Cumulative distribution function for geometric random variable

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For geometric random variable $f(k)=(1-p)^k-1p$.

Then $$F(k) = P(Xleq x)= 1-P(x>k) = 1 - sum_i=k+1p(1-p)^i-1 = 1 - (1-p)^k sum_i=1^ infty p(1-p)^i-1= 1 - (1-p)^k$$

I would appriciate a breakdown of these steps. Especially why do we take $1-P(x>k)$ and what operations are preformed on the summation sign?
Stating the obvious is very welcone, my mathematical background is quite limited.

statistics geometric-probability

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edited Aug 1 at 7:40

asked Jul 31 at 14:53

user1607

608

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up vote
0
down vote

favorite

For geometric random variable $f(k)=(1-p)^k-1p$.

Then $$F(k) = P(Xleq x)= 1-P(x>k) = 1 - sum_i=k+1p(1-p)^i-1 = 1 - (1-p)^k sum_i=1^ infty p(1-p)^i-1= 1 - (1-p)^k$$

I would appriciate a breakdown of these steps. Especially why do we take $1-P(x>k)$ and what operations are preformed on the summation sign?
Stating the obvious is very welcone, my mathematical background is quite limited.

statistics geometric-probability

share|cite|improve this question

edited Aug 1 at 7:40

asked Jul 31 at 14:53

user1607

608

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

For geometric random variable $f(k)=(1-p)^k-1p$.

Then $$F(k) = P(Xleq x)= 1-P(x>k) = 1 - sum_i=k+1p(1-p)^i-1 = 1 - (1-p)^k sum_i=1^ infty p(1-p)^i-1= 1 - (1-p)^k$$

I would appriciate a breakdown of these steps. Especially why do we take $1-P(x>k)$ and what operations are preformed on the summation sign?
Stating the obvious is very welcone, my mathematical background is quite limited.

statistics geometric-probability

share|cite|improve this question

edited Aug 1 at 7:40

asked Jul 31 at 14:53

user1607

608

For geometric random variable $f(k)=(1-p)^k-1p$.

Then $$F(k) = P(Xleq x)= 1-P(x>k) = 1 - sum_i=k+1p(1-p)^i-1 = 1 - (1-p)^k sum_i=1^ infty p(1-p)^i-1= 1 - (1-p)^k$$

I would appriciate a breakdown of these steps. Especially why do we take $1-P(x>k)$ and what operations are preformed on the summation sign?
Stating the obvious is very welcone, my mathematical background is quite limited.

statistics geometric-probability

share|cite|improve this question

edited Aug 1 at 7:40

asked Jul 31 at 14:53

user1607

608

share|cite|improve this question

share|cite|improve this question

edited Aug 1 at 7:40

edited Aug 1 at 7:40

edited Aug 1 at 7:40

asked Jul 31 at 14:53

user1607

608

asked Jul 31 at 14:53

user1607

608

asked Jul 31 at 14:53

user1607

608

608

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

The main idea is using the geometric series $$sum_i=0^inftya , r^i = fraca1-r$$
$P(X leq k)$ is a finite sum, while $P(X>k)$ is an infinite series, specifically a geometric series. So to utilize the geometric series expression, instead of looking at $P(X leq k)$ one looks at the equivalent $1-P(X>k)$.

Then $P(X>k)=sum_i=k+1^inftyp(1-p)^i-1$ simplifies as done in your expression. I personally find it easier to look at the first few terms written out:

beginalignsum_i=k+1^inftyp(1-p)^i-1
&= p(1-p)^k + p(1-p)^k+1 + p(1-p)^k+2 + dots \
&= p(1-p)^k left1+(1-p)+(1-p)^2+ldotsright\
&= p(1-p)^k frac11-(1-p) = (1-p)^kendalign

---

Added to answer the questions in the comments:

$P(X>k)$ is the probability of $X$ taking values greater than $k$ so:

beginalign
P(X>k) & = P(X=k+1)+P(X=k+2)+P(X=k+3)+dots \
& = p(1-p)^k+1-1 + p(1-p)^k+2-1+ p(1-p)^k+3-1+dots
endalign

This is the same as writing $sum_i=k+1^inftyp(1-p)^i-1$.

$1+(1-p)+(1-p)^2+ldots$ is the geometric series with $a=1$ and $r=1-p$. Plugging them in the formula $fraca1-r$ to get $frac11-(1-p)$.

share|cite|improve this answer

edited Aug 1 at 21:31

answered Jul 31 at 21:28

Just_to_Answer

65317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tnanks for explanation. Could you maybe also break down how to utilize the geometri series to get to $P(X>k)=sum_i=k+1p(1-p)^i-1$ ?
  – user1607
  Aug 1 at 9:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also how is $1+(1-p)+(1-p)^2+...=frac11-(1-p)$? I appologize if my questions are elementary, by mathematical background is not great.
  – user1607
  Aug 1 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Please see the added explanations.
  – Just_to_Answer
  Aug 1 at 21:32
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The main idea is using the geometric series $$sum_i=0^inftya , r^i = fraca1-r$$
$P(X leq k)$ is a finite sum, while $P(X>k)$ is an infinite series, specifically a geometric series. So to utilize the geometric series expression, instead of looking at $P(X leq k)$ one looks at the equivalent $1-P(X>k)$.

