Derived pushforward of exceptional divisors Let $f: X \to Y$ be a birational morphism between projective, smooth varieties. Suppose $E$ is an effective exceptional divisor of $f$. Then it is well known that $f_*(\mathcal{O}_X(E)) = \mathcal{O}_Y$ (see for example Debarre's "Higher-dimensional algebraic geometry" Page 177, section 7.12). 
My question is how does derived pushforward $\mathbf Rf_* \mathcal{O}_X(E)$of $\mathcal{O}_X(E)$ look like? Is it still $\mathcal{O}_Y$?
 A: I guess the answer is no. Let $X$ be the blow-up of $\mathbb{P}^2$ at the point $p$, and let $E$ be the exceptional curve. Also, let $f$ be the blow-down to the projective space. Then, $\mathcal{O}_X(E)|_E \cong \mathcal{O}_{\mathbb{P}^1}(-1)$.
Consider the short exact sequence
$$
0 \rightarrow \mathcal{O}_X (E) \rightarrow \mathcal{O}_X (2E) \rightarrow \mathcal{O}_{\mathbb{P}^1}(-2) \rightarrow 0.
$$
The higher direct images of the map $E \rightarrow \lbrace p \rbrace$ compute cohomology on $\mathbb{P}^1$. Also, we have $R^1 f_* \mathcal{O}_X(E) = R^2 f_* \mathcal{O}_X(E)=0$ by chasing the exact sequence
$$
0 \rightarrow \mathcal{O}_X \rightarrow \mathcal{O}_X (E) \rightarrow \mathcal{O}_{\mathbb{P}^1}(-1) \rightarrow 0.
$$
Therefore, taking higher direct images of the above short exact sequence, we get
$$
0 \rightarrow R^1 f_* \mathcal{O}_X(2E) \rightarrow H^1(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(-2)) \rightarrow 0.
$$
Thus, we get $R^1 f_* \mathcal{O}_X(2E) \cong H^1(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(-2))=k$, where $k$ is the ground field.
