unbiased estimators Suppose that $W_1, W_2,\ldots,W_n$ is a random sample from a population with density function $$f(w) = \frac{e^{\frac{-w}{\theta+4}}}{\theta+4}$$ for $w>0$ and $\theta >-4$ Find an unbiased estimator for $\theta$
I believe all I should have to do if find the $\operatorname{E}[W]$ and then set it equal to $\theta$ and then find the constant that makes them equal? 
However, when I take $$\int_0^\infty w\frac{e^{\frac{-w}{\theta+4}}}{\theta+4} \, dw $$ it doesn't converge. Why doesn't this work? 
 A: The value of an unbiased estimator of $\theta$


*

*depends on $W_1,\ldots,W_n$;

*does not depend on $\theta$, i.e. its dependence on $(W_1,\ldots,W_n,\theta)$ is only through $W_1,\ldots,W_n$;

*has expected value $\theta$ regardless of the value of $\theta$, i.e. if you change $\theta$, the expected value changes, but will change in such a way that it is still equal to $\theta$.


What does "find the constant that makes them equal" mean?  Does it mean find the value of $\theta$ that makes them equal?  That won't work because, as stated above, you need unbiasedness to hold for every value of $\theta$; otherwise it is not an unbiased estimator.
In fact, your integral does converge:
\begin{align}
\int_0^\infty we^{-w/(\theta+4)} \, \frac{dw}{\theta+4} & = (\theta+4) \int_0^\infty \frac w {\theta+4} e^{-w/(\theta+4)} \, \frac{dw}{\theta+4} \\[10pt]
& = (\theta+4) \int_0^\infty u e^{-u} \, du.
\end{align}
And then:
\begin{align}
\int_0^\infty u e^{-u}\,du & = \int u \quad \underbrace{\quad e^{-u} \, du\quad}_{dv} = \overbrace{\int u\,dv = uv - \int v\,du}^\text{integration by parts} = -ue^{-u} - \int -e^{-u}\,du \\[10pt]
& = \left[-ue^{-u} -e^{-u} \vphantom{\frac 1 1} \right]_0^\infty = 1 \quad \left(\text{You can use L'Hopital's rule to find } \lim_{u\to\infty} -ue^{-u}. \right)
\end{align}
Hence $\operatorname{E}(W) = \theta + 4$, so


*

*$W_1-4$ is an unbiased estimator of $\theta$; and

*$\dfrac{W_1+\cdots+W_n} n - 4$, is an unbiased estimator of $\theta$; and

*so are many other functions of $W_1,\ldots,W_n$; and

*Probably $\operatorname{median}(W_1,\ldots,W_n) - 4 + \text{some constant}$ is an unbiased estimator of $\theta$. (You can take it to be an exercise to find that "constant".); and

*There are yet others.

