We are given the probability mass function (PMF) of a random variable $X$, which is defined as:

$f(i) = \frac{c \lambda^i}{i!}, \quad i = 0, 1, 2, \dots$

where $\lambda$ is a positive constant. We are asked to:

1. Find the value of $c$.
2. Compute $P(X = 0)$.
3. Compute $P(X > 2)$.

Part (a) Finding $c$

Since this is a probability mass function, the sum of the probabilities over all possible values of $i$ must equal 1. That is:

$\sum_{i=0}^{\infty} f(i) = 1$

Substitute the expression for $f(i)$:

$\sum_{i=0}^{\infty} \frac{c \lambda^i}{i!} = 1$

Recognize that the sum $\sum_{i=0}^{\infty} \frac{\lambda^i}{i!}$ is the Taylor series expansion of $e^\lambda$. Thus, the above equation becomes:

$c e^\lambda = 1$

Solving for $c$:

$c = \frac{1}{e^\lambda}$

Part (b) Finding $P(X = 0)$

Using the PMF, we can compute the probability $P(X = 0)$:

$P(X = 0) = f(0) = \frac{c \lambda^0}{0!} = c$

From part (a), we know that $c = \frac{1}{e^\lambda}$, so:

$P(X = 0) = \frac{1}{e^\lambda}$

Part (c) Finding $P(X > 2)$

We want to calculate $P(X > 2)$, which is the sum of the probabilities for $i = 3, 4, 5, \dots$:

$P(X > 2) = 1 - P(X \leq 2)$

First, calculate $P(X \leq 2)$, which is the sum of the probabilities for $i = 0, 1, 2$:

$P(X \leq 2) = f(0) + f(1) + f(2)$

We already know $f(0) = \frac{1}{e^\lambda}$. Now calculate $f(1)$ and $f(2)$:

$f(1) = \frac{c \lambda^1}{1!} = \frac{1}{e^\lambda} \cdot \lambda$

$f(2) = \frac{c \lambda^2}{2!} = \frac{1}{e^\lambda} \cdot \frac{\lambda^2}{2}$

Thus, $P(X \leq 2)$ is:

$P(X \leq 2) = \frac{1}{e^\lambda} + \frac{\lambda}{e^\lambda} + \frac{\lambda^2}{2e^\lambda}$

Factor out $\frac{1}{e^\lambda}$:

$P(X \leq 2) = \frac{1}{e^\lambda} \left( 1 + \lambda + \frac{\lambda^2}{2} \right)$

Finally, subtract from 1 to get $P(X > 2)$:

$P(X > 2) = 1 - \frac{1}{e^\lambda} \left( 1 + \lambda + \frac{\lambda^2}{2} \right)$

Summary:

1. $c = \frac{1}{e^\lambda}$
2. $P(X = 0) = \frac{1}{e^\lambda}$
3. $P(X > 2) = 1 - \frac{1}{e^\lambda} \left( 1 + \lambda + \frac{\lambda^2}{2} \right)$

Do you have any questions or need further details?

Relative questions:

1. What is the significance of the parameter $\lambda$ in this PMF?
2. How would $P(X = k)$ be calculated for any arbitrary $k$?
3. What type of distribution does this PMF resemble?
4. How can we interpret $P(X > 2)$ in terms of real-world applications?
5. How would the PMF change if we were dealing with a continuous random variable instead?

Tip: In many problems involving sums like this, recognizing series expansions (like $e^\lambda$) can simplify your work.