# Tail inequalities for multivariate normal distribution

There exists an closed expression for univariate normal CDF, together with simpler upper-bounds under the form, $$\Pr\big[X > c\big] \leq \frac{1}{2}\exp\Big(\frac{-c^2}{2}\Big)~,$$ $$\text{where } X \sim \mathcal{N}(0,1)~.$$ Even if there are some algorithms to compute the CDF for multivariate normal distribution, there is no analytical formula for multivariate CDF (as mentioned in this thread). Though, is there some equivalent upper bounds for multivariate normal distribution, $$\Pr\big[X_1>C_1 \land \dots \land X_k>C_k\big] \leq ~?$$ $$\text{where } (X_1, \dots, X_n) \sim \mathcal{N}(\mu, \Sigma)~.$$

• The expression of the univariate normal standard CDF in terms of the error function is not an analytical formula, because the error function is itself an integral. Commented Oct 16, 2013 at 12:19
• Even if the error function can not be expressed in terms of elementary functions, the properties of this integral are sufficient to derive upper bounds as stated in my question. Commented Oct 16, 2013 at 13:25
• Certainly, but the statement "there exists an analytical formula for univariate normal CDF" is an incorrect statement. Commented Oct 16, 2013 at 13:30
• Oh I see, I've just edited my statement which is "a bit" more correct now :) Commented Oct 16, 2013 at 13:36

Let $M=\Sigma^{-1}$ with $M = (m_{ij})$. Assuming that for all $1\leq i\leq d$, we have $\Delta_i := \sum_{j=1}^d C_j m_{ji} > 0$, then, $$F(C,\Sigma) = \frac{ |M|^{\frac 1 2}}{(2\pi)^{\frac d 2}} \int_0^\infty\cdots\int_0^\infty \exp\Big[-\frac 1 2 (X+C)^\top M(X+C) \Big] dX_1\dots dX_d$$ $$< \Big(\prod_{i=1}^{d} \Delta_i\Big)^{-1} \frac{ |M|^{\frac 1 2}}{(2\pi)^{\frac d 2}} \exp \Big[ - \frac 1 2 C^\top M C \Big]~.$$