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$L$ is the upper limit of the sample distribution $[0, L]$ which is uniform and normal. how can I show that $L=\frac{(n+1)*max(X_i)}{n}$ is unbiased. and also has a lower MSE than MLE?

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I'll give you a hint for the first part of your question by telling you two useful results.

Theorem. Suppose $X_1,\ldots,X_n$ are i.i.d. random variables with cumulative distribution function $F$. Let $X=\max\{X_1, \ldots, X_n\}$, and let $G$ be the cumulative distribution function of $X$. Then $$ G(x) = F(x)^n $$ for all $x \in \mathbb{R}$.

Proof. Observe that $$ \max\{X_1, \ldots, X_n\} \leq x $$ if and only if $$ X_1 \leq x, \ldots, X_n \leq x. $$ Using this observation, it follows that $$ \begin{aligned} G(x) &= P(X \leq x) \\ &= P(X_1 \leq x, \ldots, X_n \leq x) \\ &= P(X_1 \leq x) \cdots P(X_n \leq x) &&\text{(by independence)}\\ &= \underbrace{F(x) \cdots F(x)}_{\text{$n$ times}} = F(x)^n. \end{aligned} $$

You can use this theorem to derive the cumulative distribution function of your estimator $$ \frac{n+1}{n} \max\{X_1, \ldots, X_n\}. $$ Once you have the cumulative distribution function, you can compute the expectation using the following result.

Theorem Let $X$ be a non-negative absolutely continuous random variable with density $g$ and cumulative distribution function $G$. Then $$ E[X] = \int_0^\infty (1 - G(x)) \, dx $$

Proof. By the definition of expectation and the Fubini-Tonelli Theorem of calculus, we have $$ \begin{aligned} E[X] &= \int_{\mathbb{R}} x g(x) \, dx &&\text{(definition of expectation)} \\ &= \int_0^\infty x g(x) \, dx &&\text{(since $X$ is non-negative)} \\ &= \int_0^\infty \left(\int_0^x 1 \, dy\right) g(x) \, dx \\ &= \int_0^\infty \int_0^x g(x) \, dy \, dx \\ &= \int_0^\infty \int_y^\infty g(x) \, dx \, dy &&\text{(changing the order of integration)} \\ &= \int_0^\infty P(X > y) \, dy \\ &= \int_0^\infty (1 - G(y)) \, dy \end{aligned} $$

This theorem can be applied to compute the mean of your estimator to show that it is unbiased.

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