I want to find the expectation of $\hat\theta$. I have the cumulative distribution of $\hat\theta$: $$\Pr{(\hat\theta>t)} = e^{n(\theta-t)}\quad \text{for $t>\theta.$}$$ Now to find the expectation I need the probability density of $\hat\theta$, i.e. $$\Pr{(\hat\theta=t)}=\frac{\rm d}{{\rm d}t}\Pr{(\hat\theta>t)}=\frac{\rm d}{{\rm d}t}e^{n(\theta-t)}$$
However my lecturer finds $$\Pr{(\hat\theta=t)}=-\frac{\rm d}{{\rm d}t}\Pr{(\hat\theta>t)}=-\frac{\rm d}{{\rm d}t}e^{n(\theta-t)}.$$ Why is this?
Note that $\Pr(\hat\theta>t)=1-\Pr{(\hat\theta\le t)}$ so $$\Pr{(\hat\theta=t)}=\frac{d}{dt}\Pr{(\hat\theta\le t)}=\frac{d}{dt}[1-\Pr(\hat\theta>t)]=-\frac{d}{dt}\Pr(\hat\theta>t).$$