Sufficient statistic for normal distribution (not iid) I am a bit confused with this exercise, since I never worked with samples of this type. I would appreciate if you can help me. The exercise is as follows:

Let $\{Xi\} \sim N(iθ, 1)$ for $i = 1, .... , n$ be an independent, but not identically distributed sample. Check that $T = \sum_iX_i$  it is a sufficient statistic for $θ$.

What I need is to verify that the $T$ statistic is sufficient for the $\theta$ parameter. I know how to do it with the Fisher and Neymann factorization theorem, but always with a identically distributed sample of random variables. In this case, the sample is not identically distributed. Therefore, I don't know how to verify it.
 A: We have that the probability density function of $X_i$ is
$$f_{X_i}(x_i) = \dfrac{1}{\sqrt{2\pi}}\exp\left[ -\dfrac{1}{2}\left(x_i-i\theta \right)^2\right]\text{.}$$
By independence, the joint density is given by
$$f_{X_1, \dots, X_n}(x_1, \dots, x_n) = \prod_{i=1}^{n}f_{X_i}(x_i) = \dfrac{1}{(2\pi)^{n/2}}\exp\left[-\dfrac{1}{2}\sum_{i=1}^{n}(x_i - i\theta)^2 \right]\text{.}$$
Now we expand the sum:
$$\sum_{i=1}^{n}(x_i - i\theta)^2 = \sum_{i=1}^{n}x_i^2 - 2\theta\sum_{i=1}^{n}ix_i + \theta\sum_{i=1}^{n}i^2\text{.}$$
Recalling the sum
$$\sum_{i=1}^{n}i^2 = \dfrac{n(n+1)(2n+1)}{6}$$
we obtain
$$\sum_{i=1}^{n}(x_i - i\theta)^2 = \sum_{i=1}^{n}x_i^2+\theta\left[\dfrac{n(n+1)(2n+1)}{6}- 2\sum_{i=1}^{n}ix_i\right]\text{.}$$
The joint density may thus be written as
$$\underbrace{\dfrac{1}{(2\pi)^{n/2}}\exp\left[-\dfrac{1}{2} \sum_{i=1}^{n}x_i^2\right]}_{h(\mathbf{x})}\underbrace{\exp\left\{ \theta\left[\dfrac{n(n+1)(2n+1)}{6}- 2\sum_{i=1}^{n}ix_i\right]\right\}}_{\varphi(\sum_{i=1}^{n}ix_i, \theta)}\text{.}$$
By the factorization criterion, $\sum_{i=1}^{n}iX_i$ is sufficient for $\theta$.
Unless there's something I'm missing here, I don't think $\sum_{i=1}^{n}X_i$ is sufficient for $\theta$.
