TITLE: mean of the posterior distribution of $\theta$
QUESTION [0 upvotes]: Let $X_1, X_2,\dots, X_n$ be i.i.d. $N(\theta, \sigma^2)$, and let $\theta$ have double exponential distribution, i.e. $\pi(\theta)=e^{|\theta|/a}/(2a)$, $a$ is known. Find the mean of the posterior distribution of $\theta$.

The posterior distribution, $p(\theta)=c\cdot \exp\Big[-\dfrac{n\theta^2-n\theta \bar{x}}{2\sigma^2}-\dfrac{|\theta|}{a}\Big]$, $c$ is a constant.
To find posterior mean, I need to find $c$. But don't know how to get that because of $|\theta|$. Thanks.

REPLY [1 votes]: Following along with Ross' comment, writing out:
\begin{equation}
p(\theta)\propto \textrm{Exp}(-\frac{n\theta^2}{2\sigma^2}+\theta(\frac{n\bar{X}}{2\sigma^2}-\frac{\textrm{Sign}(\theta)}{a}))
\end{equation}
and completing the square could also help you understand what the solution is. 
There is an intimate relationship with the LASSO and soft thresholding operators, which you can see by viewing the MAP estimate as the maximizer of $\log(p(\{X_i\}_{i=1}^n|\theta,\sigma^2)p(\theta_)$, which is the minimizer of:
\begin{equation}
\sum_{i=1}^n\frac{1}{2\sigma^2}(X_i-\theta)^2 + \frac{|\theta|}{a}
\end{equation}
Which is the LASSO solution, that you can read about the soft thresholding in that link.
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