$\int_{-\infty}^{\infty} x^3 f(x) dx < \infty$ then $Cov(X,X)<\infty$ ? TRUE OR FALSE $x  \in  R$ 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.


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.
