Derivative of $\int_{-\infty}^{x}(z-x)f(z)\, dz$ with respect to $x$

by Richard Hardy   Last Updated September 11, 2019 09:20 AM

I am trying to apply the Leibnitz integral rule to the following problem: $$ \frac{d}{dx}\int_{-\infty}^{x}(z-x)f(z)\, dz $$ where $f(z)$ is the density function of random variable $Z$ (in contrast to the reference above, I use $z$ in place of $t$). I get \begin{aligned} \frac{d}{dx}\int_{-\infty}^{x}(z-x)f(z)\, dz &= (x-x)f(x)\cdot 1-(-\infty-x)\cdot 0+\int_{-\infty}^{x}(-f(z))\, dz \\ &= 0-\color{red}{(-\infty-x)\cdot 0}-F(x) \end{aligned} where $F(\cdot)$ is the cumulative density function of random variable $Z$.


  1. Is the derivation correct?
  2. How do I deal with the expression in red?

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