If X has a Poisson distribution with parameter λ > 0. Find (i) P(X is even) (ii) P(X is odd).

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?

---

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.

Explore the probabilities of even and odd values in a Poisson distribution with parameter λ > 0. Understand the Poisson distribution PMF and its generating function. Suitable for advanced undergraduate students and beyond.

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