independence of x and pdf

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This is probably a naive question but anyways.. If I have an iid x data, can I claim that

$$E(xf(x))=E(x)E(f(x)$$

where f(x) is an estimate of the pdf of x? Since x is random and its pdf some distribution with a potential functional form, can I assume they are formally independent?

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There is a straight-forward way to find out. Consider:

$$ E(X \cdot f_X(x)) = \int_\mathbf{R} x f_X(x) f_X(x) \text{d}x = \int_\mathbf{R} x (f_X(x))^2 \text{d}x $$

and

$$ E(X)E(f_X(x)) = \left( \int_\mathbf{R} x f_X(x) \text{d}x\right) \left( \int_\mathbf{R} (f_X(x))^2 \text{d}x\right) $$

Prove or disprove that the are equal. I would suggest using integration by parts on $E(X\cdot f_X(x))$.