Expectation of an estimator

Question

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?

$P(\hat\theta > t) = 1 - P(\hat\theta \leq t).$ – Ehsan M. Kermani Jan 08 '15 at 00:48 — Ehsan M. Kermani, Jan 08 '15 at 00:48
Hi, thanks, I just noticed as I posted this. – user2850514 Jan 08 '15 at 00:49 — user2850514, Jan 08 '15 at 00:49

score 0 · Answer 1 · answered Jan 08 '15 at 00:53

0

Note that $\Pr(\hat\theta>t)=1-\Pr{(\hat\theta\le t)}$ so $$\Pr{(\hat\theta=t)}=\frac{d}{dt}\Pr{(\hat\theta\le t)}=\frac{d}{dt}[1-\Pr(\hat\theta>t)]=-\frac{d}{dt}\Pr(\hat\theta>t).$$

answered Jan 08 '15 at 00:53

user2850514

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Expectation of an estimator

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