I am reading the research paper [A New Bayesian Lasso], where $u$ has the distribution

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The expectation of $u_j$ is given by


I know that the term $\frac{1}{\lambda}$ is the expectation of the exponential distribution but where did the term $|\beta_j|$ come from?

I have tried to calculate $$\int _{|\beta_j|}^\infty u_j \lambda \exp(-\lambda u_j)du_j=\exp(-\lambda |\beta_j| )\left(\frac{1}{\lambda}+|\beta_j|\right)$$

which gives $\frac{1}{\lambda}+|\beta_j|$ multiplied by $\exp(-\lambda |\beta_j| )$.


1 Answer 1


This is correct: \begin{align} \int_\alpha^\infty u\lambda\exp\{-u\lambda\}\,\text du &= \int_\alpha^\infty u\frac{\text d[-\exp\{-u\lambda\}]}{\text du}\,\text du\\ &= \left[ -u\exp\{-u\lambda\}\right]_\alpha^\infty + \int_\alpha^\infty \exp\{-u\lambda\}\,\text du\\ &= \alpha\exp\{-\alpha\lambda\}-0+\frac{1}{\lambda}\exp\{-\alpha\lambda\}\\ &= (\alpha+\frac{1}{\lambda})\times \exp\{-\alpha\lambda\} \end{align} However, the density of the truncated Exponential distribution is $$\dfrac{\lambda\exp\{-u\lambda\}}{\exp\{-\alpha\lambda\}}$$ rather than $\lambda\exp\{-u\lambda\}$. This explains for the discrepancy.

Note also that a different explanation for the result is the memoryless property of the Exponential distribution. Once known that the Exponential variate $U$ is larger than $\alpha$, the excess above $\alpha$, $U-\alpha$ is again an Exponential variate with the same parameter.


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