Norm of random vector plus constant

Suppose that $w$ is a multivariate standard normal vector and $c$ a real vector of the same size. I know that for positive x $$P(||w+c||^2\geq x)\ \geq \ P(||w||^2\geq x)$$ but I cannot prove it. We use the euclidean norm , in dimension 2 or 3 drawings show that the inequality above holds, but I need a proof.

• can you not just compute the integrals? you know the density for the standard normal vector... Commented Nov 3, 2013 at 19:07
• it is not that easy Commented Nov 3, 2013 at 20:49

Very rough sketch for a general case: Perhaps the equivalent $P(\|w+c\|^2\leq x)\ \leq \ P(\|w\|^2\leq x)$ is more intuitive to deal with. For both probabilities, one evaluates the probability that $w$ is in a subset of $R^d$ of $d$-dimensional Lebesgue measure (let us call this measure $\mu$) $m$, where $m$ is the volume of a disk of radius $\sqrt{x}$, i.e. $m = \int_{\|w\|^2 \leq x} dw$. Consider the problem, $$\sup_{\substack{\mathcal{A}\subset\mathbb{R}^d \\ \mu(\mathcal{A}) \leq m}}\int_{\mathcal{A}}f(w)dw,$$ where $f(\cdot)$ is the PDF of our Gaussian. The solution is given by starting with the points of $\mathbb{R}^d$ where $f(w)$ takes its maximum values, and keep adding points until $\mu(\mathcal{A}) = m$. But since $f(w) = \mathtt{const}\exp(-\mathtt{const}\|w\|^2)$, the solution is of the form $\mathcal{A} = \{w\in\mathbb{R}^d:\|w\|^2 \leq y\}$ for some $y$ that satisfies $\mu(\mathcal{A}) = m$. We immediately have $y = x$, and therefore, $$P(w\in\mathcal{A}) \leq P(\|w\|^2 \leq x)$$ for any $\mathcal{A}$ with $\mu(\mathcal{A}) = \mu(w\in\mathbb{R}^d:\|w\|^2 \leq x)$. In particular, $P(\|w+c\|^2\leq x)\ \leq \ P(\|w\|^2\leq x)$.