A random variable V is has a mixed distribution as follows:
$P(V=0)=\rho$
$P(V>v|V>0)=e^{-\lambda v}, v\ge0$
with $\rho=\lambda/\mu$ and $1>\mu>\lambda$
How to find the expectation $E[V$] and how to prove the integration of that pdf is 1? I have no clue. Thanks.
If we're told somewhere that $P(V\geq 0) = 1$ and $\lambda > 0,$ then we can conclude $$V = \rho\cdot \delta_0 + (1-\rho)\cdot X \quad \text{almost surely}$$ where $X$ is an exponential with mean $\lambda$.
By linearity of the expectation, $$E[V] = \rho\cdot E[\delta_0] + (1-\rho)\cdot E[X] = (1-\rho)\lambda$$
As for your second question, we're only told $P(V=0)$ and a distribution for $V | V>0$ that you already recognized as being an exponential, so there's no other pdf to integrate. The only thing we need to check for the above argument to go through is that $0 < \rho < 1$.