# Define Geometric Binomial Distribution and prove that the pdf of this distribution satisfies the following conditions: p(k) >= 0 and ∑p(k) = 1.

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A geometric distribution is a special case of the binomial distribution. Here, we look at how many trials will occur before our desired result comes about. For example, this distribution would describe the number of throws of die that would be necessary in order to get a 6. More in the real world, it may describe how many times you might be able to show up late before the boss takes notice.

Here, we have a certain formula for the probability distribution. Given a probability `p` of an event occurring, the probability that it occurs in exactly `k` trials is found by the following equation:

`P(k) = (1-p)^(k-1)p`

If you think about this, all we're doing is looking at `k` trials and recognizing each as an independent event. We are saying that for `k-1` trials, the event does not occur (with probability `1-p`), and then, at the kth trial the event does occur. This is very important, because once we put parameters on our numbers, it is clear that all `P(k) > = 0`.

Notice that `k>0` because we cannot have a negative number of trials, and if `k = 0` , then we have a trivial case with no trials. Of course, because `p` is a probability, it must be true that `0<p<=1`.

If `p=1`, then clearly, `P(k) = 0` for all `k > 1`. Anywhere in between 0 and 1, `P(k) > 0` because `(1-p) > 0` and `p > 0`, therefore, any product of the two will also be positive. Therefore, `P(k) >= 0`.

Now, we want to prove that the sum over the sample space for this distribution is equal to 1. In our case, the sample space will be over the entire set of possible numbers of trials. In other words, we need to prove the following:

`sum_(k=1)^oo P(k) = 1`

Again, if `p = 1`, then the event will occur on the first trial, and the probability if our requiring more events will be zero. Therefore, the sum clearly becomes 1 if `p = 1`. To do this for values of `p` less than 1, let's simply substitute the above relation that we said would define this particular distribution:

`sum_(k=1)^oo (1-p)^(k-1) p`

Based on the rules of sums, we can actually rewrite this sum in the following way because we're summing to infinity. We can adjust the lower bound however we please!

`sum_(k=0)^oo (1-p)^kp`

Because p is a constant with respect to k, we can actually remove it from the summation:

`p sum_(k=0)^oo (1-p)^k`

Because `0<=p<=1`, we also know that `0<=1-p<=1`. The summation term is actually a simple infinite geometric series, and because it is a geometric series of a base that is less than 1, we can actually use the following relation (`r<1`):

`sum_(k=0)^oo r^k = 1/(1-r)`

Let's use the above relation to prove the case for distribution:

`p sum_(k=0)^oo(1-p)^k = p(1/(1-(1-p)))`

Simplifying the denominator:

`=p/(1-1+p) = p/p = 1`

Here is where it becomes important that `p>0`. Of course, the case in which `p = 0` is trivial because that event will never occur.

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