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$X_1,X_2,X_3$ form a random sample from the following p.d.f:

$$ f(x\mid \theta) = \begin{cases} {e^{-(x-\theta)}} & \text{for } x>\theta\\[10pt] 0 &\text{for $x\le \theta$} \end{cases} $$

there are three estimator of $\theta$, $\hat\theta_1=X_3-1$, $\hat\theta_2=Y_1=\min\{X_1,X_2,X_3\}$, $\hat\theta_3 = {1\over3}(X_1+X_2+X_3)-1$. find $E(\hat\theta_i)$ and $\operatorname{Var}(\hat\theta_i)$ ($i=1,2,3$),which estimator should be used for estimating $\theta$.

I have tried to calculate the E and Var by the p.d.f, and the following answer.(I am not sure if my method of solving and the answer is correct)

$$E(\hat\theta_1)=\theta,\quad E(\hat\theta_2)=\theta+1, \quad E(\hat\theta_3)=\theta$$

$$\var(\hat\theta_1)=e, \quad \var(\hat\theta_2)={5\over9},\quad \var(\hat\theta_3)={1\over3}$$

so.the $\hat\theta_2$ is obviously a biased estimator,$\hat\theta_3$ seem the best choice. but, I also calculate the M.L.E ,and find the following fact. ${d \over d\theta}\ln L(\theta)=3$, so, $L(\theta)$ is monotone increasing, so $\hat\theta=Y_1=\min\{X_1,X_2,X_3\}$, $\hat\theta_2$ is the maximum likelihood estimator.

which should be used? and if possible, could anyone, please, tell if my calculation about the E and Var is correct?

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The variance you have for $\hat\theta_3$ is correct, and that makes me wonder if the variance you report for $\hat\theta_1$ is a typo, since that should be $1$.

But you should have $\var\hat\theta_2=\dfrac19$.

Notice that for $x\ge\theta$, $$\Pr(\min>x)=\Pr(X_1>x\ \&\ X_2>x\ \&\ x_3>x)=(\Pr(X_1>x))^3 = (e^{\theta-x})^3 =e^{3(\theta-x)}.$$

So $$f_\min(x)=\frac{d}{dx}\Pr(\min\le x)=\frac{d}{dx}(1-e^{3(\theta-x)}) = 3e^{3(\theta-x)}\text{ for }x>\theta.$$ You can use that probability density function to find the variance.

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  • $\begingroup$ sorry, I calculated agian,and the variance of $\hat\theta_1$ is e. what the p.d.f of $\hat\theta_1$ do you think?is it $$ f(x) = \begin{cases} {e^{-(x-\theta)}} & \text{for } x>\theta-1\ and 0 &\text{for $x\le \theta-1$} \end{cases} $$ ? $\endgroup$ – nina li Jan 27 '13 at 16:08
  • $\begingroup$ and I use the formula $\displaystyle\int_\theta^{\infty} (x-\theta-1)^2*3*e^{3(\theta-x)}$dx , get the 5/9... $\endgroup$ – nina li Jan 27 '13 at 16:33
  • $\begingroup$ The expected value of the minimum is not $\theta+1$, but $\theta+(1/3)$. $\endgroup$ – Michael Hardy Jan 27 '13 at 16:35
  • $\begingroup$ oh,yes,I made a stupid mistake of calculation,you are right.how about the p.d.f and variance of $\theta_1$? $\endgroup$ – nina li Jan 27 '13 at 16:47
  • $\begingroup$ You need $f(x)=e^{-(x-(\theta-1))}$ for $x\ge\theta-1$ and $0$ for $x<\theta-1$. $\endgroup$ – Michael Hardy Jan 27 '13 at 16:52

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