# Thread: Geometric Distribution question

1. ## Geometric Distribution question

probability of raining each day = p = 0.01
q = 0.99

What is the expected number of days until we see a rainy day?

I think the answer is

E(X) = 0.99 / 0.01 = 99

But the correct answer is

1 + E(X) = 1 + 99 = 100

Why do you add 1?

2. Originally Posted by shrek
probability of raining each day = p = 0.01
q = 0.99

What is the expected number of days until we see a rainy day?

I think the answer is

E(X) = 0.99 / 0.01 = 99

But the correct answer is

1 + E(X) = 1 + 99 = 100

Why do you add 1?
Shrek: In general, for a Geometric Distribution, the expected value and variance of the associated random variable X is:

E[X] = 1 / p

Var[X] = q/p^2.

So, the answer to your question is E[X]= 1/p = 1/0.01 = 100.

3. ## Re: Geometric Distribution question

There are two formulae use in proofing the Expected Value and Variance of a geometric distribution.

[1] fx= pq^x.
[2] fx= pq^(1-x).
I'm used to the first formula, using the first formula:
E(X)= q/p and
Var(X)= q/p^2. Why 1/p?

4. ## Re: Geometric Distribution question

You are referring to two common versions of Geometric distribution, in which one begins at 0 and the other begins at 1, i.e. one is a shift of the other by 1, like X = Y + 1.

One is use to model "the number of non-rainy days before the first rainy day" while the other is "the number of days until the first rainy day (inclusive)". So the question asking the later while you are answering the former. You may further read the article from wikipedia.

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