We let $X_1,...,X_n$ be i.i.d random variables from the pdf:
$f(x)=e^{-(x-\theta)}$ for $x\geq \theta$ and $f(x)=0$ otherwise.
Derive the method of moments estimator of $\theta$ and find its bias and variance. Is it a consistent estimator?
now ive worked out the expectation to be $-(\theta+1)$ and the variance to be $-2\theta^2-4\theta-3$
From the formula ive seen for the method of moments estimator ive tried to work it out and i got it to be $1-\frac{1}{n}\sum_{k=1}^{n}(x_i)$ but i am not sure, and im not quite sure how to work out the bias and variance or check for consistency.
I attempted to compute the mean of the estimator here, (taking into account the observation in the comments in the original question regarding the sign of the estimator being reversed), from which you can determine that the bias is zero from the definition $Bias=E\left[\hat\theta\right]-\theta$.
I believe the rest of the answer can follow along these lines. I hope this helps.
$E\left[\hat{\theta}\right]=E \left[ \frac{1}{n}\Sigma_{i=1}^{n}x_{i}-1 \right]$
$E\left[\hat{\theta}\right]=\frac{1}{n}\Sigma_{i=1}^{n}E\left[x_{i}\right] -E\left[1 \right]$
$E\left[\hat{\theta}\right]=\frac{1}{n}\Sigma_{i=1}^{n}\int_{x_{i}=\theta}^{\infty}x_{i}e^{-(x_{i}-\theta)}d{x_{i}}-1$
$E\left[\hat{\theta}\right]=\frac{1}{n}\Sigma_{i=1}^{n}\left(-(x_{i}+1)e^{\theta-x_{i}}\right)|_{x_{i}=\theta}^{\infty}-1$
$E\left[\hat{\theta}\right]=\frac{1}{n}\Sigma_{i=1}^{n}\left(\theta+1\right)-1=\theta$