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If

$$E[f(x)]=0$$

can we derive that

$$E[f'(x)]=0?$$

For example $f(x)$ means some noise with zero mean, gaussian distribution.

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    $\begingroup$ One wonders just what $f'(x)$ even means in the example. The usual sense of "noise" is an uncorrelated process, but that's almost surely not differentiable. $\endgroup$
    – whuber
    Jun 10, 2011 at 13:39

2 Answers 2

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With your definitions no. Suppose we have a random variable $X$, what you are asking if it is possible to derive

$$Ef'(X)=0$$

from

$$Ef(X)=0.$$

Take $f(x)=x$. Then $Ef(X)=EX=0$ and this means that variable $X$ has zero mean. Now $f'(x)=1$, and

$$Ef'(X)=E[1]=1,$$

hence the original statement does not hold for all functions $f$.

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Agree with Mpiktas ..

way 1 to think it:

In generalized way is that $E(f(x)) = \int f(x)p(x) = \int(f(x)p(x)dx)$ while $E(f'(x) ) = \int ( \frac{df(x)}{dx}p(x)dx)$..

Thinking it mathematically also a $d/dx$ operator comes inside the integral to cancel some part of the integral effect. It makes sense then to think that they are not equal.

way 2:

if integral is zero then $f(1)p(1)+ f(2)p(2)+ ... = 0$ meaning that the function is rising-falling..The slope of that function will then not rise fall the same way.

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