# Interpret this probability distribution

1. Oct 28, 2009

### hoffmann

i took an exam today and was sorta stumped by this question. pls take a look, thx!

how do i interpret this probability distribution:

$$\sum_{k=r}^\infty \binom{k}{r}p^k(1-p)^{k-r}$$

where r is the number of successes, p is the probability, k trials.

by looking at it, it seems like it's similar to a negative binomial distribution once you pull out a k/r. if you do some math after pulling out the k/r, it seems like it is the expected value of a geometric distribution. is this distribution saying that a negative binomial divided by the number of successes r means there is only one success, which is geometric?

2. Oct 28, 2009

### hoffmann

as in, what does this distribution equal?

3. Oct 28, 2009

### miqbal

It's just summing up binomial random variables. We're summing over k - the number of trials. r is fixed so we are summing the probabilities that we have r successes for k = r,..., infinity trials.

Hmm, let's take an analytic stab.

Using the binomial theorem,

$$(p + (1 - p))^{k} = \sum_{i=0}^k \binom{k}{i}p^{k}(1-p)^{k-i}$$

so

$$1 = 1^{\infty} = (p + (1 - p))^{\infty} = \sum_{i=0}^\infty \binom{k}{i}p^{k}(1-p)^{k-i} = \sum_{i=0}^{r-1} \binom{k}{i}p^{k}(1-p)^{k-i} + \sum_{i=r}^\infty \binom{k}{i}p^{k}(1-p)^{k-i}$$

finally we get

$$\sum_{i=r}^\infty \binom{k}{i}p^{k}(1-p)^{k-i} = 1 -\sum_{i=0}^{r-1} \binom{k}{i}p^{k}(1-p)^{k-i} = 1 - P({X < r})$$

Where $$X$$ is a binomial random variable with parameters k, p.

Last edited: Oct 28, 2009
4. Oct 28, 2009

### hoffmann

i understand the first part of your explanation. where did the "Hmm" part come from? is this an alternative explanation?

5. Oct 28, 2009

### miqbal

Yeah sorry I was wasn't finished editing it. Its confusing no doubt though . I guess it's the complement that we will have less than r successes in k trials, but i'm not sure.

6. Oct 28, 2009

### hoffmann

7. Oct 28, 2009

### miqbal

Yeah i thought that made sense too but "doing the math" gave a different answer.

Think about the same scenario again (a fair coin, look for 8 successes, k trials). Sum up the probabilities that we get 8 successes after k flips from k = 8 to infinity flips. Should this sum be 1? I have no idea so I'll resort to computation:

8 trials:
$$\sum_{i = 8}^8 \binom{8}{i}\left(\frac{1}{2}\right)^{8} = 0.003906$$

9 trials:
$$\sum_{i = 8}^9 \binom{9}{i}\left(\frac{1}{2}\right)^{9} = 0.019531$$

20 trials:
$$\sum_{i = 8}^{20} \binom{20}{i}\left(\frac{1}{2}\right)^{20} = 0.868412$$

50 trials:
Sum is .999999

Turns out it does come out to 1 and I was initially right.

8. Oct 28, 2009

### miqbal

Well my math up there is right too. In the special case where r = k (the number of trials). P(X < r) is P(we have r successes in less than r trials) = 0. So the math gives the same result of 1. Hope that helped!

9. Oct 28, 2009

### hoffmann

let me sum it up analytically:

$$\sum_{k=r}^\infty \binom{k}{r}p^k(1-p)^{k-r} = (p + (1 - p))^{n} = 1^n = 1$$

where r = 0,1,2...n

10. Oct 28, 2009

### hoffmann

the first step is a direct result of the binomial theorem.

11. Oct 29, 2009

### miqbal

Not exactly. You did not use the binomial theorem correctly. The summation must start at 0. Also n = $$\infty$$.