Last modified: December 24, 2022
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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 $\lambda = np$:
$$\text{Binomial}(n, p) \approx \text{Poisson}(np)$$