Derivative of an integral with change of variable

Question

I want to calculate the following:

$$ \dfrac{\partial}{\partial x} \int_{\tilde{\tau}=t}^{\infty} \tilde{\tau}\dfrac{1}{\sqrt{2 \pi}\sigma}e^{-\frac{(\tilde{\tau}-x)^2}{2\sigma^2}} d \tilde{\tau} $$

where $t$ is a constant and $\tilde{\tau}$ is a random variable with mean $x$ and variance $\sigma^2$. I have that $\tilde{\tau} = x + \sigma \varepsilon$, so I use change of variables and redefine the integral with $u = \dfrac{\tilde{\tau}-x}{\sigma}$

$$ \dfrac{\partial}{\partial x} \int_{\tilde{\tau}=t}^{\infty} (u \sigma + x)\dfrac{1}{\sqrt{2 \pi}}e^{-\frac{u^2}{2}} d u $$

I break the derivative of the integral into two parts:

$$ \dfrac{\partial}{\partial x} \left( \sigma \int_{u \sigma + x=t}^{\infty}u\dfrac{1}{\sqrt{2 \pi}}e^{-\frac{u^2}{2}} d u + x\int_{u \sigma + x=t}^{\infty} \dfrac{1}{\sqrt{2 \pi}}e^{-\frac{u^2}{2}} d u \right)$$

$$ \dfrac{\partial}{\partial x} \sigma \int_{u \sigma + x=t}^{\infty}u\dfrac{1}{\sqrt{2 \pi}}e^{-\frac{u^2}{2}} d u + \dfrac{\partial}{\partial x} x(1-F(t)) $$

$$ \sigma \dfrac{\partial}{\partial x} \int_{u \sigma + x=t}^{\infty}u\dfrac{1}{\sqrt{2 \pi}}e^{-\frac{u^2}{2}} d u + 1-F(t) $$

According to a colleague the first term will vanish. The reason is that apparently after doing the substitution of variable I now have $u$ as a dummy variable and that taking the derivative with respect to $x$ will be zero. But I defined $u=\dfrac{\tilde{\tau}-x}{\sigma}$. How can the derivative with respect to x be zero?

Answer 1

If your bounds were constant your coworker would be correct (the integral is essentially $\int_a^b f(u)\mathrm{d}u$, which won't have dependence on $x$ for constant $a,b$.). They aren't constant though. We have the lower bound is $u\sigma+x = t$ for $t,\sigma$ constant. So, we can rewrite this as $u = \frac{t-x}{\sigma}$. So, your lower bound is a function of $x$, so the derivative isn't necessarily $0$.

As an easy example of how variable bounds can lead to nonzero derivative, consider: $$F(x) = \int_x^\infty e^{-t}\mathrm{d}t = e^{-x}$$ We get that $\frac{\partial}{\partial x}F(x) = -e^{-x}\neq 0$, with reasoning similar to why your derivative is likely nonzero.

Answer 2

$$S={\sigma\over\sqrt{2\pi}}\int_{t-x\over \sigma}^\infty ue^{-{u^2\over 2}}du={\sigma\over\sqrt{2\pi}}\int_{\left({t-x\over\sigma}\right)^2}^\infty e^{-{y\over2}}d\left({y\over 2}\right)\\={\sigma\over\sqrt{2\pi}}\left[e^{-{y\over 2}}\right]_\infty^{\left({t-x\over\sigma}\right)^2}={\sigma\over\sqrt{2\pi}}\left[e^{-{{(t-x)^2\over2\sigma^2}}}\right]$$

So $${\delta S\over \delta x}\neq 0$$.

Answer 3

If we use the Leibniz rule for integration we find,

$\frac{\partial}{\partial x} \int_{a(u(x))}^{\infty}f(u(x))du=\big[\int_{a(u(x))}^{\infty} \frac{\partial}{\partial x}(f(u(x))du\big]-f(a(u(x))\cdot\frac{d}{dx}(a(u(x))$

The general form of the Leibniz rule for integration is,

$\frac{d}{d t} \int_{a(t)}^{b(t)}f(x,t)dx=\big[\int_{a(t)}^{b(t)} \frac{\partial}{\partial t}(f(x,t)dx\big]+f(b(t))\cdot \frac{db(t)}{dt}-f(a(t)\cdot\frac{da(t)}{dt}$

I believe the first equation I wrote is correct. This is just application of the chain rule to integral limits which depend on the variable of integration. You should be able to use this to find out whether the integral is zero or not. But as another commenter said, in general it isn't.