Calculating probability of floor'ed log-normal distribution I want to programmatically draw a discrete random sample from a log-normal distribution. In order to do so, I calculate a $x$-value from uniform random source, e.g. some pseudo random number generator. This I use as input to the log-normal $PDF$ for a random variable $X \sim \ln\mathcal{N}(\mu, \sigma)$ with some parameters $\mu$ and $\sigma$.
To get discrete values from this I round down the result, therefore I would have a new a random variable:
$Y \sim \lfloor \ln\mathcal{N}(x;\mu,\sigma)\rfloor$ 
My question is now, how to calculate the probability that an observed result $y_0$ came from this distribution. 
My attempt is as follows:
In my understanding the floor function of the value has the effect of equaling the probabilities in an open interval $[y, y+1)$, therefore the probability could be calculated as 
$P(Y=y_0) = P(y_0 \le X < y_0+1) = P(X < y_0+1) - P(X < y_0)$.
As the log-normal distribution is continuous, $\forall y: P(X=y) = 0$. 
Therefore
$P(X < y_0+1) - P(X < y_0) = P(X \leq y_0+1) - P(X \leq y_0)$. 
Hence, the probability of the observed value coming from the floor'ed variable $Y$ could be calculated as 
$P(Y=y_0) = P(X \leq y_0 + 1) - P(X \leq y_0)$.
This could be easily done by using the log-normal $CDF$. 
Am I looking at this the right way? Is this the correct solution? 
 A: The OP has implicitly used a step-length equal to $1$ for the support of its discrete random variable. This in practice may be too large a step, especially if the standard deviation is small. More generally, and staying in the framework of non-negative random variables we can define the support of the discrete r.v. as
$$S_Y = \{0, h, 2h , 3h,...\},  h \in \mathbb Q_+$$
... where to keep things  manageable, we designated a support equally spaced (although this is by no means necessary). 
Since the OP rounds down the values he obtains, we have 
$$\text {Prob}(Y = kh) = \text {Prob}(X < (k+1)h) - \text {Prob}(X < kh), \;\;k=0,1,...$$
and due to the continuity of $X$
$$\text {Prob}(Y = kh) = \text {Prob}(X \leq (k+1)h) - \text {Prob}(X \leq kh)$$
which in our case becomes
$$\text {Prob}(Y = kh) = \Phi\left(\frac {\ln [(k+1)h] - \mu}{\sigma}\right)-\Phi\left(\frac {\ln [kh] - \mu}{\sigma}\right)$$
where $\Phi()$ is the standard normal CDF.
Note that $k$ should not necessarily follow the Naturals one by one - we could have for example again an equally spaced support like  $S_Y = \{0,  2h , 4h, 6h,...\}  $ etc.  
Looking at coding, the above formulation provides flexibility in exploring the structure of the support that best fits the needs.
