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In my lecture slides there was an optimization problem involving a random variable $w$, that we can call "wage". Part of the maximization problem was taking the derivative of the expected value of $E(w)$ which it doesn't explain how the answer was derived but it said was the p.d.f of the random variable $w$ as such:

\begin{eqnarray*} \frac{dE(w)}{dw} & = \frac{1}{dw}& \int_\bar{w}^\infty wdF(w) =\frac{1}{dw}\int_\bar{w}^\infty wf(w)=f(w) \end{eqnarray*}

Is this true??

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2 Answers 2

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No.   Not at all.

$\mathsf E(w)$ would be a constant, and the derivative of a constant is zero.

Further $\displaystyle \mathsf E(w) = \int_{-\infty}^\infty \psi\operatorname d F(\psi)$, where $\psi$ is the variable of integration - a token whose scope is bound within the integral.   You cannot take the derivative of this token from outside the integral.

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No.

$$\frac{dE(w)}{dw} = \frac{d}{dw} \int_{-\infty}^\infty wdF(w) = \int_{-\infty}^\infty \frac{d}{dw} wdF(w)$$

How would you evaluate, even if w had a pdf,

$$\frac{d}{dw} wdF(w)$$

?

Actually,

$$\frac{dE(w)}{dw} = \frac{d}{dw} \int_{-\infty}^\infty wdF(w) = \frac{d}{dw} (a) = 0$$

where $a$ is the real (or complex or extended real) number s.t.

$$\int_{-\infty}^\infty wdF(w) = a$$


Your thinking might apply to situations like:

  1. mgf

$$M_{X}'(w) = \frac{d}{dw} E[e^{Xw}] = E[\frac{d}{dw} e^{Xw}] = E[Xe^{Xw}]$$

$$= \int_{\mathbb R} xe^{xw} f_{X}(x) dx$$

  1. $$\frac{d}{dw} E[wX] = E[\frac{d}{dw} wX] = E[X]$$

  2. $$\frac{d}{dw} F_X(w) = \frac{d}{dw} \int_{-\infty}^{w} dF_X(t)$$

$$= \frac{dF_X(w)}{dw}$$

If X has a pdf:

$$= \frac{d}{dw} \int_{-\infty}^{w} f_X(t) dt$$

$$= f_X(w)$$

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