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A discrete random variable $U$ follows a geometric distribution with $p = \frac{1}{4}$. Find $F(u)$, the cumulative distribution question of $U$, for $u = 1, 2, 3 ...$

$$F(u)=P(U<u)=\sum_{n=1}^{\infty} \frac{1}{4}(\frac{3}{4})^{n-1}$$

Using the infinite series formula $S_n = \frac{u_1}{1-r}$,

$$\sum_{n=1}^{\infty} \frac{1}{4}(\frac{3}{4})^{n-1}=\frac{\frac{1}{4}}{1-\frac{3}{4}}=1$$

I used $u_1 = \frac{1}{4}$ as the first term (when $n = 1$) is substituted into the expression in the summation, we get $\frac{1}{4}$.

However, my book uses another expression, which seems to have come out of no where:

enter image description here

Any ideas?

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StopReadingThisUsername
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  • The limit of summation is incorrect. The upper limit should be $u$ not $\infty$. Using $\infty$ as upper limit in your case is like summing the probabilities corresponding to each event (in your case each value of random variable U) which will, ofcourse, come out to be 1. – Dhruv Kohli Nov 19 '16 at 13:53
  • With the upper limit as $u$, the first term is still $0.25$, right? – StopReadingThisUsername Nov 19 '16 at 13:55
  • Yes. The first term will always be $\frac{1}{4}$. Please refer to the definition of CDF and read about the geometric distribution from wiki or your book. – Dhruv Kohli Nov 19 '16 at 13:59
  • I read my book and don't understand where the $1-(\frac{3}{4})^u$ came from. Could you please explain? – StopReadingThisUsername Nov 19 '16 at 14:02
  • Once you set the upper limit of summation to $u$ and compute the sum of the resulting geometric progression, you'll get the result as shown in the snapshot of book. $\sum\limits_{r=1}^{n} ap^{r-1} = \frac{a(1-p^n)}{1-p}$ – Dhruv Kohli Nov 19 '16 at 14:06

1 Answers1

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It is the formula of the partial sum of a geometric series

$$S_n=\sum_{i=0}^{n-1}c\cdot x^{i}=c\cdot \frac{1-x^n}{1-x}$$

It is equal to

$$S_n=\sum_{i=1}^{n}c\cdot x^{i-1}=c\cdot \frac{1-x^{n}}{1-x}$$

In your case $x=\frac34$. And $c=\frac14$

Thus $S_n=\frac14\cdot \frac{1-\left(\frac34\right)^n}{1-\frac34}$$

$S_n=\frac14\cdot \frac{1-\left(\frac34\right)^n}{\frac14}$$

$\frac14$ is cancelling out.

$s_n=1-\left(\frac34\right)^n$

What happens if n goes to infinity ?

$$\lim_{n \to \infty} \left(\frac34\right)^n=0$$

Thus $S_n=1$

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