Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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
add comment

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.