# What is the expectation of the exponential distribution multiplied by indicator function?

I am reading the research paper [A New Bayesian Lasso], where $$u$$ has the distribution The expectation of $$u_j$$ is given by

$$\frac{1}{\lambda}+|\beta_j|$$.

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 $$\exp(-\lambda |\beta_j| )$$.

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.