Question
If $X_{0}$ has the Poisson distribution $\pi(\alpha)$, then$$\lim _{\alpha \rightarrow \infty} P\left\{\frac{X_{\alpha}-\alpha}{\sqrt{\alpha}} \leq u\right\}=\Phi(u)$$for every $u$. [Hint: use the Laplace transform $E\left(e^{-\lambda\left(X_{\alpha}-\alpha\right) / \sqrt{\alpha}}\right)$, show that as $\alpha \rightarrow \infty$ it converges to $e^{\lambda^{2} / 2}$, and invoke the analogue of Theorem 9 of $\S$ 7.5.]
Step 1
If \( X_0 \) follows a Poisson distribution with parameter \( \alpha \), then the probability mass function is given by: \[ P(X_0 = k) = \frac{e^{-\alpha} \alpha^k}{k!}, \quad k = 0, 1, 2, \ldots \] Show more…
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Let $X$ have a Poisson distribution with parameter $\lambda$ a. Show that the moment-generating function of $Y=(X-\lambda) / \sqrt{\lambda}$ is given by $$ m_{Y}(t)=\exp \left(\lambda e^{t / \sqrt{\lambda}}-\sqrt{\lambda} t-\lambda\right) $$ b. Use the expansion $$ e^{t / \sqrt{\lambda}}=\sum_{i=0}^{\infty} \frac{[t / \sqrt{\lambda}]^{i}}{i !} $$ to show that $$ \lim _{\lambda \rightarrow \infty} m_{Y}(t)=e^{t^{2} / 2} $$ c. Use Theorem 7.5 to show that the distribution function of $Y$ converges to a standard normal distribution function as $\lambda \rightarrow \infty.$
Sampling Distributions and the Central Limit Theorem
Summary
The zero-truncated Poisson distribution has probability function P(X = x) = (e^(-A) * A^x) / (x! * (1 - e^(-A))) and is denoted as X ~ ZTP(A). If U ~ ZTP(A1) independently of V ~ ZTP(A2), using the convolution formula, show that their sum W = U + V has probability function P(W = w) = (e^(-A1 - A2) * (A1 + A2)^w) / (w! * (1 - e^(-A1 - A2)))
Show that, similar to the de Moivre-Laplace approximation, the normal distribution is the limiting case for the Poisson distribution as the expected value goes to infinity.
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