# Estimator problems

I am stuck with some parts of a problem in my textbook, and the solutions in my textbook do not seem to help me. The problem goes:

Two independent observations $$X_1$$ and $$X_2$$ are made of continuous random variables with pdf: $$f(x)=\frac{1}{k}; 0\le x \le k$$

i) How do I find a probability distribution of M which stands for the larger of $$X_1$$ and $$X_2$$?

ii) how do I show that M is an unbiased estimator of $$k?$$

iii) how does $$X_1 + X_2$$ for an unbiased estimator of $$k?$$

My thoughts:

I think I know how to do the unbiased estimator problems as you just have to put in the estimator in to $$E(X)$$ and show it equals to whatever the parameter may be. However I am confused to what the parameter here is to begin with, are they assuming $$k$$is the parameter? So for the problem iii), do I go $$E(X_1+X_2)=E(X_1)+E(X_2)$$ and use the mean results from continuous uniform distributions?

The question i) and ii), I am completely stuck, I do not know where to go with this problem.

I would appreciate the help offered.

• How about starting by writing the likelihood function of the observation Feb 26, 2020 at 6:24

i) Your variables are uniform on $$[0,k]$$. You can see this from the density you wrote.
ii) $$M = max\{X_1, X_2\}$$ is not an unbiased estimator for $$k$$. To see this you need to obtain $$M$$'s density and from there calculate E(M). You can derive the density from the fact that $$F_M(x) = P(X_1 \leq x, X_2 \leq x) = \frac{x}{k}\frac{x}{k}$$. $$F_M(x) = \frac{x^2}{k^2} \Rightarrow f_M(x) = 2\times \frac{x}{k^2} \Rightarrow E(M) = \int_0^k2\times \frac{x^2}{k^2}dx= \frac{2}{3}k.$$
iii) $$E(X_1 + X_2) = E(X_1) + E(X_2) = \frac{k}{2} + \frac{k}{2} = k$$ I.e. $$X_1 + X_2$$ is an unbiased estimate for $$k$$.
• Thank you, may I have a more clear explanation for i)? Also why can two independent distributions be multiplied to form a new CDF when it is $\max (X_1 ,X_2)$? this is the part I do not get. I appreciate it. Feb 26, 2020 at 7:15
• @AuroraBorealis i)the density of a uniform distribution on $[a, b]$ is $f(x) = \frac{1}{b-a}$. Your density is $f(x) = \frac{1}{k}$ on $[0, k]$. I.e. for k =b, 0=a the two functions are identical. ii) The event $x \geq max\{X_1, X_2\}$ is equivalent to the event $X_1 \leq x, X_2 \leq x$. Because $X_1$ and $X_2$ are independent, $P(X_1 \leq x, X_2 \leq x) = P(X_1 \leq x)P(X_2 \leq x) = F_{X_1}(x) \times F_{X_2}(x)$. Feb 26, 2020 at 8:16
• I also have one more question, when in the mark scheme it states that $Var(\frac{3}{2}M) = \frac{k^2}{8}$. I mean I understand why $Var(X_1+X_2)=\frac{k^2}{6}$ Could you please show me why $Var(\frac{3}{2}M) = \frac{k^2}{8}$ this holds? thank you Feb 26, 2020 at 10:33