# Derivative of a Parameter Integral (involving a Gaussian density)

I have a question about the derivative of a certain parameter integral. To fix notation, let $\phi_{\mu,\sigma}(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp{\left(-\frac{(x-\mu)^2}{2 \sigma^2}\right)}$ denote the density function of a Gaussian distribution with mean $\mu\in\mathrm{R}$ and standard deviation $\sigma>0$. I am interested in the derivative $$\frac{\partial}{\partial \sigma}\int_{0}^{\infty}\exp{(x)}\phi_{\mu,\sigma}(x)\,\mathrm{d}x.$$ In particular, is this derivative always positive? If so, how can I show that (rigorously)? Any help, suggestions, or the like, are highly welcome. Many thanks in advance!

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There is a general result: Let $(\Omega,\mathcal{A},\mathbb{P})$ a measure space. Let $u:(a,b) \times \Omega \to \mathbb{R}$ such that

1. $x \mapsto u(\sigma,x) \in L^1(\mathbb{P})$ for all $\sigma \in (a,b)$
2. $\sigma \mapsto u(\sigma,x)$ is differentiable for all $x \in \Omega$
3. $\exists w \in L^1(\mathbb{P}) \, \forall (\sigma,x) \in (a,b) \times \Omega: |\partial_\sigma u(\sigma,x)| \leq w(x)$

Then $V(\sigma) := \int u(\sigma,x) \, d\mathbb{P}(x)$ is differentiable, $\partial_\sigma V(\sigma) = \int \partial_\sigma u(\sigma,x) \, d\mathbb{P}(x)$.

So in this case we choose $$u(\sigma,x) := e^x \cdot \varphi_{\mu,\sigma}(x).$$ We have to check the last property (the other ones are clearly fulfilled). By calculating the derivative one can show that

$$|\partial_\sigma u(\sigma,x)| \leq e^x \cdot \exp \left(- \frac{(x-\mu)^2}{a^2} \right) \cdot \left( \frac{1}{\sqrt{2\pi \cdot a^4}} + \frac{2}{\sqrt{2\pi a^2}} \cdot \frac{(x-\mu)^2}{a^3} \right)=:w(x) \in L^1$$

for all $\sigma \in (a,b)$ where $0<a<b$. Thus we can apply the theorem and obtain a formula for the derivative. By using the moments of the Gaussian distribution it is possible to calculate the integral.

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thanks for the help! Following your suggestions, I managed to find the desired result. –  s_2 Nov 17 '12 at 14:47