Estimator of $\theta$, uniform distribution $(\theta, \theta +1)$

Question

I would like to ask you, if somebody helps me to solve one statistics problem.

I have $X_1, ..., X_n$ a sample of independent random variables with uniform distribution $(\theta, \theta +1)$

I have an estimator $\hat{\theta}_n=\max$ $X_i -1$ for $1 \le i \le n$, and I should find if it is unbiased and consistent, I found, that it is consistent but I think that it is biased. I find density of my estimator: $f_Y(y)=n(y+1-\theta)^{(n-1)}$, where $y=\max$ $X_i$, but I am not sure, if it is correct, because if I calculate $E(\hat{\theta}_n)$, the result was quite strange.

The last step, which I should do, is following: how should I change $\hat{\theta}_n$ to make it unbiased. I think that I have to substract 1, but then I am not sure.

Please, can somebody check if I calculate the density correctly and how I can find an unbiased estimator (I cannot use MLE)?

Thank you

Answer 1

Comment: It seems there are several logical candidates. Let $A$ be the average of the Max and Min (the midrange). Intuitively it seems $E(A) = \theta + .5,$ and you could let $\hat \theta_1 = A - .5$ be one candidate.

Intuitively, on average The $n$ uniform observations in $(\theta, \theta+1)$ tend to divide the interval into $n+1$ equal intervals, so the Min has expected value $\theta + \frac{1}{n+1}.$ Thus if we let $\hat \theta_2 = Min - \frac{1}{n+1},$ it should be unbiased. (This disagrees with @leonbloy's current Answer, and you should check your math to see what is right.)

Two other obvious unbiased candidates are the adjusted sample mean $\hat \theta_3 =\bar X - .5$ and the adjusted sample median $\hat \theta_4 = \tilde X -.5.$

Let's try a simulation in R statistical software to see if these proposed estimates are unbiased, and which is the "best" estimator (having the smallest variance among them). In the simulation $n = 10$ and $\theta = 5.$ Based on a million samples of size $n,$ results should be accurate to two significant digits (or better).

 m = 10^6;  n = 10;  th = 5.
 x = runif(m*n, th, th + 1)
 DTA = matrix(x, nrow=m) # each row a sample of size n
 mx = apply(DTA, 1, max); mn = apply(DTA, 1, min)
 a = (mx + mn)/2;  th.1 = a - .5
 th.2 = mn - 1/(n+1) 
 th.3 = rowMeans(DTA) -.5
 th.4 = apply(DTA, 1, median) - .5

 mean(th.1);  mean(th.2);  mean(th.3);  mean(th.4)
 ## 4.999984  # all four consistent with unbiasedness
 ## 4.999985
 ## 5.000001
 ## 5.000006
 sd(th.1); sd(th.2); sd(th.3);  sd(th.4)
 ## 0.06153263 # among these four--unbiased midrange is the winner
 ## 0.08287243
 ## 0.09123998
 ## 0.1376457

You should be able to find means of the various proposed estimators to verify unbiasesdness analytically. Also perhaps variances to judge which is the best of the three. Also, please be sure you understand why the usual method for finding an MLE does not produce an answer to this problem.

Answer 2

Your result is correct

Letting $Y=\hat{\theta}_n=\max (X_i) -1$ we have

$$P(Y \le y) = P(\max (X_i) \le y +1)=\prod P(X_i \le y+1) = (y +1 - \theta)^n $$

Hence $$f_Y(y) = n (y+1-\theta)^{n-1} \hspace{1cm } \theta-1\le y \le \theta$$

And $E(Y)=\theta - \frac{1}{n+1}$

Hence the estimator is biased (but also asymptotically unbiased)

(Both results, and the sign of the bias are intuitively obvious : for one thing, note that always $X < \theta+1 \implies \hat{\theta_n} < \theta$. Also, it's easy to see that for large $n$ the maximum will be very near $\theta+1$, hence we should expect $E(\hat{\theta}_n) \to \theta$)

To make it unbiased, you can try some linear transformation $Z=a(Y+b)$