# On Expected Value

• Sep 16th 2009, 01:40 PM
zling110
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]
• Sep 18th 2009, 08:05 AM
Moo
Hello,

Quote:

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
• Sep 18th 2009, 09:14 AM
Laurent
Quote:

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)