How to find PDF and Expected value of max(x,0), for a random variable x

[user_01]

Let $a,b,c, \tau$ be positive constants and $x$ is an exponentially distributed variable with parameter $\lambda = 1$, i.e. $x\sim\exp(1)$.

\begin{equation}
E = \tau\Big[a\frac{1+a}{1+e^{-bx+c}} - 1 \Big]^+
\end{equation}


where $[z]^+ = \max(z,0)$

How can I find

1. The PDF for $E$
2. The expected value of E.

 

[Valkarie]

The PDF for $E$ can be found by using the probability density function of the exponential distribution and transforming it into a function for $E$. The PDF of $E$ can then be expressed as:$$f_E(x) = \tau\Big[a\frac{1+a}{1+e^{-bx+c}} - 1 \Big]^+ \cdot \lambda e^{-\lambda x}$$The expected value of $E$ can then be found by integrating the PDF of $E$ over its support. $$E[E] = \int_{-\infty}^{\infty} \tau\Big[a\frac{1+a}{1+e^{-bx+c}} - 1 \Big]^+ \cdot \lambda e^{-\lambda x}\,dx $$This integral can be solved using integration by parts.