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.

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

1 Answer 1


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 $k$.

  • $\begingroup$ 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. $\endgroup$ Feb 26, 2020 at 7:15
  • 1
    $\begingroup$ @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, $P(X_1 \leq x, X_2 \leq x) = P(X_1 \leq x)P(X_2 \leq x) = F_{X_1}(x) \times F_{X_2}(x)$. $\endgroup$ Feb 26, 2020 at 8:16
  • $\begingroup$ Thank you for the clarification $\endgroup$ Feb 26, 2020 at 9:31
  • $\begingroup$ 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 $\endgroup$ Feb 26, 2020 at 10:33

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