Monte Carlo simulation for fitting distributions (Weibull and log-normal)

Question

I am working on a computer project which needs statistical analysis and I am not much of a statistic person. I have a Bluetooth (BT) detector device which detects passing Bluetooth devices (i.e one BT passed at time 0, one BT passed at time 3,....) and I used AIC to select the best distribution which describes the interarrival times. Now, I would like to do a Monte Carlo simulation to see what is the effect of detectable BT ratio (shown by lambda below and increased by 0.05 at each step) on the outcome. I have two options:

1. D = distribution with lowest AIC which describes interarrival times 
   theta = paramters of D which obtained by mle
   z = number of detected BT 
    for (lambda in 0.05,0.1,0.15,..., 1)
   {
       P=generate z*lambda random numbers with D and theta
       (like Lnorm(z*lambda,meanlog=10,sdlog=2) in r)
       Find the best distribution and its paramters which fits P for further processing
   }
   

2. this second approach is suggested by a statistician fiend:

   Assume that you have fitted a log-Normal distribution to the BT data, and found parameters $\mu$ and $\sigma^2$ for the BT interarrival times:

   $$ \newcommand{\Var}{{\rm Var}} \newcommand{\LogNormal}{{\rm LogNormal}} \newcommand{\shape}{{\rm shape}} \newcommand{\scale}{{\rm scale}} Y \sim \LogNormal(\mu, \sigma^2) $$

   with mean and variance of the BT interarrival times are

   \begin{align} E[Y] &= \exp(\mu + 0.5*\sigma^2) \\ \Var[Y] &= (\exp(\sigma^2)-1) \exp(2*\mu + \sigma^2) \end{align} However you know that only a proportion lambda of devices are actually detected. That means that the true BT interarrival times are shorter than this. If X is the true interarrival time random variable then X is approximately $\lambda * Y$ on average, so you expect

$E[X] = \lambda * E[Y]$ and $\Var[X] = \lambda^2 * Var[Y]$

i.e., $E[X] = \lambda*\exp(\mu + 0.5*\sigma^2) = \exp((\mu+\delta) + 0.5*\sigma^2)$ and $\Var[X] = \lambda^2 * (\exp(\sigma^2)-1) \exp(2*\mu + \sigma^2) = (\exp(\sigma^2)-1) \exp(2*(\mu+\delta) + \sigma^2)$
where $\delta = \log(\lambda)$.
This suggests that
$X \sim \LogNormal(\mu+\delta, \sigma^2)$

Now, I have three questions and appreciate any help.

1. I cannot understand how he concluded the last line (i.e $X \sim \LogNormal(\mu+\delta, \sigma^2)$) (SOLVED)

2. How would the second approach work for the Weibull distribution, I know that:

   \begin{align} E[\ln(W)] &= \ln(\scale)- (\gamma /\shape), \\ \Var[\ln(W)] &= (\pi^2)/(6 * \shape^2), \end{align} where $\gamma$ is the Euler constant, but I do not know how to extend this and find the effect of lambda on scale and shape in a similar way to the second approach. (SOLVED)

3. Why did my friend suggest the second approach, and what is wrong with the first approach?