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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.

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    $\begingroup$ Isn't the expectation of the distribution $+(\theta+1)$ and so the MoM estimator of $\theta$ would be $\bar{X}-1$? Your variance statement looks potentially negative $\endgroup$
    – Henry
    Commented Dec 6, 2019 at 15:24
  • $\begingroup$ Where is the trouble finding bias and variance? (This is after you get the correct estimator of course.) What do you know about consistency of an estimator? $\endgroup$ Commented Dec 6, 2019 at 19:03
  • $\begingroup$ @Henry can you explain why its positive because ive done this damn intergral a million times and i dont see why its that, can i just simplify it through and divide by negative 1? $\endgroup$
    – user655883
    Commented Dec 9, 2019 at 13:49
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    $\begingroup$ $\int\limits_\theta^\infty x e^{-(x-\theta)}\, d x = \left. - (x+1) e^{-(x-\theta)}\right|_\theta^\infty = 0 -(-(\theta+1)e^0) = \theta+1$ $\endgroup$
    – Henry
    Commented Dec 9, 2019 at 13:58
  • $\begingroup$ @henry thank you man, for some reason I just left out the infinity substitution because it equalled 0 and just had the second part. $\endgroup$
    – user655883
    Commented Dec 9, 2019 at 14:19

1 Answer 1

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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.

$E\left[\hat{\theta}\right]=E \left[ \frac{1}{n}\Sigma_{i=1}^{n}x_{i}-1 \right]$

$E\left[\hat{\theta}\right]=\frac{1}{n}\Sigma_{i=1}^{n}E\left[x_{i}\right] -E\left[1 \right]$

$E\left[\hat{\theta}\right]=\frac{1}{n}\Sigma_{i=1}^{n}\int_{x_{i}=\theta}^{\infty}x_{i}e^{-(x_{i}-\theta)}d{x_{i}}-1$

$E\left[\hat{\theta}\right]=\frac{1}{n}\Sigma_{i=1}^{n}\left(-(x_{i}+1)e^{\theta-x_{i}}\right)|_{x_{i}=\theta}^{\infty}-1$

$E\left[\hat{\theta}\right]=\frac{1}{n}\Sigma_{i=1}^{n}\left(\theta+1\right)-1=\theta$

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