Let $v=(v_1,\dotsc,v_n)$ be a vector with length in $\mathbb{R}^{n-1}$, uniformly distributed over a unit sphere. I want to show $E[\exp(\alpha_n v_1)] \sim \exp (\alpha_n^2/n)$ as $n->\infty.$ If v was a multivariate Gaussian, the result can be proven using MGF of multivariate Gaussian.

Edit: $\alpha_n$ can be a function of $n$ as well. In particular, suppose as $n\to\infty$, $\alpha_n\to \infty$


1 Answer 1


for large $n$, any finite set $\{\nu_1,\nu_2,\ldots \nu_k\}$ becomes a set of i.i.d. Gaussians with zero mean and variance $1/n$, so $$E\left[e^{\alpha\nu_1}\right]=\sqrt{\frac{n}{2\pi}}\int_{-\infty}^\infty dx\, e^{\alpha x}e^{-nx^2/2}=e^{\alpha^2/2n}.$$ (there is a factor of two difference in the exponent with respect to the OP)

to see the Gaussian limit, start from $$ P(\nu_1,\nu_2,\ldots \nu_n)\propto\delta\left(1-\sum_{i=1}^{n}\nu_i^{2}\right). $$ Integrate out $n-k$ elements, to arrive at the distribution $$P(\nu_1,\nu_2,\ldots \nu_k)\propto\left(1-\sum_{i=1}^{k}\nu_i^{2}\right)^{(n-k-2)/2} \theta\left(1-\sum_{i=1}^{k}\nu_i^{2}\right),$$ with $\theta(x)$ the unit step function. In the large-$n$ limit at fixed $k$ this tends to $$P(\nu_1,\nu_2,\ldots \nu_k)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^k \nu_i^2\right)$$

The new version of the question now asks for an $n$-dependent $\alpha$; then one will just have to work with the exact expression, $$P(\nu_1)=C_n\left(1-\nu_1^{2}\right)^{(n-3)/2} \theta\left(1-\nu_1^{2}\right),\;\;C_n=\frac{2 \Gamma \left(\frac{n}{2}\right)}{\sqrt{\pi } \,\Gamma \left(\frac{n-1}{2}\right)}$$ and hence $$E\left[e^{\alpha_n\nu_1}\right]=2^{\frac{n}{2}-1} \alpha_n^{1-\frac{n}{2}} \Gamma \left(\frac{n}{2}\right) \left(L_{\frac{n}{2}-1}(\alpha_n)+I_{\frac{n}{2}-1}(\alpha_n)\right)$$ with $I$ a Bessel function and $L$ a Struve function.

  • $\begingroup$ Thanks Carlo. For a fixed alpha, I agree. But what if \alpha is a function of n? As n->\infinity, \alpha_n ->\infinity as well. $\endgroup$ May 3, 2017 at 16:11
  • $\begingroup$ OK, so you've changed the question, I'll see if I can answer that as well (but it will be necessarily much more messy, since if $\alpha_n$ grows in an unspecified way with $n$ all subdominant terms can contribute.) $\endgroup$ May 3, 2017 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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