# Derivatives of Gaussian measure with respect to (co)variance

For $$a > 0$$, let $$G_{a}$$ denote a centered Gaussian random variable with variance $$a$$. That is, the density of $$G_a$$ is $$\frac{e^{\frac{-x^2}{2a}}}{\sqrt{2\pi a}}.$$

Fix $$t > 0$$ and define a function $$f:\mathbb{R}^+\to \mathbb{R}$$ by $$f(a) = \mathbb{P}(G_a\geq t).$$ It is straightforward to verify $$f(a) = \mathbb{P}\left(G_1\geq \frac{t}{\sqrt{a}}\right) = 1 - \Phi\left(\frac{t}{\sqrt{a}}\right),$$ where $$\phi$$ is the standard Gaussian CDF. Using this we can calculate the derivative of $$f$$ by $$\frac{d}{da}f(a) = \frac{d}{da}\Phi\left(\frac{t}{\sqrt{a}}\right)\cdot\frac{-t}{2a^{3/2}} = -t\frac{e^{\frac{-x^2}{2a}}}{2\sqrt{2\pi}a^{2}}.$$ One may now notice that $$|\frac{d}{da}|$$ is bounded by some constant.

I am now interested in the multivariate case, but the 2-dimensional case seems equally intresting. For $$\Sigma$$ a positive $$2 \times 2$$ positive definite matrix let $$G_\Sigma$$ be the a centered $$2$$ dimensional Gaussian with covariance $$\Sigma$$.

As before we now define the function $$f(\Sigma) = \mathbb{P}\left(G_\Sigma \in [t, \infty) \times [t, \infty)\right).$$

What can be said about $$\nabla f$$ in this case, does it have any nice representation? Can one show that $$||\nabla f||$$ is also bounded?