I have seen similar questions to this but none have answered my question entirely. I was working on a problem with $f(x,y)=6(x^2+y^2)$. It was then stated that the pdf was symmetric in x and y. I was unsure what this meant or how it helped me so I carried on as normal. As I progressed I realized that $E(X) = E(Y)$ and $V(X) = V(Y)$
I then came to the conclusion that this may be what symmetric entails. Despite all my research I have only fount information that $E(X) = E(Y)$ come with symmetry but nothing about variance. Logically it seems if their expectation is the same than their variance would be as well. Although this makes sense to me some posts I have seen seem to go against this idea.
My question is with symmetry of a joint pdf is the $V(X)$ always equal to $V(Y)$ or was this merely coincidence of my specific problem.
if it is helpful both of my expectations were 1/2 and my variances are 7/60
The statement that the joint pdf $f(x,y)$ is symmetric in $x$ and $y$ means $f(x,y) = f(y,x)$. In particular this tells us that marginals $X$ and $Y$ have the same distribution and therefore $X$ and $Y$ have the same mean and variance.
Compute the marginal pdf of $X$ by integrating the joint pdf over $y$: $$f_X(x):=\int_y f(x,y)\,dy\tag1$$ while the marginal pdf of $Y$ is found by integrating the joint pdf over $x$: $$f_Y(y):=\int_x f(x,y)\,dx\tag2$$ To see that $f_X$ and $f_Y$ are the same function, write: $$f_X(t)\stackrel{(1)}=\int_y f(t,y)\,dy \stackrel{(a)}= \int_y f(y,t)\, dy\stackrel{(b)} = \int_x f(x,t)\,dx\stackrel{(2)} = f_Y(t);$$ in step (a) we use symmetry of the joint pdf, while in step (b) we rename the dummy variable of integration from $y$ to $x$.
This statement isn't true; two random variables can have the same expectation yet different variance. However, in the symmetric case as we have here, all properties of $X$ and $Y$ are the same.