# Maximum likelyhood of distribution

$$L$$ 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?

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.