# expectation of an exponential function over a unit sphere

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 $n\to\infty$, $\alpha_n\to \infty$

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.
• 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.) May 3, 2017 at 19:45