The Statement of the Problem:
Let $\{P_{\theta}: \theta \in \Theta \}$ be a statistical model. Suppose that $\hat \theta$ is an estimator for a parameter $\theta$ and $E_{\theta}(\hat \theta) = a\theta + b$ for given non-zero constants $a$ and $b$.
(a) In terms of $a, b,$ and $\theta$, what is the biased $b(\hat \theta)$?
(b) Use the estimator $\hat \theta$ to construct an unbiased estimator for $\theta$.
Where I Am:
This problem just seems a little too easy...
For part (a), all I did was the following:
$$b(\hat \theta) = E[\hat \theta] - \theta \quad \text{(by definition)} $$
so
$$ b(\hat \theta) = (a\theta + b) - \theta \quad \text{(by what was provided)}$$
which, I guess, can be simplified to
$$ b(\hat \theta) = \theta(a-1) + b $$
but, aside from that, I'm not sure if there's any more to it?
Now, for part (b), we need to satisfy the following condition:
$$ E[\hat \theta] = \theta. $$
Ok. Well, that simply means the following, I guess:
$$ a\theta + b = \theta $$
and then solving for $\theta$:
$$ \hat \theta = \frac{-b}{a-1}. $$
Is that it? Am I missing something?
If $E(\hat \theta) = a\theta + b,$ then $\tilde \theta = \frac{\hat \theta - b}{a}$ is unbiased.
I can't immediately think of a simple example that has both $a \ne 1$ and $b \ne 0$. So here are two separate examples where $b = 0$, since $a$ is the part that seems to be giving you trouble.
Consider using independent observations $X_1, X_2, \dots X_n$ from $Unif(0, \theta)$ to estimate $\theta.$ You might consider estimator $\bar X.$ Because $E(\bar X) = \theta/2$, an unbiased estimator is $2\bar X.$
Alternatively, you might consider the maximum observation $X_{(n)}$ as an estimator. Because $E(X_{(n)}) = \frac{n}{n+1}\theta$, the estimator $\frac{n+1}{n}X_{(n)}$ is unbiased for $\theta.$
Between these two, the estimator $\frac{n+1}{n}X_{(n)}$ has the smaller variance, so it is preferable. Indeed, one can show that, among unbiased estimators, this one has the smallest possible variance. You have many interesting ideas coming if this course is at the level I suspect.