Changing the measure of a stochastic process I am trying to understand how to change the measure of a stochastic process using Girsanov's theorem.
In particular, I have the process $dX_t = a dt + dB_t$ for $t \in [0,T]$, and some arbitrary, well-behaved function $v(X)$, where $X$ denotes the path of $X_t$ up to $T$.
I have the quantity $\mathbb{E}_\tilde{a}[v(X)]$, where the subscript $\tilde{a}$ denotes that the expectation is taken with respect to the measure associated with $dX_t = \tilde{a} + dB_t$. (Hopefully this statement does make sense.)
Now I would like to differentiate this quantity with respect to $\tilde{a}$, and evaluate it at $\tilde{a} = a$.
My first interpretation of Girsanov's theorem is that I can write
$\mathbb{E}_\tilde{a}[v(X)] = \mathbb{E}_0[v(X)e^{\tilde{a}B_T-\frac{1}{2}\tilde{a}^{2}T}]=\mathbb{E}_a[v(X)e^{(\tilde{a}-a)B_T-\frac{1}{2}(\tilde{a}^{2}-a^2)T}]$
Differentiating this with respect to $\tilde{a}$ and evaluating the derivative at $\tilde{a}=a$ gives
$\frac{d}{da}\mathbb{E}_{\tilde{a}}\left[v(X)\right]=\mathbb{E}_{a}\left[v\left(B_{T}-aT\right)\right]$
My second interpretation of Girsanov's theorem is that I can write 
$\mathbb{E}_\tilde{a}[v(X)] = \mathbb{E}_a[v(X)e^{(\tilde{a}-a)B_T-\frac{1}{2}(\tilde{a}-a)^2T}]$
in which case I get
$\frac{d}{da}\mathbb{E}_{\tilde{a}}[v(X)]=\mathbb{E}_{a}[vB_{T})]$
Clearly, only one (or possibly) none of these are correct, and I would like to understand which one, and why. Thank you! :) 

FYI - figured out the answer. The first interpretation is incorrect, while the second interpretation is correct. 
When I change the measure in two steps (as in the first interpretation), in the second change of measure, the Brownian motion $B_T$ is with respect to the measure $P^0$, not $P^{\tilde{a}}$. Once I write it properly, the two-step approach gives the same answer as the one-step approach, as it should.
