Problem:
$K_1, K_2,…$ is an independent, identically distributed sequence of Poisson random variables with $E[K] = 1$. $W_n = K_1 +…+ K_n$. Use the approximation:
$P[k_1 \leq K \leq k_2] = \phi{\frac{(k_2+0.5-E[K])}{\sigma_K}}-\phi{\frac{(k_1-0.5-E[K])}{\sigma_K}}$
to estimate $P[W = n]$ for $n = 4, 25, \text{and } 64$. Compare your approximations with the exact value of $P[W = n]$.
Attempt at Solution:
The equation for the probability mass function of a Poisson random variable is
$P_K(k) = \begin{cases}
\frac{\alpha^ke^{-\alpha}}{k!}, & k = 0, 1, 2, …;\\
0, & \text{otherwise.}
\end{cases}$
I know that for a Poisson random variable, $E[K] = Var[K] = \alpha$, so in this case, $\alpha = E[K] = Var[K] = \sigma_K = 1$. Further, by substitution, the probability mass function for any $K_i$ in the sequence is
$P_K(k) = \begin{cases}
\frac{1}{e(k!)}, & k = 0, 1, 2, …;\\
0, & \text{otherwise.}
\end{cases}$
This is the point at which I am unsure of how to go about solving the problem. I have a table of moment generating functions from which I can tell that the moment generating function for any $K_i$ is $M_K(k) = e^{e^s – 1}$, and since all $K_i$ are independent and identically distributed, the moment generating function for $W$ is $M_W(w) = (e^{e^s – 1})^n = e^{n(e^s – 1)}$. However, moment generating functions provide expected values, not probabilities, so I'm not sure if it is of any use for me to have derived this in the first place.
Also, I assume that to find an estimate for $n = 4$ (as an example), one would just substitute 3 for $k_1$ and 5 for $k_2$ in the approximation equation, but this yields $z$-values for $\phi(z)$ that exceed the values given in the standard normal table (which provides numerical values of $\phi(z)$ up to $\phi(3.49)$).
Best Answer
Personally I would use $\Phi(x)$ for the cumulative distribution function of a standard normal distribution and $\phi(x)$ for its density function.
You might find it easier if you knew that the sum of independent Poisson random variables is a Poisson random variable with parameter/mean/variance equal to the sum of the original parameters. So $n$ in this case, with $\Pr(W_n=n)=e^{-n}\frac{n^n}{n!}$.
You can see this with moment generating functions. The moment generating function of a Poisson distribution with parameter $\lambda$ is $e^{-\lambda(e^s-1)}$. So if $\lambda=1$ this is as you say $e^{-(e^s-1)}$ and the sum of $n$ i.i.d. examples of this has moment generating function $\left(e^{-(e^s-1)}\right)^n=e^{-n(e^s-1)}$, and so is a Poisson distribution with parameter $n$.
For example, with $n=64$ you need to compare $e^{-64}\frac{64^{64}}{64!}$ with $\Phi\left(\frac{64.5-64}{\sqrt{64}}\right)-\Phi\left(\frac{63.5-64}{\sqrt{64}}\right)$. They are both about $0.0498$. $\frac{1}{\sqrt{64}}\phi(0)$ is also close, and the central limit theorem leads to $e^{-n}\frac{n^n}{n!} \approx \frac{1}{\sqrt{2\pi n}}$ for large $n$, giving an alternative approach to Stirling's approximation to the factorial.