probability distributionsstatistics
For a random sample $X_1,X_2,…X_n$ from a uniform $[0,\Theta]$ distribution, with probability density function
$$f(x;\Theta) = \left\{ \begin{array} \ \frac{1}{\Theta} & 0\le x \le\Theta,\\ 0 & \text{otherwise}.\end{array}\right.$$
What is the value of $k$ such that $\hat{\Theta}=k\bar{X}$ is an unbiased estimator of $\Theta$?
I've done some questions similar to this but I'm not sure how to go about this one. I have a test in 3 hours so help is really appreciated!
I will not use the quoted theorem, but instead derive the distribution of the maximum $Y$ of the $X_i$. The same argument works more generally, and can be used to prove the max part of the theorem you quoted.
The density function of the $X_i$ is $12x^2(1-x)$ on $(0,1)$ and $0$ elsewhere. We first find the cdf $F_Y(y)$ of $Y$.
The only interesting values of $y$ are $0\lt y\lt 1$, so we assume $y$ is in this interval. We have $Y\le y$ if and only if all the $X_i$ are $\le y$. The probability any individual $X_i$ is $\le y$ is $$\int_0^y 12x^2(1-x)\,dx.$$ This is $4y^3-3y^4$. So by independence the probability all the $X_i$ are $\le y$ is $(4y^3-3y^4)^n$. Now we know $F_Y(y)$ and can differentiate to find the density function of $Y$. We get $12ny^2(1-y)(4y^3-3y^4)^{n-1}$.
Remarks: 1) The proof of the min part of the theorem you quoted goes along similar lines. For the probability the min of the $X_i$ is $\gt z$ is the probability all the $X_i$ are $\gt z$.
2) The probability that the maximum is exactly $1$ is $0$. That does not require the above machinery. For any random variable $W$ with continuous distribution, and any real $a$, we have $\Pr(W=a)=0$.
(1) The 'general method' is to set the sample mean $\bar X$ equal to the population mean $\theta/2$ to get the method of moments estimator (MME) $\hat \theta = 2\bar X$ of $\theta.$
(2) Yes. By definition, the standard error of the estimator $\hat \theta$ is $SD(\hat \theta) = \sqrt{Var(\hat \theta)}.$
Notes: The multiple $(n+1)/n$ of the maximum $X_{(n)}$ of the data $X_1, X_2, \dots, X_n$ is also an unbiased estimator of $\theta$, and it has a smaller variance than the MME. Maybe showing that is a later exercise in your text.
It is not difficult to find the PDFs of these two estimators and thus to derive these relationships analytically (see this Q & A), but the following simple simulation (using the sampling method) in R provides a preview of the results (for $n = 5$ and $\theta = 10$).
Notice that the distribution of the MME is roughly normal after averaging only five uniform observations. Even though the second estimator is less variable, the mode of its PDF lies to the right of $\theta = 10.$
m = 10^5; n = 5; th = 10
x = runif(m*n, 0, th)
DTA = matrix(x, nrow=m) # each row is sample of size n
a = rowMeans(DTA) # vector of m means
b = apply(DTA, 1, max) # vector of m maximums
mme = 2*a; mean(mme); sd(mme)
## 10.01064 # approx. of E(MME) = 10 (unbiased)
## 2.588815 # approx. of SE(MME)
est2 = ((n+1)/n)*b; mean(est2); sd(est2)
## 10.00017 # approx. of E(EST2) = 10 (unbiased)
## 1.69188 # indicates SE(EST2) < SE(MME)
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