# Sampling distribution of $n(\hat{\theta}-\theta)$

I have $X_1, X_2,...,X_n$ i.i.d on the discrete uniform distribution: $$f(x;\theta) = \frac{1}{\theta+1},~\theta \in Z_+,~ x=0,1,...,\theta$$ and derived the MLE estimator for $\theta$ as: $$\hat{\theta}=\max\{X_1, X_2,...,X_n\}$$ Now, ordinarily, to determine the sampling distribution of $n(\hat{\theta}-\theta)$, I would begin with the CDF of $\hat{\theta}$ and then defining $W=n(\hat{\theta}-\theta)$, use the change-of-variables formula to find the density.

However, in this case, because the distribution is discrete, I have run into a roadblock. Is anyone able to point me in the right direction?

My attempt thus far:

Since $X_1,...,X_n$ are i.i.d., $$P(\hat{\theta} \leq x) = P(X_1 \leq x,...,X_n \leq x) = \prod^n_{i=1} P(X_i \leq x) = [F_X(x)]^n$$ With $F_X(x)=\frac{1+x}{1+\theta}$, the CDF of $\hat{\theta}$ follows.

• You aren't very clear about what the difficulty is. Did you work out the cdf of $\hat{\theta}$? – Glen_b May 19 '18 at 14:16
• I have edited in my attempt at finding the cdf of $\hat{\theta}$. I am only just beginning to learn this content so am not quite sure how to proceed from there/whether the cdf is correct to begin with. – euqin May 19 '18 at 15:57
• Now you can find the probability mass function by the finite difference $P(\hat{\theta}\le x) - P(\hat{\theta}\le x-1$ and unless $theta$ is enormous you can find everything from there. – kjetil b halvorsen May 19 '18 at 16:06
• For continuous 'probability density function' $$f(x) = \frac{d F(x)}{dx}$$ For discrete 'probability mass function' $$f(x_n) = F(x_n) - F(x_ {n-1})$$ – Martijn Weterings May 22 '18 at 19:33