If $x_1, ..., x_n$ are an iid sample from the double-exponential (Laplace) distribution with density:

$$p_{\theta}(x) = (2 \theta)^{-1} exp (-|x|/\theta)$$

then from the factorization theorem it is obvious that $$t(x) = \sum_i |x_i|$$ is sufficient.

We want to show that the conditional distribution of the data given T = t is free of $\theta$. According to the following discussion, it seems that we can infer that the [distribution of T](https://math.stackexchange.com/questions/199460/distribution-of-sum-of-absolute-value-of-random-variable) is given by $T \sim Gamma(n \theta, n)$. Then using Bayes theorem we have:


$$
\begin{align}
P(X_1 = x_1, ..., X_n = x_n | T = t) & = \frac{P(X_1 = x_1, ..., X_n = x_n, T = t)}{P(T = t)} \\
& = \frac{P(X_1 = x_1, ..., X_n = x_n)}{P(T = t)} \\
& = \frac{(2 \theta)^{-n} exp (-t/\theta)}{\Gamma (n)^{-1} (n \theta)^{-n} t^{n - 1 } exp (-t/\theta)} \\
& = \frac{\Gamma (n)}{t^{n - 1}} \bigg( \frac{2}{n} \bigg)^{-n}
\end{align}
$$

which is free of $\theta$ as expected.

93fb9ca3dfd3161ee11c1acae32b7007a469918a
ipynb
Jupyter Notebook
python/chapter-3/exercises/EX3-5.ipynb
covuworie/in-all-likelihood
6638bec8bb4dde7271adb5941d1c66e7fbe12526
MIT
python/chapter-3/exercises/EX3-5.ipynb
covuworie/in-all-likelihood
6638bec8bb4dde7271adb5941d1c66e7fbe12526
MIT
2020-03-24T17:53:04.000Z
2021-08-23T20:16:17.000Z
python/chapter-3/exercises/EX3-5.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