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?


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