Decide which estimator is "better" (in terms of MSE - "Mean Square Error") than the other I was given the following question in my exam and i wonder if you can find an efficient way to solve this.
We take a sample $X_1,...,X_n, (n>1),$ of iid random variables, with the following PDF:
$$
f(x, \theta) =
\begin{cases}
\frac{2x}{\theta^2}  && 0 \leq x \leq \theta \\
0 && else
\end{cases}
$$
where $\theta$ is the unknown parameter.
We mark:
$$
X_{(n)}=max\{X_1,...,X_n\}
$$
Now, given the following estimators of parameter $\theta$:
$$
T_1 = X_{(n)} \\
T_2 = \frac{n+1}{n} \cdot X_{(n)}
$$
we need to $\underline{\text{decide whether $MSE(T_1) > MSE(T_2)$ or $MSE(T_1) < MSE(T_2)$}}$ . Plus, we need to $\underline{\text{decide whether $T_2$ is biased or not.}}$
$$$$
This was my method so far:

*

*Find the CDF of $X$:
$$F_x(t) = \int_0 ^t f(x,\theta)dx = \int_0 ^t \frac{2x}{\theta^2}dx=\frac{t^2}{\theta^2} \implies
F_x(t) = 
\begin{cases}
\frac{t^2}{\theta^2}  && 0 \leq x \leq \theta \\
0 && else
\end{cases}
$$


*Find the CDF of $X_{(n)}$:
$$
F_{X_{(n)}}(t) = 
\prod _{i=1} ^{n} F_{x_i}(t) = 
\left( \frac{t^2}{\theta^2} \right)^n = 
\frac{t^{2n}}{\theta^{2n}} 
$$


*Find the PDF of $X_{(n)}$ (derivation):
$$
f_{X_{(n)}}(t) = 2n \cdot \frac{t^{2n-1}}{\theta ^ {2n}}
$$


*Find the expected value of $X_{(n)}$:
$$
E(X_{(n)}) = \int_0^\theta t \cdot f_{X_{(n)}}(t)dt = \int_0^\theta t \cdot 2n \cdot \frac{t^{2n-1}}{\theta ^ {2n}}dt = \frac{2n}{2n+1} \cdot \theta
$$


*Find the second moment of $X_{(n)}$:
$$
E \left( X_{(n)}^2 \right) =
\int_0^\theta t^2 \cdot f_{X_{(n)}}(t)dt = 
\int_0^\theta t^2 \cdot 2n \cdot \frac{t^{2n-1}}{\theta ^ {2n}}dt =
\frac{n}{n+1} \cdot \theta^2
$$


*Find $MSE's$:
$$
MSE(T_1) = MSE(X_{(n)}) = E\left( X_{(n)}^2 \right) - 2 \theta \cdot E\left( X_{(n)} \right) + \theta^2 = ...^{\text{long calculation}}... = \frac{\theta^2}{(n+1)(2n+1)}
$$
$$
MSE(T_2) = MSE\left( \frac{n+1}{n} \cdot X_{(n)} \right) =
\left( \frac{n+1}{n} \right)^2 E\left( X_{(n)}^2 \right) - \left( \frac{n+1}{n} \right) \cdot 2 \theta \cdot E\left( X_{(n)} \right) + \theta^2 =  ...^{\text{long calculation}}... =
\frac{\theta^2}{n(2n+1)}
$$
And now we can finally conclude that:

