# Find an unbiased estimate of $\hat{\theta}$ which is a function of $T$

$(X,Y)$ is uniform over the triangular region with vertices at $(0,0)$,$(\theta, 0)$, and $(0,\theta)$, $\theta$ is unknown. Let $(X_i,Y_i)$ be iid as $(X,Y)$. Find a one dimensional sufficient statistic $T$ for $\theta$, and prove it's sufficient. Find an unbiased estimate of $\hat{\theta}$ which is a function of $T$.

## Attempt

$f_{X,Y}(x,y) = \frac{2}{\theta^2}$

$L(\theta| \mathbf{X},\mathbf{Y}) = (\frac{\theta^2}{2})^{-n} \prod I_{x_i+y_i < \theta}$

Sufficient statistic: $T = max_i (X_i+Y_i)$

$\hat{\theta} = ?$

Denote $Z =T= \max \limits_i (X_i + Y_i)$. Note that the joint pdf is $$f_{X,Y}(x,y) = \frac{2}{\theta^2} \mathbf{1}_{0 \le x \le \theta} \cdot \mathbf{1}_{0 \le y \le \theta} \cdot \mathbf{1}_{0 \le x+y \le \theta}$$ and so we find the CDF of $Z$: \begin{align*} P(X + Y \le z) = \begin{cases} \int_0^{z } \int_0^{x} \frac{2}{\theta^2} \; dy \; dx = \frac{z^2}{\theta^2}& 0 \le z \le \theta \\ 1 & z \ge \theta \end{cases} \end{align*} From the fact that $X_i + Y_i$ is iid, we get \begin{align*} P(Z \le z) = P(\max_i (X_i + Y_i) \le z) &= P(X_1 + Y_1 \le z,\dots,X_n + Y_n \le z) \\ &= P(X_1 + Y_1 \le z) \cdots P(X_n + Y_n \le z) \\ &= P(X+Y \le z)^n\\ &= \left( \frac{z^2}{\theta^2} \right)^n \cdot \mathbf{1}_{0 \le z \le \theta} \end{align*}
Finally, calculate $E[Z]$ . . . I'm getting $\frac{2n}{2n+1} \theta$, and therefore $\frac{2n+1}{2n} Z$ is unbiased for $\theta.$