sep 14 at 11:45
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    $\begingroup$ Hint: where does $|x|³>x²$ occur? Remark: the assumption should be $\int |x|^3 f(x)\,\text dx<\infty$ $\endgroup$
    – Xi'an
    Sep 14 at 11:56
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    $\begingroup$ stats.stackexchange.com/questions/244202 and stats.stackexchange.com/questions/251431 show several techniques to address this. $\endgroup$
    – whuber
    Sep 14 at 15:17
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Is it true that IF $\int_{-\infty}^{\infty} x^3 f(x) dx < \infty$ then both $\int_{-\infty}^{\infty} x^2 f(x) dx < \infty$ and $\int_{-\infty}^{\infty} x f(x) dx < \infty$?.

It looks False to me. But I am not sure how to prove it.

Instead it sound me true. There is a theorem that warrant us that if moment of order $k$ exist then all lower order moments exists too. If third order moment exist, variance must exist.

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