Let $X_1, X_2, ...X_n$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $a = max(X_i)$. Argue intuitively why aˆ cannot be an unbiased estimator for a. b. Suppose that E(a) = na/(n + 1). Is it reasonable that aˆ consistently underestimates a? Show that the bias in the estimator approaches zero as n gets large. c. Propose an unbiased estimator for a. d. Let $Y = max(X_i)$. Use the fact that Y ≤ y if and only if each $X_i ≤ y$ to derive the cumulative distribution function of Y . Then show that the probability density function of Y is. $f(y) = [ny^n - ^1/a^n 0$, 0 ≤ y ≤ a otherwise, Use this result to show that the maximum likelihood estimator for a is biased. e. We have two unbiased estimators for a: the moment estimator $a_1=2\overline{\mbox{X}}$ and $a_2 = [(n + 1)/n] max(X_i)$, where max $(X_i)$ is the largest observation in a random sample of size n. It can be shown that $V(a_1) = a^2/(3n)$ and that $V(a_2) = a^2/[n(n + 2)]$. Show that if n > 1, aˆ2 is a better estimator than aˆ. In what sense is it a better estimator of a?

For this case the value for [tex]\hat a[/tex] is always smaller than the value of a, assuming [tex] X_i \sim Unif[0,a][/tex] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

[tex] E(a) - a= 0 [/tex] and that's not our case.

b) [tex] E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}[/tex]

Since is a negative value we can conclude that underestimate the real value a.

[tex] \lim_{ n \to\infty} -\frac{1}{n+1}= 0[/tex]

c) [tex] P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)[/tex]

And assuming independence we have this:

[tex]P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n[/tex]

[tex] f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a][/tex]

e) On this case we see that the estimator [tex]\hat a_1[/tex] is better than [tex] \hat a_2[/tex] and the reason why is because:

[tex] V(\hat a_1) > V(\hat a_2) [/tex]

[tex] \frac{a^2}{3n}> \frac{a^2}{n(n+2)}[/tex]

[tex] n(n+2) = n^2 + 2n > n +2n = 3n [/tex] and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming [tex] X_1, X_2 , ..., X_n \sim U(0,a)[/tex]

And we are are ssuming the following estimator:

[tex] \hat a = max(X_i)[/tex]

For this case the value for [tex]\hat a[/tex] is always smaller than the value of a, assuming [tex] X_i \sim Unif[0,a][/tex] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

[tex] E(a) - a= 0 [/tex] and that's not our case.

Part b

For this case we assume that the estimator is given by:

[tex] E(\hat a) = \frac{na}{n+1}[/tex]

And using the definition of bias we have this:

[tex] E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}[/tex]

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

[tex] \lim_{ n \to\infty} -\frac{1}{n+1}= 0[/tex]

Part c

For this case we the followng random variable [tex] Y = max (X_i)[/tex] and we can find the cumulative distribution function like this:

[tex] P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)[/tex]

And assuming independence we have this:

[tex]P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n[/tex]

Since all the random variables have the same distribution.

Now we can find the density function derivating the distribution function like this:

[tex] f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a][/tex]

Now we can find the expected value for the random variable Y and we got this:

[tex] E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}[/tex]

And the bias is given by:

[tex]E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}[/tex]

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

[tex] V(\hat a_1) = \frac{a^2}{3n}[/tex]

[tex] V(\hat a_2) = \frac{a^2}{n(n+2)}[/tex]

On this case we see that the estimator [tex]\hat a_1[/tex] is better than [tex] \hat a_2[/tex] and the reason why is because:

[tex] V(\hat a_1) > V(\hat a_2) [/tex]

[tex] \frac{a^2}{3n}> \frac{a^2}{n(n+2)}[/tex]

[tex] n(n+2) = n^2 + 2n > n +2n = 3n [/tex] and that's satisfied for n>1.