# Maximum Likelihood Estimator and Cumulative distribution for an i.i.d distribution $P_θ(x; θ)$

Assume the standard situation, that is, let $$X_1, . . . , X_n$$ be independent and identically distributed with $$X_k ∼ Pθ(x; θ)$$ , where $$P_θ(x; θ) = 2x/θ^2$$ if $$0 ≤ x ≤θ$$ and $$0$$ otherwise

It is required to estimate θ . Show that the maximum likelihood estimator for $$θ$$ is $$\hat{θ} = max{[X_1, . . . , X_n]}$$ and then show that the cumulative distribution function of $$\hat{θ}$$ is $$F_θ(z) = z^{2n}/θ^{2n}$$

Here's what i did so far:

Maximum Likehood estimator: $$L_x(θ) = \prod_{i = 1}^{n} P_θ(x_k)$$ Here we have $$P_θ(x_1,...,x_n;θ) = P_θ(x_1,θ).P_θ(x_2,θ)...P_θ(x_n,θ)$$

Likehood = $$L_{x,θ}(θ) =P_θ(x_1,θ).P_θ(x_2,θ)...P_θ(x_n,θ)= 2x_1/θ^2. 2x_2/θ^2.. 2x_n/θ^2 = [2^n.\prod_{i = 1}^{n}x_i]/θ^{2n}$$

Log-likehood: $$\sum_{i = 1}^{n}log(P_θ(x_1,...x_n;θ))= \sum_{i = 1}^{n}log(2x_i/θ^2)$$

Is this correct so far? Im still not sure how to get to $$\hat{θ} = max{[X_1, . . . , X_n]}$$

As for the the cumulative distribution part to show $$F_θ(z) = z^{2n}/θ^{2n}$$:

$$F(z) = P(max({x_k})

Not sure if this is correct. Would really appreciate some help.

Edit: From the answers below, we can deduce the the estimator is biased. What estimator would be unbiased? How can i find it?

• Argue that the likelihood is a decreasing function of $\theta$, so it is maximised for the minimum possible value of $\theta$. For the second part, you have the right idea but the probabilities are not correct. – StubbornAtom Feb 14 at 19:52

Given a sample $$x\equiv \{x_i\}_{i=1}^n$$, the likelihood is $$L(\theta\mid x)=\left(\frac{2}{\theta^{2}}\right)^n\prod_{i=1}^n x_i \times1\{\theta\ge M(x),m(x)\ge 0\},$$ where $$M(x):=\max_{1\le i\le n}x_i$$ and $$m(x):=\min_{1\le i\le n}x_i$$. The indicator suggests that $$\hat{\theta}_n(x)\ge M(x)$$ ($$\because$$ $$L=0$$ otherwise). However, taking values larger than $$M(x)$$ decreases $$L$$ because of the first term (assuming that $$m(x)> 0$$). Thus, $$\hat{\theta}_n(x)= M(x)$$.

As for the distribution of $$\hat{\theta}_n$$, for $$z\in [0,\theta]$$, $$F(z)=\mathsf{P}(\hat{\theta}_n\le z)=\prod_{i=1}^n\mathsf{P}(X_i\le z)=\prod_{i=1}^n \left(\frac{z}{\theta}\right)^{2}=\left(\frac{z}{\theta}\right)^{2n}.$$

Since $$\mathsf{E}X_i=2\theta/3$$, examples of an unbiased estimator are $$\hat{\theta}_n'=\frac{3}{2}\times \frac{1}{n}\sum_{i=1}^n x_i, \quad \hat{\theta}_n''=\frac{2n+1}{2n}\hat{\theta}_n.$$

• Very helpful. Thank you – Neels Feb 14 at 20:17
• I have another question if you dont mind answering, how can it be deduced that the estimator is biased, and what would be an unbiased estimator(how do i find it)? – Neels Feb 14 at 21:06
• The first part im assuming is quite obvious; its simply that M(x) does not equal θ. Is this correct? – Neels Feb 14 at 21:06
• Using the expression for $F(z)$: $$\mathsf{E}\hat{\theta}_n=\int_0^{\theta}(1-F(z))dz=\frac{2n\theta}{2n+1}\ne\theta.$$ However, it's asymptotically unbiased since $\mathsf{E}\hat{\theta}_n\to \theta$. – d.k.o. Feb 14 at 21:09
• Thanks a lot. May i ask, is there are formula that suggest E(θ^n) = the integral of (1-F(x))? And is this the standard method to find unbiased estimators? – Neels Feb 14 at 21:13