Given that $X$ has a Poisson distribution with parameter $\lambda > 0$, the probability mass function (PMF) of $X$ is given by:

$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots$

(i) $P(X \text{ is even})$

Let $P(X \text{ is even}) = P(X = 0) + P(X = 2) + P(X = 4) + \dots$.

This can be expressed as: $P(X \text{ is even}) = \sum_{k=0}^{\infty} P(X = 2k)$ $P(X \text{ is even}) = \sum_{k=0}^{\infty} \frac{\lambda^{2k} e^{-\lambda}}{(2k)!}$

(ii) $P(X \text{ is odd})$

Let $P(X \text{ is odd}) = P(X = 1) + P(X = 3) + P(X = 5) + \dots$.

This can be expressed as: $P(X \text{ is odd}) = \sum_{k=0}^{\infty} P(X = 2k + 1)$ $P(X \text{ is odd}) = \sum_{k=0}^{\infty} \frac{\lambda^{2k+1} e^{-\lambda}}{(2k+1)!}$

Simplified Solution Using Generating Functions

To find $P(X \text{ is even})$ and $P(X \text{ is odd})$ more simply, we use the fact that the sum of probabilities for even and odd values should equal 1:

$P(X \text{ is even}) + P(X \text{ is odd}) = 1$

Also, using the identity: $e^{\lambda} = \sum_{k=0}^{\infty} \frac{\lambda^k}{k!}$

Consider: $e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{2k}}{(2k)!} \quad \text{and} \quad e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{2k+1}}{(2k+1)!}$

By using the generating function for the Poisson distribution, you can derive: $P(X \text{ is even}) = \frac{1}{2} \left(1 + e^{-2\lambda}\right)$ $P(X \text{ is odd}) = \frac{1}{2} \left(1 - e^{-2\lambda}\right)$

Thus, the probabilities are: $P(X \text{ is even}) = \frac{1}{2} \left(1 + e^{-2\lambda}\right)$ $P(X \text{ is odd}) = \frac{1}{2} \left(1 - e^{-2\lambda}\right)$

Summary

1. $P(X \text{ is even}) = \frac{1}{2} \left(1 + e^{-2\lambda}\right)$
2. $P(X \text{ is odd}) = \frac{1}{2} \left(1 - e^{-2\lambda}\right)$

Would you like more details on the derivation, or do you have any questions?

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1. How do you derive the generating function for a Poisson distribution?
2. Can the result be applied to any specific value of $\lambda$?
3. What happens when $\lambda$ is very large?
4. How does this relate to the expectation of the distribution?
5. How would the probabilities change if we consider $X$ being divisible by 3 instead?

Tip: The Poisson distribution is widely used in scenarios where events occur independently with a constant mean rate.