I want to calculate the derivative of

$$\varphi(\mu) = \int_{-\infty}^{\mu} r(x-\mu) f(x)dx,$$ wrt to $\mu$, where $r$ is a function and $f$ is a density function. How can I account for the presence of $\mu$ in the integration limit and in the function inside the expectation?


Consider any differentiable function $f:\mathbb{R}^2\to \mathbb{R}.$ It is a theorem that the derivative of $f$, written $Df,$ is given by its partial derivatives,

$$Df(x,y) = \left(\frac{\partial f}{\partial x}(x,y), \frac{\partial f}{\partial y}(x,y)\right).$$

Let $\iota:\mathbb{R}\to\mathbb{R}^2$ be given by


and note that this is differentiable with

$$D\iota(x) = (1,1).$$

To differentiate the expression $f(x,x),$ compose $f$ with $\iota$ and apply the Chain Rule thus:

$$Df(x,x) = D(f\circ \iota)(x) = Df(\iota(x))\circ D\iota(x) = \frac{\partial f}{\partial x}(x,x) + \frac{\partial f}{\partial y}(x,x).$$

To obtain the answer, apply this result to the function

$$\varphi(\mu,\nu) = \int_{-\infty}^{\mu} r(x-\nu) f(x)dx.$$

The derivative with respect to $\mu$ is (of course) obtained with the Fundamental Theorem of Calculus. The derivative with respect to $\nu$ can be computed by differentiating under the integral sign provided $r$ is continuously differentiable.

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