# expected value of the dot product of normalized random vector and its mean

Suppose $U$ is a random vector satisfying $\mathbb E[U] = \mu$ and $\mathrm{var}(\|U\|_2) \le V$. Let $\bar{U} = U / \|U\|_2$ and $\bar\mu = \mu / \|\mu\|$. What is a lower bound on $\mathbb E[\bar U^\intercal\bar\mu]$ in terms of $V$?

(For what it's worth, I have in mind a setting where $V$ is quite small compared to $\|\mu\|_2$---seems like there should be some non-trivial bound in this case.)

The best possible general bound is the trivial one, $-1$. (This deserves to be called "trivial" because $\bar U^\prime \bar \mu \ge -1$ in any event.) To demonstrate this, I offer examples whose bounds come arbitrarily close to $-1$.

The intuition is this: the mean of the normalized vector $\bar U$ can be the opposite of the mean of $U$ itself because the normalization can greatly expand values close to zero. If we pick a vector $U$ that is far from zero in one direction a small fraction of the time and otherwise is very close to zero but in the opposite direction the rest of the time, then (a) the mean of $U$ can lie in this privileged direction but (b) the mean of $\bar U$ can be arbitrarily close to a unit vector in the opposite direction. This will cause $\bar U^\prime \bar \mu$ generally to be close to $-1$.

This construction works in any number of dimensions. I will provide an explicit example in one dimension.

Let $0\lt\epsilon \ll 1/3$. This makes $p=2\epsilon/(1+\epsilon)$ a probability because $0 \lt p \lt 1$. Let $U$ take on the value $1$ with probability $p$ and the value $-\epsilon$ with probability $1-p$. Straightforward computations yield

$$\mu=\mathbb{E}(U) = \epsilon \gt 0,$$

$$\bar\mu = 1,$$

and

$$V = \operatorname{Var}(||U||_2) = \frac{2\epsilon(1-3\epsilon+3\epsilon^2-\epsilon^3)}{(1+\epsilon)^2} \approx 2\mu.$$

The random variable $\bar U$ takes on the value $1$ with probability $p$ and $-1$ with probability $1-p$. Therefore

$$\mathbb{E}(\bar U^\prime \bar\mu)=\mathbb{E}(\bar U)=2p-1 = -1 + \frac{4\epsilon}{1+\epsilon} \lt 0.$$

This value remains unchanged even when we scale $U$ by $\sigma \gt 0$. That scaling changes $\mu$ to $\mu\sigma$ and $V$ to $V\sigma^2$, so by taking $\sigma^2$ small we may make $V\sigma^2$ as small as we like compared to $\mu\sigma$. Moreover, the greatest lower bound of $\mathbb{\bar U^\prime \bar \mu}$ is $-1$, QED.

• Thank you for the thoughtful response! What if I added the condition that $V \ll \|\mu\|_2^2$ ? Is there a non-trivial bound then? – Jeff Jan 22 '17 at 20:27
• Yes, that inequality makes more sense. When it holds, Chebyshev's Inequality implies there is a high chance that $U$ is close to $\mu$ and therefore $\bar U$ has to be nearly parallel to $\bar \mu$. – whuber Jan 22 '17 at 21:18
• Whew! That's a relief. I'll use Chebyshev's Inequality then. Thanks for your help. – Jeff Jan 22 '17 at 21:36