$\newcommand{\sgn}{\operatorname{sgn}}$Good evening everyone. I might need some help on something. Suppose we have $n$ independent variables from $U[0,1+θ]$ and suppose also that $λ=1+θ$. The estimator of $θ$ is $\bar{θ}=2 \bar{X} - 1$. Ι need to find the bias of $\bar{θ}$ knowing that the distribution of $\bar{X}$ of $n$ variables, which they are from $U[0,1]$ , is the Bates distribution, which has the following probability density function: $$f(x)=\frac{n}{2(n-1)!} \sum_{k=0}^n (-1)^k \binom{n}{k}(nx - k)^{(n-1)} \sgn(nx -k)$$
where $\sgn(nx-k) = \begin{cases} -1,\ & \text{if } nx<k, \\ \phantom{-}0, & \text{if } nx=k, \\ \phantom{-}1, &\text{if } nx > k. \end{cases}$
If I understand correctly, you have $X_1,\ldots,X_n\sim\text{ i.i.d.} \operatorname{Uniform}[0,1+\theta]$ and $\bar X = (X_1+\cdots+X_n)/n.$ If you want the bias of $2\bar X-1$ as an estimator of $\theta,$ you do not need to know much about the distribution of $\bar X.$ You only need its expected value, which can easily be found without knowing much about its distribution. Its expected value can be shown to be the same as that of $X_1,$ as follows:
\begin{align} & \operatorname E\left( \frac {X_1+\cdots+X_n} n \right) = \frac 1 n \operatorname E(X_1+\cdots+X_n) \\[10pt] = {} & \frac 1 n\left( \operatorname E(X_1) + \cdots + \operatorname E(X_n) \right) = \frac 1 n \Big( n \operatorname E(X_1)\Big) = \operatorname E(X_1) = \frac{\theta+1} 2. \end{align}
(Note that $2\bar X -1$ is a lousy estimator, not only because there are others with substantially smaller mean squared errors, but also because in some cases $2\bar X-1$ is smaller than $\max\{X_1,\ldots,X_n\},$ and you know that $\theta$ cannot be smaller than that maximum. The best unbiased estimator is a constant multiple of the aforementioned maximum observed value. There are biased estimators that are also constant multiples of the maximum and that have smaller mean squared errors than the best unbiased estimator.)