*

*$MSE\left(T_1\right) < MSE\left(T_2\right)$

*$E\left(T_2\right) = E\left( \frac{n+1}{n} \cdot X_{(n)} \right) = 
\frac{n+1}{n} \cdot E\left( X_{(n)} \right) = 
\frac{n+1}{n} \cdot \frac{2n}{2n+1} \cdot \theta = 
\frac{2(n+1)}{2n+1} \theta  \neq 
\theta  \implies 
\text{$T_2$ is $\mathbf{Biased}$}
$
$$$$
My exam included 10 questions and we were given 2.5 hours to solve them all. This question took me almost 30 minutes, which is a lot!! I know it's not a difficult one, but it is very confusing. You can easily miscalculate one step and you'll reach the wrong answer.
Usually, all those questions hides a "trick", something that can ease the question and spare you a lot of time. Tricks like, i don't know, maybe "modifying" $X_{(n)}$ a little bit (adding or multiplying it with some constant) and reach some known distribution, and then the next steps are much more easier to handle. For example, maybe $\alpha \cdot X_{(n)} \sim U(0, \theta)$ for some $\alpha$. Then, we can say that $E\left( X_{(n)} \right) = E\left( \frac{1}{\alpha} \cdot \alpha \cdot X_{(n)} \right) = \frac{1}{\alpha} E\left( \alpha \cdot X_{(n)} \right) = \frac{1}{\alpha} \cdot \frac{\theta}{2}$, all thanks to that fiding $E\left( \alpha \cdot X_{(n)} \right)$ is straightforward.
Another thought crossed my mind- Maybe i don't need to fully-calculate the MSEs? After all, the question only asked which one is bigger. Is there a way to determined such a thing without actually calculate the whole MSEs? $T_1$ and $T_2$ are both biased, so i can't even guess if one is "better" (in terms of MSE) than the other.
Any ideas? How would you solve it? I'd love to hear your thoughts.
 A: Indeed you don't have to calculate both $MSE$'s. You are only intreseted in which one is bigger, so you want to know the sign of
$$ MSE(T_2) - MSE(T_1) = \mathbb E [(T_2 - \theta)^2 - (T_1 - \theta)^2 ] \\= 
\mathbb E [(T_2  + T_1 - 2\theta) (T_2 - T_1) ] = 
\mathbb E [(\tfrac {2n + 1} n X_{(n)} - 2\theta) (X_{(n)} / n) ] \\
=1/n(\tfrac {2n + 1} {n} \mathbb E [ X_{(n)}^2] - 2\theta\mathbb E [X_{(n)}] ).$$
Using your calculation we have in parenthesis $\tfrac {2n + 1} {n + 1}\theta^2 - \tfrac {4n} {2n + 1}\theta^2$ which is positive.
Personally I would use some "standardization" for this problem as you suggested. The steps would be all in all the same, but calculations would be easier.
Note that given distribution family is one with scale parameter (with $\theta$ being it: its PDF is in the form of $f(x/\theta) /\theta$, with $f(t) = 2t\mathbf{1}_{[0, 1]}(t)$). So it is resonable to consider $Y_k = X_k / \theta$. Then $Y_k$ has CDF $F(x) = x^2$ for $0<x<1$ which is easier to deal with. Since we only use multiplication we have $X_{(n)} = \theta Y_{(n)}$, so finding distribution of $Y_{(n)}$ will give us distribution of $X_{(n)}$ for free. Let's see:
CDF of $Y_k$ is $F_{Y_k}(t) = t^2$ for $0 < t < 1$, and therefore CDF of $Y_{(n)}$ is $F(t) = t^{2n}$. PDF of the latter is $f(x) = 2nx^{2n-1}$, which yields $$\mathbb E Y_{(n)} = \int_0^1 2nx^{2n} dx = \tfrac {2n}{2n + 1}, \\
\mathbb E Y_{(n)}^2 = \int_0^1 2nx^{2n + 1} dx = \tfrac {2n}{2n + 2}.$$
Now we can again plug-in our results:
$$ MSE(T_2) - MSE(T_1) = \ldots \text{like before} \ldots = 
1/n(\tfrac {2n + 1} {n} \mathbb E [ X_{(n)}^2] - 2\theta\mathbb E [X_{(n)}] )\\ = 
\theta^2/n(\tfrac {2n + 1} {n} \mathbb E [ Y_{(n)}^2] - 2\mathbb E [Y_{(n)}]
)$$
and get the same answer.
We could use standardized version to check for $T_2$ being biased again using linearity of expected value:
$$ \mathbb E T_2 = \mathbb E\tfrac {n + 1} n X_{(n)}= \theta\tfrac {n + 1} n \mathbb EY_{(n)}.$$
However those steps provide only some computational aid, and would make solution only a bit shorter and harder to pick up a mistake. But on exam it could make a real difference.
