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geometric distribution is memoryless

A random variable X that follows a geometric distribution satisfies:

P(X > s + t \mid X > s) = P(X > t)

This means:

The probability you still have to wait t more trials does NOT depend on how long you’ve already been waiting.

Your past failures don’t change the future.

🎯 Intuition

Imagine flipping a coin until you get heads.

If you’ve already flipped 10 tails in a row, what’s the probability the next flip is heads?

Still 0.5.

The coin doesn’t care about your suffering.

The geometric distribution models exactly this kind of “fresh start every time” process.

📌 Why it’s memoryless

Let X be the number of trials until the first success, with success probability p.

We want to show:

P(X > s + t \mid X > s) = P(X > t)

Left side:

P(X > s + t \mid X > s)= \frac{P(X > s + t)}{P(X > s)}

But:

P(X > k) = (1 - p)^k

So:

\frac{(1 - p)^{s+t}}{(1 - p)^s}= (1 - p)^t= P(X > t)

Done.
The past cancels out.

⭐ Example

A basketball player makes each free throw with probability p = 0.7.

Let X = number of shots until the first make.

Suppose she has already missed 3 shots.

What is the probability she will miss the next 2 shots?

Memorylessness says:

P(X > 5 \mid X > 3) = P(X > 2)

Compute:

P(X > 2) = (1 - 0.7)^2 = 0.3^2 = 0.09

Even after missing 3 in a row, the chance she misses the next 2 is still 9%.

Her past misses don’t change the future.

⭐ Only two distributions are memoryless

Geometric distribution (discrete)
Exponential distribution (continuous)

These are the only memoryless probability distributions.

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