Negative Binomial question

Hello,

I have this question which I am uncertain about. The question is:
If X is a geometric (negative binomial) random variable with probability mass function
P(X = x) = (1 − p)p^(x−1) (where ^(x-1) indicates "to the (x-1) power")
for x = 1, 2, ...
What range of k will ensure that P (X ≤ k) ≥ α ∈ (0,1)?

options are 1. k ≥ log(1-alpha) / log(p) 2. k < log(1-alpha) / log(p) 3. k ≥ log(p) / log(1-alpha) 4. k < 5 or 5. k = log(p)

I first tried 1 and 2 but those do not work. I am not sure how to approach this question and if anyone can help, that would be great

thanks!

You want \(\displaystyle P(k)= (1- p)p^{k-1}\ge \alpha\). Taking logarithms of both sides, \(\displaystyle ln(1- p)+ (k-1)ln(p)\ge ln(\alpha)\). \(\displaystyle (k-1)ln(p)\ge ln(\alpha)- ln(1- p)= ln\left(\frac{\alpha}{1- p}\right)\). Solve for k.

I'm not that great at doing this ln stuff so I got (x is the alpha since I don't know how to use the special math characters)
k ln p - ln p >= ln (x/1-p)
k ln p >= ln (x/1-p) + ln p
k >= [ln (x/1-p) + ln p] / ln p

now this is where I am stuck because I am not sure what to do next

funnybabe said:

I'm not that great at doing this ln stuff so I got
(x is the alpha since I don't know how to use the special math characters)

k ln p - ln p >= ln (x/1-p) <------

k ln p >= ln (x/1-p) + ln p <------

k >= [ln (x/1-p) + ln p] / ln p <------

now this is where I am stuck because I am not sure what to do next

Click to expand...

The denominators have to be inside grouping symbols when you write it out
horizontally that way.

To cut down on your steps, I would not distribute ln(p) on the left-hand side.
Instead, I would divide each side right away by ln(p), but note the following:

ln(p) and ln(1 - p) are made equal to 0 and undefined, respectively, if p could equal 1.

If it is the case that 0 < p < 1, then ln(p) is negative.

Then when both sides of the inequality are divided by ln(p), the direction of
the inequality sign would change at that step.

I'm going to type "alpha" instead of "x."


(k - 1)ln(p) >= ln[alpha/(1 - p)]


\(\displaystyle \dfrac{(k - 1)ln(p)}{ln(p)} \ \le \ \dfrac{ ln\bigg(\frac{alpha}{1 - p}\bigg)}{ \ ln(p) \ }\)



\(\displaystyle k - 1 \ \le \ \dfrac{ ln\bigg(\frac{alpha}{1 - p}\bigg)}{ \ ln(p) \ }\)


\(\displaystyle k \ \le \ \dfrac{ ln\bigg(\frac{alpha}{1 - p}\bigg)}{ \ ln(p) \ } \ + \ 1\)


or


\(\displaystyle k \ \le \ \dfrac{\ ln(alpha) \ - \ ln(1 - p) \ }{ln(p)} \ + \ 1\)