# Thread: On Expected Value

1. ## On Expected Value

Hi everyone, is the following statement true, if so, why is it true?

dE[f(x)]/dx=E[df(x)/dx]?

Thanks

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After some thought, I come up with a "heuristic" proof as the following, any idea for a rigorous proof?

dE[f(x)]/dx={E[f(x+dx)]-E[f(x)]}/dx={E[f(x)+dx*df(x)/dx]-E[f(x)]}/dx={E[f(x)]+E[dx*df(x)/dx]-E[f(x)]}/dx=dx*E[df(x)/dx]/dx=E[df(x)/dx]

2. Hello,

dE[f(x)]/dx={E[f(x+dx)]-E[f(x)]}/dx
this relationship is false, it's an equality when dx goes to 0.

And how does df/dx appear in the third member (after the second equality) ?

And... where does this relationship come from ?
Because E(f(x)) is a constant, so its derivative is 0. And E(df(x)/dx) is not necessarily 0.
Also, I think taking x instead of X, a random variable, is inconsistent, as it makes the problem "trivial" or "weird" lol

3. Originally Posted by zling110
Hi everyone, is the following statement true, if so, why is it true?

dE[f(x)]/dx=E[df(x)/dx]?
I guess you mean $\displaystyle \frac{d}{dx}E[f(X,x)]=E[\frac{\partial}{\partial x}f(X,x)]$, where $\displaystyle f:\mathbb{R}^2\to\mathbb{R}$ is a fixed function and $\displaystyle X$ is a random variable. Don't you?

If so, the result is not always true, but one can give sufficient condition for that. For instance: this is true if ($\displaystyle x\mapsto f(t,x)$ is differentiable for all $\displaystyle t$, and)
- $\displaystyle X$ takes only finitely many different values (then the expectation is a finite sum)
- there exists a random variable $\displaystyle Y\geq 0$ such that $\displaystyle E[Y]<\infty$ and, for all $\displaystyle x$, $\displaystyle \left|\frac{\partial}{\partial x}f(X,x)\right|\leq Y$ almost surely. (this is the theorem of derivation under the integral symbol)