Expectation of Conditional Expression for Exponentially Distributed RV

[user_01]

Given an Exponentially Distributed Random Variable $X\sim \exp(1)$, I need to find $\mathbb{E}[P_v]$, where $P_v$ is given as:$$ P_v=
\left\{
\begin{array}{ll}
a\left(\frac{b}{1+\exp\left(-\bar \mu\frac{P_s X}{r^\alpha}+\varphi\right)}-1\right), & \text{if}\ \frac{P_s X}{r^\alpha}\geq P_a,\\
0, & \text{otherwise}.
\end{array}
\right.
$$

**My Take:**

First, let's solve the equation for $P_v$. For that, let's assume g(x) to be:

$$g(x) = \frac{P^0}{\exp(\overline{\mu}P_{th} + \varphi)}\left( \frac{1+\exp(-\overline{\mu}P_{th} + \varphi)}{1 + \exp(-\overline{\mu}P_s x r^{-\alpha} + \varphi)} - 1\right),$$Then,

$$
P_v = \begin{cases}
g(x) & x \geq \frac{P_{th}}{P_s}r^\alpha\\
0 & x < \frac{P_{th}}{P_s}r^\alpha
\end{cases}
$$Then, with the knowledge that the PDF for Exponentially distributed RV is $f(x) = e^{-\lambda x}$ (with $\lambda = 1$ for our case), we can find $\mathbb{E}[P_v]$.

$$ \mathbb{E}[P_v]= \int_Q^\infty g(x)f(x)dx \ \ \ \ \ \ \ \ \ \ \ (1) $$
where $Q = \frac{P_{th} r^{\alpha}}{P_s}$.

Is this method correct or am I making any mistake?

 

[Valkarie]

**Answer:**Yes, this method is correct. To calculate $\mathbb{E}[P_v]$, you need to integrate $g(x)f(x)$ as you have done in (1).