# inner product of two gaussian random vectors?

Suppose that $x, y\sim N(0,I_n)$ are independent. Consider the inner product $\langle x, y\rangle$. Intuitively, $y$ behaves like a random vector of length $\sqrt n$, so $\langle x, y\rangle$ is close to $\langle x, z\rangle$, where $z$ is uniform on $\sqrt{n}S^{n-1}$. Now, $\langle x,z\rangle \sim N(0,\|z\|^2)$, that is, $\langle x,z\rangle \sim N(0, n)$. Therefore $\langle x,y\rangle$ is close to an $N(0,n)$ variable.

I want to show that the density function of $\langle x,y\rangle$ is close to the density function of an $N(0,n)$ variable, without resorting to the explicit formula of the density function of $\langle x,y\rangle$ (which can be obtained by characteristic function/Fourier transform and involves Bessel function).

Basically I'd like to formalize the intuition above. Write $\langle x,y\rangle = Z + W$, where $Z\sim N(0,n)$ and $W$ is small most of the time but not independent with $Z$. This gives that the cumulative density function of $\langle x,y\rangle$ and that of $Z$ are close, but this is not enough to argue that the p.d.f's are close. Is it possible to argue that following this outline?

• Something which might simplify the calculations slightly (apologies if you already know this): by the rotational invariance of the standard multivariate Gaussian distribution, you can WLOG assume (since you only care about distribution) that Y=(R, 0, ... , 0) where R is the square root of a chi-squared on n degrees of freedom. Then the random variable you care about has the same distribution as RZ where Z is a standard Gaussian independent of R. – Yemon Choi Jun 16 '13 at 2:09
• The only situation where the cumulative density functions are close but the probability density functions are far is if at least one of the probability density functions oscillates a lot. But since both probability density functions are clearly monotonic on the positive and negative numbers, they cannot oscillate very much. – Will Sawin Jun 16 '13 at 3:52
• Will Sawin, I thought about this. But even within a small band, a function could be relatively flat for a range, then increase sharply, then become flat again... – user14432 Jun 16 '13 at 4:40

## 1 Answer

$S_n:=\langle x,y\rangle =\sum_{i=1}^nx_iy_i$ is the $n$th partial sum of a sequence of standardized and i.i.d. random variables, and the characteristic function of each summand, namely the function $t$ $\mapsto$ $$\int \mathrm{e}^{-t^2x^2/2} \frac{\mathrm{e}^{-x^2/2}}{\sqrt{2\pi}}\,\mathrm{d}x = \frac{1}{\sqrt{1+t^2}}$$
is square integrable. So a standard form of the CLT for densities, see Feller II (1971, p. 516, Theorem 2 and the remark following it), shows the uniform convergence of the density of $S_n/\sqrt{n}$ to the standard normal density.

• I'd like to know something like $g(x) = f(x)(1\pm 1/\sqrt{n})$, basically the density function of $g(x)$ is $f(x)(1+o(1))$, where the $o(1)$ is an inverse polynomial in $n$. Is this possible? – user14432 Jun 16 '13 at 17:09
• The next chapter in Feller II gives you an an asymptotic expansion for densities as $n\rightarrow\infty$, in your case in particular the Gaussian density plus $O(1/n)$ uniformly in $x$ (as the third moment occuring in the $1/\sqrt{n}$ term is zero), but you can't even get the factor $(1+ o(1))$ uniformly for all $x$. – Lutz Mattner Jun 16 '13 at 17:47