Q.If for a Poisson variate X, $P(X = k) = P(X = k + 1)$, then the variance of X is : (A) $k - 1$ (B) $k$ (C) $k + 1$ (D) $k + 2$

Poisson Distribution: The Art of Counting Rare Events

Imagine you're watching a busy highway from a bridge. Cars pass by at random moments — sometimes two come together, sometimes there's a gap. You want to answer: How many cars will pass in the next minute? That's a counting problem. But if the traffic is light, most minutes will see 0 or 1 car, and occasionally 2 or 3. This is exactly where the Poisson distribution lives.

The Poisson distribution models the number of times a rare event happens in a fixed interval of time (or space, volume, area) when events occur independently at a constant average rate.

The Intuition

Think of a call centre. On average, you get 5 calls per hour. Some hours you get 3, some 7, rarely 12. The Poisson distribution tells you the probability of each possible count — 0 calls, 1 call, 2 calls, and so on — given that average rate.

The key assumptions are:

• Events happen one at a time (no simultaneous arrivals)
• The rate is constant over the interval
• What happens in one interval doesn't affect the next (independence)

The Precise Statement

Let $X$ be the number of events occurring in a fixed interval. Let $\lambda$ (lambda) be the average number of events in that interval. Then $X$ follows a Poisson distribution with parameter $\lambda$, written as:

The probability of observing exactly $k$ events (where $k = 0, 1, 2, \dots$) is:

Here:

• $e \approx 2.71828$ is Euler's number
• $\lambda$ is the average rate (must be positive)
• $k!$ is factorial: $k! = k \times (k-1) \times (k-2) \times \dots \times 1$, and $0! = 1$

What This Formula Says

Let's test it with the call centre example ($\lambda = 5$ calls per hour):

• Probability of exactly 0 calls: $P(0) = e^{-5} \cdot 5^0 / 0! = e^{-5} \approx 0.0067$ — very unlikely
• Probability of exactly 5 calls: $P(5) = e^{-5} \cdot 5^5 / 120 \approx 0.175$ — the most likely outcome
• Probability of exactly 10 calls: $P(10) = e^{-5} \cdot 5^{10} / 10! \approx 0.018$ — quite rare

The distribution is unimodal (one peak) and skewed right when $\lambda$ is small, becoming more symmetric as $\lambda$ grows.

Two Key Properties

The Poisson distribution has a remarkable feature: its mean and variance are equal.

This equality is a quick check: if you're analysing data and the sample mean and variance are very different, the Poisson model probably doesn't fit.

When to Use It — And When Not To

Use Poisson when:

• You're counting occurrences (not measuring continuous quantities)
• Events are independent …