I believe, by (w'x) you mean the inner product of two vectors, am I right?

Now, remark that if w and u are fixed (not random) vectors of the same norm, then by symmetry  E{abs(w'x)} = E{abs(u'x)}. Taking u = ||w|| (1,0,...,0)  we get for a fixed (not random) vector w the formula E{abs(w'x)}= ||w||E{|x_1|}.

Consider now the case of random vector w. Since x and w are independent, we can consider them defined on the product of two probability spaces T and S, with x being dependent only on t \in T, and w  dependent only on s \in S.

By the Fubini theorem, E{abs(w'x)} (which is a double integral over T \times S) can be computed integrating abs(w(s)'x(t))  first over T with  s \in S fixed, and then integrating the result over S. But, when s is fixed, w(s) is a fixed vector, and we know that \int_T abs(w(s)'x(t)) = E{abs(w(s)'x)} = ||w(s)||E{|x_1|}. Finally, the integration over S gives the desired formula.

Why should E{abs(w'x)} = E{abs(u'x)} be true for fixed vectors of the same norm? Take x=(0, 1)', w=(0,1)', u=(1, 0)'. The LHS is 1 and the RHS is 0. What did I miss?

To my understanding the answer from Vladimir Kadets works, if we apply the following modifications:

Let us only fix w and still consider x to be the random variable having standard normal distribution. Then E{abs(w'x)} = E{abs(u'y)}, where u=Nw and y=Nx for a orthonormal matrix N such that u = ||w|| (1,0,...,0)'. Notice (or lookup) that y has as well standard normal distribution, i.e. the same distribution as x. Therefore we still get E{abs(w'x)}= ||w||E{|y_1|} = ||w||E{|x_1|}.

Since N is not present in the equation, we can still continue as before examining the double integral.

(With these fixes: Nice straight argument, indeed.)

Dear Daniel,

Please, read carefully my answer. I have written that u and v are fixed vectors. About x I did not say this!   x is, of course, a random vector, which has a standard normal distribution (N(0,I_n)). Your further explanation is correct. It is exactly what I meant when I wrote that E{abs(w'x)} = E{abs(u'x)}  "by symmetry".

Best regards, Vladimir

Dear Vladimir, thank you very much!

Dear Daniel, thank you very much!

Hi all respondents,

The validity of the equation depends on the norms applied on the RHS.

For a counterexample for the \ll_1-norm, one may take  n=2,  w_1 = 1 w_2 = 1 (these are independent of anythig:)  If so, then  for  y :=  w' x   =  x_1  +  x_2  the desired expectation is equal to

E|y| = E |x_1 + x_2|  = \sqrt{2} \cdot E(|x_1|)

whereas

E( || w ||_1 ) = 2   and   therefore      E(|y|) \ne E( || w||_1 ) \cdot E(|x_1|)

Moreover, for the  \ell_2-norm one can complete the Vladimir proof by making the observation that for any deterministic coefficients  w  the probability distribution of \sum_j  w_j  x_j  and   ||w||_2 \cdot  x_j  are equal (due to stability of the normal standard distribution of order  2 ).  Continuing with other stable p.d. we can get e.g. the following.

If  w  and  x  are independent, and if all  x-s  are  iid with centered Cauchy p.d., then

E |w'  x| =  E (|| w||_1 ) \cdot E|x_1|  

(since then p.d. of  w' x  and   || w||_1 \cdot  x_1 are equal centered Cauchy distributions).

These are IMPORTANT complementary remarks (possibly interesting for someone:)

1. The  Cauchy p.d. does not possess expectation unless concentrated at a point, so the only true claim is better if formulated as follows:

If  w  and  x  are independent, and if all  x-s  are  iid with centered  Cauchy p.d., then the p.d.s  of  w' x  and   || w||_1 \cdot  x_1 are equal centered Cauchy distributions, in particular  E |w'  x| =  E (|| w||_1 ) \cdot E|x_1|  (note: then E|x_1|  equals 0 or  \infty, and so, the equality  looks useless)

2. The extension to other stable p.d.s holds similarly with   the  exponent   p  of stability within  the interval (1, 2].  The proposition reads as follows:

If w and x are independent, and if all x-s are iid with centered p-stable p.d. where1

Thank you Joachim!