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?

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.