commutation of operator product expansion In CFT, when we have an OPE:
$$O_1(z)O_2(w)=\frac{O_2(w)}{(z-w)^2}+\frac{\partial O_2(w)}{(z-w)}+...$$
this holds inside a time-ordered correlation function, so $O_1(z)O_2(w)=O_2(w)O_1(z)$. Does it mean that
$$O_1(z)O_2(w)=\frac{O_1(z)}{(w-z)^2}+\frac{\partial O_1(z)}{(w-z)}+...$$
?
 A: I am not an expert in 2d CFT. However I hope following manipulations are valid. 
Assume that your second equation follows from first one. 
Then on RHS of your first equation Taylor expansion of $O_2(w)$ at point $z$ gives :
$O_2(w)=O_2(z)+(w-z)\partial_z O_2(z)+ ...$
taking derivative wrt $w$ on both sides we get 
$\partial_wO_2(w)=\partial_zO_2(z)+...$
Using these two results in your first equation we get 
$O_1(z)O_2(w)= \displaystyle\frac{O_2(z)}{(z-w)^2}+regular\:terms$
Subtracting it from your second equation, multiplying with $(z-w)^2$ and taking limit $w\rightarrow z$ we conclude that $O_2$ and $O_1$ should be equal. Since to begin with we didn't assume any such thing regarding fields $O_2$ and $O_1$ so in general your second equation shouldn't follow from the first one. 
I think equality of $O_2(w)O_1(z)$ and $O_1(z)O_2(w)$ (assuming fields are 'bosonic') within time ordered product only implies that their OPE should be symmetric under exchange of z and w. So if your first equation for OPE can be realized for some (bosonic) fields, then by exchanging z with w on RHS you should get the same result within a regular term.  
