Suppose $Y$ is an algebraic variety and $\mathcal{E}$ a coherent sheaf on $Y$. Suppose $f:X=\mathrm{Proj}(\mathrm{Sym}(\mathcal{E})) \to Y$ is a morphism of algebraic varieties with all fibres scheme theoretically projective spaces.


If the fibres all had the same dimension, I would have $\mathrm{R}f\_\* \mathbb{C}\_X = \mathrm{H}^\*(\mathbb{P}^n) \otimes \mathbb{C}\_Y$. 



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> In the case that the fibre dimension varies, let $Y\_k$ be the locus where the fibre dimension is at least $k$. Then is it true that $\mathrm{R}f\_\* \mathbb{C}\_X = \bigoplus \mathbb{C} \_{Y\_k}\[-2k\](k)$?
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(and if not in general, are there any reasonable assumptions which make it true?) 



Cohomology of projective space bundles
I think what you want is true if $X$ and all the $Y\_k$ are smooth (or have some very mild singularities e.g. quotient singularities) but I don't know many such examples. In general it appears to be false as shown by the following example:


Let $Y$ be the quadric cone given by $x\_1x\_2 - x\_3x\_4 = 0$ in $\mathbb{A}^4$. If we blow up the vertex the exceptional divisor is isomorphic to the quadric in $\mathbb{P}^3$ given by the same equation. $Y$ has a small resolution $f:X \to Y$ which is given (in the fibre over the vertex) by projecting the quadric onto one of its factors. The fibre of $f$ over the vertex is $\mathbb{P}^1$ and $f$ is a morphism of the type you want.


Since $f$ is birational and $X$ is smooth, it follows from Verdier duality that $Rf\_\*({\mathbb{C}}\_X)$ is self dual (with a shift, depending on your conventions). However, one can see that the object in the derived category given by your formula is not self dual.


