Poisson Process Convergence in Distribution CLT 
For part 1, is the CLT applicable here? Since there's no square root of $k$ on the bottom. Does this mean the variance of $Y_k$ is $n$?
Part 3, how do you find limit in distribution when there are two variables in the function?
Thank you very much!
 A: For the Poisson distribution, the expected value is the same as the variance.  Since the rate is $1$, we have
$$
\operatorname{E}(N_k) = k
$$
and so that is also the variance.  Therefore the denominator is
$$
\sqrt{\operatorname{var}(N_k)} = \sqrt k.
$$
Observe that
$$
N_k = N_1 + \Big(N_2 - N_1\Big) + \Big(N_3-N_2\Big) + \cdots + \Big( N_k - N_{k-1}\Big)
$$
and


*

*every $\Big(\text{term between Big parentheses}\Big)$ has a $\mathrm{Poisson}(1)$ distribution; and

*all the $\Big(\text{terms between Big parentheses}\Big)$ are mutually independent.


These facts in bullet points follow from the nature of the Poisson process, and together they justify the application of the central limit theorem.
(However, it appears that you are being asked to prove this by using moment-generating functions rather than by using the central limit theorem.)
Your last question is a big oddly phrased: "how do you find limit in distribution when there are two variables in the function?" It's as if you think you shouldn't work on the problem until you know how you're going to do it.  But often you find out how you're going to do it only by working on it.  Before thinking about finding the limit in distribution, you might first think about finding the distribution.  Independence of increments implies $N_k$ is independent of $N_{2k}-N_k$, and that can tell you that
$$
N_{2k} - 2N_k = \Big( N_{2k}-N_k\Big) + \Big(-N_k\Big)
$$
is a sum of two independent random variables.  That enables you to find not only the expected value (for which you don't need to cite independence) but also the variance (for which you do).
