Given $g^a, Y$, is it hard to distinguish $e(g,g)^{ab}$ from a random value? where
$g$ is a group element in bilinear group $G$

$Y = M.e(g,g)^{ab}$ 

$M$ is a message 

Does anyone know the answer or suggest some material for reference?
I'll assume that the attacker has the value $g$ as well.

If so, we can rewrite $M \cdot e(g, g)^{ab} = M \cdot e(g^a, g)^b = M \cdot h^b$ (where the attacker knows the value $h = e(g^a, g)$.

Assuming that $a$ is relatively prime to the size of the subgroup generated by $e(g, g)$, and that $M$ is a member of that subgroup,then, yes, it is indistinguishable; for any message $M'$ within that subgroup, there a $b'$ for which $Y = M' \cdot h^{b'}$