We have shown that if $x_1, ..., x_n$ are an iid sample from $N(\theta, 1)$ then $\bar{x}$ is sufficient for $\theta$.

We want to show that the conditional distribution of $X_1, ... X_n$ given $T = t = \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is free of $\theta$. First we can show, for example using the [moment generating function technique](https://newonlinecourses.science.psu.edu/stat414/node/172/), that:

$$\sum_{i=1}^n x_i \equiv n \bar{x} \equiv nt \sim N(n \theta, n)$$

so $T \sim N(\theta, 1)$. Then, using the Bayes theorem we have 

\begin{align}
P(X_1 = x_1, ..., X_n = x_n | T = t) & = \frac{P(X_1 = x_1, ..., X_n = x_n, T = t)}{P(T = t)} \\
& = \frac{P(X_1 = x_1, ..., X_n = x_n)}{P(T = t)} \\
& = \frac{\sim N(\theta, 1)}{\sim N(\theta, 1)} \\
& = 1
\end{align}

which is free of $\theta$.

61cc1b49b045ed0067b6c5cb933fe0905e4a2ead
ipynb
Jupyter Notebook
python/chapter-3/exercises/EX3-3.ipynb
covuworie/in-all-likelihood
6638bec8bb4dde7271adb5941d1c66e7fbe12526
MIT
python/chapter-3/exercises/EX3-3.ipynb
covuworie/in-all-likelihood
6638bec8bb4dde7271adb5941d1c66e7fbe12526
MIT
2020-03-24T17:53:04.000Z
2021-08-23T20:16:17.000Z
python/chapter-3/exercises/EX3-3.ipynb
covuworie/in-all-likelihood
6638bec8bb4dde7271adb5941d1c66e7fbe12526
MIT
2021-11-21T10:24:59.000Z
2021-11-21T10:24:59.000Z
Qwen/Qwen-72B
1. YES
2. YES