It depends...

Let s_i be Rademacher variables (+/-1 with P=1/2) and x_i real numbers. We want to calculate the expectation and investigate whether or not it is true that

(*) E[ | sum_i s_i * x_i |] = max{ | x_i | }.

First consider the case of 2 terms. Without loss of generality let 0 < x_2 < x_1. Note that

(i) prob(s_1=1,s_2=1) = 1/4, (ii) prob(s_1=1,s_2=-1) = 1/4,

(iii) prob(s_1=-1,s_2=1) = 1/4, (iv) prob(s_1=-1,s_2=-1) = 1/4

Then investigate |s_1*x_1 + s_2*x_2| for all 4 possible cases

(i) x_1 + x_2 >0, (ii) x_1 - x_2>0, (iii) -(-x_1 + x_2)>0, (iv) -(-x_1 - x_2)>0

Thus we can calculate

E(|s_1*x_1 + s_2*x_2|) = 4*x_1 * 1/4 = x_1 = max(x_1 , x_2)

and the claim (*) is proven for N=2.

Now take 3 terms and without loss of generality 0 < x_3 < x_2 < x_1 (if not, just re-label variables). We have to distinguish two cases

(A) x_1 - x_2 > x_3

In this case (A) we can boil everything down to our 2-dimensional case,

(s_1*x_1 + s_2*x_2) + s_3*x_3

because in the "worst case" we have x_1- x_2 - x_3, but that is positive (and likewise vice versa with minus sign). So here the claim also holds here.

(B) x_1 - x_2 < x_3

If we solve the |.| we have to consider 8 cases for the s_1, s_2 and s_3. After some calculations using the new probabilities of 1/8 we end up with

E(|s_1*x_1 + s_2*x_2 + s_3*x_3|) = (x_1 + x_2 + x_3)/2

which is not equal the maximum x_1 and the general claim is false here.

I think your claim holds true in a certain region of the hypercube of dimension N. All we need to do is to investigate 2^N possible cases for |s_1*x_1 + s_2*x_2 + ... + s_3*x_N| , where x_1>x_2>...>x_N. My guess is that for the hierarchy of inequalities

x_1 - x_2 > 0

x_1 - x_2 - x_3 > 0

x_1 - x_2 - x_3 - x_4 > 0

...

we are on the safe side. However, the effect of violating the inequalities further down the hierarchy seems to be only a small perturbation (looking at some numerical simulations). And maybe if we have two or three maximum outliers in the data, x_1, x_2 and x_3 sufficiently large versus the rest of the data and satisfying (A), the claim holds approximately true regardless.

Thank you for working that out. The whole problem arose in the course of a stability analysis around |x_1|>0, x_2=...=x_N=0, so it seems clear now that (*) is valid in a sufficiently small neighborhood of that point (for finite N), and that's sufficient for me. Just as an interesting aside, from your analysis and just looking at ||x||=1, my guess is that for unlimited N we can't do better than Khintchine's inequality since the neighborhood around x_1 = 1, x_2=...=x_N=0 for which x_1-x_2-...-x_N > 0 holds goes to zero.