Part BCSIR NET December 2023sufficient-statistic-by-expansion

Sufficient statistic by expansion

For n ≥ 2, let $\varepsilon_{1}, \dots, \varepsilon_{n}$ be $i.i.d. N(0, \sigma^{2})$ and $Y_{i} = i\alpha + i^{2}\alpha^{2} + \varepsilon_{i}$ , where $\sigma > 0$ and $\alpha \in \mathbb{R}$ are unknown. Which of the following is a jointly minimal sufficient statistic for $(\alpha, \sigma)$ ?

A. $(\sum{}Y_{i}^{2}, \sum$ iY $_{i}, \sum i^{2}Y_{i})$ ✓ Correct
B. $(\sum{}Y_{i}^{2}, \sum$ iY $_{i}, \sum i^{2}Y_{i}^{2})$
C. $(\sum$ iY $_{i}, \sum i^{2}Y_{i}^{2})$
D. $(\sum{}Y_{i}, \sum$ iY $_{i})$

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests what the inference means.

See pricing

50 are analysed free — try those first.

The trap it tests

What the inference means

p-values, size, power and error types say specific things. This was not one of them.

Drill statements like this

Related counterexample: A sufficient statistic is complete

Sufficient statistic by expansion

More on this topic