Recently I got asked this question during an interview: If a point (X, Y, Z) has a uniform distribution on the plane X+Y+Z = 1, what's E[X] ?
I tried assuming $f_X,_Y,_Z(x,y,z) = c$, and thus $f_X(x) = \int_{0}^{1}\int_{0}^{1-z-x} cdydz$. But solving this gives me $f_X(x) = c(\frac{1}{2}-x)$, which is negative when $x > \frac{1}{2}$. This is not right as x can definitley take values larger than $\frac{1}{2}$. So what did I do wrong there?
Sorry if the way I formulate the problem is not clear or precise enough but this was all the information given to me.
Update: I realized the upper bound I wrote of the integral for $f_X(x)$ is not correct. It should be $f_X(x) = \int_{0}^{1-x}\int_{0}^{1-z-x} cdydz$, where $\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-z-x} c dydzdx = 1$. Solving for c I got c = 6, and $E[X] = \int_{0}^{1}xf_X(x)dx = \frac{1}{4}$. However according to the hint from @awkward, $E[X] + E[Y] + E[Z] = 1$, and I believe due to the symmetry of the distributions of X, Y, Z we should have $E[X] = E[Y] = E[Z] = \frac{1}{3}$? This confuses me as I don't know which answer is correct and why the other one isn't.