If $X_1, X_2\ldots,X_n$ are independent variables with $X_i \sim \mathcal N(i\theta,1)$, $\theta$ is an unknown parameter. What is a one dimensional sufficient statistic $T$ of this sample?
I have a intuition guess that the answer is $\frac{1}{n}\sum_{i=1}^n \frac{X_i}{i}$, but I don't know how to prove it through definition or get it using factorization. Can anyone give me a hint on how to derive it?
Thanks!
$$f_{X_1,\ldots,X_n}(x_1,\ldots,x_n \mid \theta) = (2\pi)^{-n/2} \prod_{i=1}^n \exp\left( -\frac{1}{2} (x_i - i\theta)^2 \right) $$
$$= (2\pi)^{-n/2} \exp\left( -\frac{1}{2} \sum_{i=1}^n(x_i - i\theta)^2 \right) $$
$$= (2\pi)^{-n/2} \exp\left( -\frac{1}{2} \left(\sum_{i=1}^n x_i^2 -2\theta\sum_{i=1}^n ix_i + \theta^2\sum_{i=1}^n i^2 \right)\right) $$
$$= (2\pi)^{-n/2} \exp\left( -\frac{\displaystyle\sum_{i=1}^n i^2}{2} \left(\frac{\displaystyle\sum_{i=1}^n x_i^2}{\displaystyle\sum_{i=1}^n i^2}-\left(\frac{\displaystyle\sum_{i=1}^n ix_i}{\displaystyle\sum_{i=1}^n i^2}\right)^2 +\left(\frac{\displaystyle\sum_{i=1}^n ix_i}{\displaystyle\sum_{i=1}^n i^2}\right)^2 -2\theta\frac{\displaystyle\sum_{i=1}^n ix_i}{\displaystyle\sum_{i=1}^n i^2} + \theta^2 \right)\right) $$
$$= (2\pi)^{-n/2} \exp\left( -\frac{1}{2} \left(\displaystyle\sum_{i=1}^n x_i^2-\frac{\left(\displaystyle\sum_{i=1}^n ix_i\right)^2}{\displaystyle\sum_{i=1}^n i^2} \right)\right)\exp\left( -\frac{\displaystyle\sum_{i=1}^n i^2}{2} \left(\frac{\displaystyle\sum_{i=1}^n ix_i}{\displaystyle\sum_{i=1}^n i^2} - \theta\right)^2 \right). $$
You are interested in the last $\exp$ term, so $\displaystyle\sum_{i=1}^n ix_i$ is a sufficient statistic for $\theta$.