*Last modified: September 21, 2024*

*This article is written in: ðŸ‡ºðŸ‡¸*

## Poisson Distribution (Discrete)

A discrete random variable X follows a Poisson distribution if the events occur independently and at a constant average rate. The Poisson distribution is denoted as $X \sim \text{Poisson}(\lambda)$, where $\lambda$ is the average rate (or mean) of events occurring in a given interval.

### Probability Mass Function (PMF)

The PMF of a Poisson distribution is given by:

$$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$$

where $k \in {0, 1, 2, \dots}$.

### Cumulative Distribution Function (CDF)

The CDF of a Poisson distribution is given by:

$$F(k) = P(X \le k) = \sum_{i=0}^{k} \frac{e^{-\lambda} \lambda^i}{i!}$$

### Expected Value and Variance

The expected value (mean) and variance of a Poisson distribution are both equal to the average rate parameter $\lambda$:

$$E[X] = \text{Var}(X) = \lambda$$

### Moment Generating Functions and Moments

The moment generating function (MGF) of a Poisson distribution is:

$$M_X(t) = E[e^{tX}] = e^{\lambda(e^t - 1)}$$

To find the n-th moment, we take the n-th derivative of the MGF with respect to t and then evaluate it at t=0:

$$E[X^n] = \frac{d^n M_X(t)}{dt^n}\Bigg|_{t=0}$$

**First Moment (Mean):**

$$E[X] = \lambda$$

**Second Moment (Variance + Mean^2):**

$$E[X^2] = \lambda^2 + \lambda$$

### Example: Hospital Emergency Room Visits

A hospital's emergency room experiences an average of 5 visits per hour due to accidents. We can model this using a Poisson distribution.

Given:

- Average rate of success (visits per hour) $\lambda = 5$

I. What is the probability exactly 3 accidents occur in an hour?

For exactly 3 accidents:

$$ P(X = 3) = \frac{e^{-5} \cdot 5^3}{3!} = 0.1404 $$

II. What is the probability of more than 7 accidents in an hour?

For more than 7 accidents:

$$ P(X > 7) = 1 - \sum_{k=0}^{7} \frac{e^{-5} \cdot 5^k}{k!} = 0.1339 $$

III. What is the probability of at most 2 accidents in an hour?

For at most 2 accidents:

$$ P(X \leq 2) = \sum_{k=0}^{2} \frac{e^{-5} \cdot 5^k}{k!} = 0.1247 $$

### Poisson Approximation to the Binomial Distribution

If the number of trials in a binomial distribution is large (n) and the probability of success (p) is small, then the binomial distribution can be approximated by a Poisson distribution with parameter $\lambda = np$:

$$\text{Binomial}(n, p) \approx \text{Poisson}(np)$$