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Given random sample $X_1, X_2, ..., X_n$ with the distribution $$f(x|\theta) = \left \{ \begin{aligned} e^{-(x-\theta)}, \ \ 0 < x < \theta \\ 0, \text{ otherwise.} \end{aligned} \right. $$

where $\theta \in (-\infty, \infty).$ Show that the estimator $\theta_1 = \min\left\{X_1, X_2, ..., X_n \right\} $ is unbiased.

So, I don't even know where to start, should I randomly generate a minimum for a large amount of normally distributed data and then compare it to the minimum of a sample consisting some arbitrary number of variables? I need to find a solution for the problem in R and I cannot even imagine how to do this.

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    $\begingroup$ why do you think $e^{-(x-\theta)}, \ \ 0 < x < \theta$ is a density function?! $\endgroup$
    – Masoud
    Commented Mar 11, 2020 at 20:32

1 Answer 1

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I assume that you mean $X_1, X_2, ..., X_n$ with the distribution

$$f(x|\theta) = \left \{ \begin{aligned} e^{-(x-\theta)}, \ \ \theta <x \\ 0, \text{ otherwise.} \end{aligned} \right. $$ since in that case $f$ is a density function.

define $Y=\min (X_1,\cdots X_n) $

$$F_Y(y)=1-P(Y>y)=1-P(X_1>y,\cdots, X_n>y)=1-e^{-n(y-\theta)}$$ so

$$f_Y(y)=ne^{-n(y-\theta)} \hspace{1cm} \theta < y$$

so $E(Y)=\theta + \frac{1}{n}$.

so $Y$ is baised. By a simple simulation in R we show $Y$ is biased (as you asked).

we do this step $k$ times:

1) draw $n$ independent random sample $(X_1,\cdots X_n)\sim f_X$

2)calculate $Y_k=\min(X_1,\cdots X_n)$

then we have $k$ of $Y_k$. calculate mean of this $Y_k$ and it is done.

 theta<-5
 n<-10
 sim<-200
 Ymin<-c()
 for(k in 1:sim){
 set.seed(k)
 x<-rexp(n,1)+theta
 Ymin[k]<-min(x)
 }
 mean(Ymin);theta+1/n
 #[1] 5.094421   [1] 5.1
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  • $\begingroup$ Can you explain, why did you add to the random deviate the theta? Why did you choose theta to be 5? Also why did you chose rate to be equal to 1? $\endgroup$
    – user
    Commented Mar 19, 2020 at 16:54
  • $\begingroup$ It showed for $\theta=5$ and $\lambda=1$(the exponential distribution parameter) the estimator $Min(X_i)$ is biased . I choose them arbitrary, since if the estimator is unbaised, it must be unbaised for all value of parameters. on the other hands ,$\forall \theta$ $E_{\theta}(T)=\theta$. so I choose a $\theta$ and simulate form the distribution to check unbaisedness $\endgroup$
    – Masoud
    Commented Mar 19, 2020 at 18:06
  • $\begingroup$ this distribution is shifted exponential shodhganga.inflibnet.ac.in/bitstream/10603/130549/11/… . on the other hand, if $X\sim E(1)$ so $X+\theta$ is the distribution of above. $\endgroup$
    – Masoud
    Commented Mar 19, 2020 at 18:10
  • $\begingroup$ How did get from $f_Y (\theta) $ to $E(Y)$? $\endgroup$
    – user
    Commented Jun 5, 2020 at 19:30
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    $\begingroup$ For the last question, Note $F_Y(y)=P(Y\leq y)=1-P(Y>y)$ $\endgroup$
    – Masoud
    Commented Jun 5, 2020 at 20:53

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