This answer is meant to supplement this excellent analytical answer. I often like to numerically check things to rapidly visualize.
A quick simulation shows this can't be true in general. Let $X_i \stackrel{iid}{\sim} \text{Uniform}(0,1)$.
With 10000 samples, it is clear that $Z_i$ are not i.i.d.
Some MATLAB code below:
% MATLAB R2018a
n = 10000; % sample size
% Generate samples from Xi ~ U(0,1)
X1 = rand(5000,1);
X2 = rand(5000,1);
X3 = rand(5000,1);
% Get Zi
Z1 = X1;
Z2 = X1 + X2;
Z3 = X1 + X2 + X3;
If two random variables $Y_1$ & $Y_2$ are independent, their correlation will be zero. We know the reverse is not always true (e.g. zero correlation does not imply independence).
% Check correlation for Xi's (should be approx zero)
% Cx(i,j) = corr(Xi,Xj)
Cx = ones(3);
Cx(1,2) = corr(X1,X2); Cx(2,1) = Cx(1,2);
Cx(1,3) = corr(X1,X3); Cx(3,1) = Cx(1,3);
Cx(2,3) = corr(X2,X3); Cx(3,2) = Cx(2,3);
% Check correlation for Zi's (are they even close to zero?)
Cz = ones(3);
Cz(1,2) = corr(Z1,Z2); Cz(2,1) = Cz(1,2);
Cz(1,3) = corr(Z1,Z3); Cz(3,1) = Cz(1,3);
Cz(2,3) = corr(Z2,Z3); Cz(3,2) = Cz(2,3);
% Visually inspect distributions
figure
s(1)= subplot(1,3,1)
histogram(Z1,'Normalization','pdf','FaceColor','r')
s(2) = subplot(1,3,2)
histogram(Z2,'Normalization','pdf','FaceColor','b')
s(3) = subplot(1,3,3)
histogram(Z3,'Normalization','pdf','FaceColor','g')
% Cosmetics
ylabel(s(1),'PDF')
for k = 1:3
xlabel(s(k),['Z' num2str(k)])
s(k).XLim = [0 4];
s(k).YLim = [0 1.2];
end