Find $c \geq 0$ so that $c\hat{\vartheta}$ is unbiased

Question

I have found the following statistical model.

Say $\Omega:=[0,\infty)^{n}$ and $P_{m}$~$[\mathcal{U}(0,m)]^{\otimes n}$ where $m$ is the parameter $\in [0,\infty)$ and $n \in \mathbb N$

Define the estimator $\hat{\vartheta}(\omega_{1},...,\omega_{n})=\max\{\omega_{1},...,\omega_{n}\}$

Find a $c \geq 0$ so that $\operatorname{Bias}_{c\hat{\vartheta}}(P_{m})=0$

I worked out :

$\mathbb E_{P_{m}}(\hat{\vartheta})=\int_{0}^{\infty}P_{m}(\hat{\vartheta}\geq t)dt=\int_{0}^{m}P_{m}(\hat{\vartheta}\geq t)dt=\int_{0}^{m}(1-P(\hat{\vartheta}< t))dt=[t-\frac{1}{n+1}(\frac{t}{m})^{n+1}\vert_{0}^{m}]=m-\frac{1}{n+1}$

So surely, in order

$0=\operatorname{Bias}_{c\hat{\vartheta}}(P_{m})=\mathbb E_{P_{m}}[c\hat{\vartheta}]-c\vartheta(P_{m})=c\mathbb E_{P_{m}}[\hat{\vartheta}]-c\vartheta(P_{m})=c(m-\frac{1}{n+1})-cm$

But isn't $c$ superfluous here?

Answer 1

First of all, $\int_0^m 1-\left(\frac{t}{m}\right)^n \,dt = [t-\frac{m}{n+1}(\frac{t}{m})^{n+1}\vert_{0}^{m}]=m-\frac{m}{n+1}=m\frac{n}{n+1}$.

Bias of any estimator is the difference between its expectation and parameter. Parameter here is $\vartheta$, not $c\vartheta$. So you need to find $c$ s.t. $$ 0=\operatorname{Bias}_{c\hat{\vartheta}}(P_{m})=\mathbb E_{P_{m}}[c\hat{\vartheta}]-\vartheta)=c\cdot m\frac{n}{n+1}-m. $$