Counter-intuitive: Mean of absolute value of normalized Gaussian Vector

Question

Here's a simple matlab code:

    n = 10; % Any natural number  
    X = randn(n,1);
    X = X/norm(X);
    mean(abs(X))

This always comes out to be empirically around $$\frac{1}{1.25 \sqrt{n}} $$ But I intuitively feel that it should be $\frac{1}{\sqrt{n}}$. Because the entries are independent and the norm is one. Please explain how the 1.25 factor comes. Also, this is independent of the variance of the distribution.

Answer 1

This multiplicative factor is not exactly $\dfrac1{1.25}=0.8$, even taking its expected value rather than a particular simulation

• for small $n$, the factor is larger as the norm is substantially affected by the particular sample values: for example with $n=1$ it is $1$ as the norm has exactly the same magnitude as the original data, while for $n=2$ its expected value is (empirically) closer to $0.9$

• for large $n$ the expected factor approaches $\sqrt{\dfrac2\pi} \approx 0.79788$ when $n$ increases, as the mean of the absolute value of a standard normal random variable (sometimes called the half-normal distribution)