Then $P(X>k)=sum_i=k+1^inftyp(1-p)^i-1$ simplifies as done in your expression. I personally find it easier to look at the first few terms written out:

beginalignsum_i=k+1^inftyp(1-p)^i-1
&= p(1-p)^k + p(1-p)^k+1 + p(1-p)^k+2 + dots \
&= p(1-p)^k left1+(1-p)+(1-p)^2+ldotsright\
&= p(1-p)^k frac11-(1-p) = (1-p)^kendalign

---

Added to answer the questions in the comments:

$P(X>k)$ is the probability of $X$ taking values greater than $k$ so:

beginalign
P(X>k) & = P(X=k+1)+P(X=k+2)+P(X=k+3)+dots \
& = p(1-p)^k+1-1 + p(1-p)^k+2-1+ p(1-p)^k+3-1+dots
endalign

This is the same as writing $sum_i=k+1^inftyp(1-p)^i-1$.

$1+(1-p)+(1-p)^2+ldots$ is the geometric series with $a=1$ and $r=1-p$. Plugging them in the formula $fraca1-r$ to get $frac11-(1-p)$.

share|cite|improve this answer

edited Aug 1 at 21:31

answered Jul 31 at 21:28

Just_to_Answer

65317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tnanks for explanation. Could you maybe also break down how to utilize the geometri series to get to $P(X>k)=sum_i=k+1p(1-p)^i-1$ ?
  – user1607
  Aug 1 at 9:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also how is $1+(1-p)+(1-p)^2+...=frac11-(1-p)$? I appologize if my questions are elementary, by mathematical background is not great.
  – user1607
  Aug 1 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Please see the added explanations.
  – Just_to_Answer
  Aug 1 at 21:32
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

The main idea is using the geometric series $$sum_i=0^inftya , r^i = fraca1-r$$
$P(X leq k)$ is a finite sum, while $P(X>k)$ is an infinite series, specifically a geometric series. So to utilize the geometric series expression, instead of looking at $P(X leq k)$ one looks at the equivalent $1-P(X>k)$.

Then $P(X>k)=sum_i=k+1^inftyp(1-p)^i-1$ simplifies as done in your expression. I personally find it easier to look at the first few terms written out:

beginalignsum_i=k+1^inftyp(1-p)^i-1
&= p(1-p)^k + p(1-p)^k+1 + p(1-p)^k+2 + dots \
&= p(1-p)^k left1+(1-p)+(1-p)^2+ldotsright\
&= p(1-p)^k frac11-(1-p) = (1-p)^kendalign

---

Added to answer the questions in the comments:

$P(X>k)$ is the probability of $X$ taking values greater than $k$ so:

beginalign
P(X>k) & = P(X=k+1)+P(X=k+2)+P(X=k+3)+dots \
& = p(1-p)^k+1-1 + p(1-p)^k+2-1+ p(1-p)^k+3-1+dots
endalign

This is the same as writing $sum_i=k+1^inftyp(1-p)^i-1$.

$1+(1-p)+(1-p)^2+ldots$ is the geometric series with $a=1$ and $r=1-p$. Plugging them in the formula $fraca1-r$ to get $frac11-(1-p)$.

share|cite|improve this answer

edited Aug 1 at 21:31

answered Jul 31 at 21:28

Just_to_Answer

65317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tnanks for explanation. Could you maybe also break down how to utilize the geometri series to get to $P(X>k)=sum_i=k+1p(1-p)^i-1$ ?
  – user1607
  Aug 1 at 9:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also how is $1+(1-p)+(1-p)^2+...=frac11-(1-p)$? I appologize if my questions are elementary, by mathematical background is not great.
  – user1607
  Aug 1 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Please see the added explanations.
  – Just_to_Answer
  Aug 1 at 21:32
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The main idea is using the geometric series $$sum_i=0^inftya , r^i = fraca1-r$$
$P(X leq k)$ is a finite sum, while $P(X>k)$ is an infinite series, specifically a geometric series. So to utilize the geometric series expression, instead of looking at $P(X leq k)$ one looks at the equivalent $1-P(X>k)$.

Then $P(X>k)=sum_i=k+1^inftyp(1-p)^i-1$ simplifies as done in your expression. I personally find it easier to look at the first few terms written out:

beginalignsum_i=k+1^inftyp(1-p)^i-1
&= p(1-p)^k + p(1-p)^k+1 + p(1-p)^k+2 + dots \
&= p(1-p)^k left1+(1-p)+(1-p)^2+ldotsright\
&= p(1-p)^k frac11-(1-p) = (1-p)^kendalign

---

Added to answer the questions in the comments:

$P(X>k)$ is the probability of $X$ taking values greater than $k$ so:

beginalign
P(X>k) & = P(X=k+1)+P(X=k+2)+P(X=k+3)+dots \
& = p(1-p)^k+1-1 + p(1-p)^k+2-1+ p(1-p)^k+3-1+dots
endalign

This is the same as writing $sum_i=k+1^inftyp(1-p)^i-1$.

$1+(1-p)+(1-p)^2+ldots$ is the geometric series with $a=1$ and $r=1-p$. Plugging them in the formula $fraca1-r$ to get $frac11-(1-p)$.

share|cite|improve this answer

edited Aug 1 at 21:31

answered Jul 31 at 21:28

Just_to_Answer

65317

The main idea is using the geometric series $$sum_i=0^inftya , r^i = fraca1-r$$
$P(X leq k)$ is a finite sum, while $P(X>k)$ is an infinite series, specifically a geometric series. So to utilize the geometric series expression, instead of looking at $P(X leq k)$ one looks at the equivalent $1-P(X>k)$.

Then $P(X>k)=sum_i=k+1^inftyp(1-p)^i-1$ simplifies as done in your expression. I personally find it easier to look at the first few terms written out:

beginalignsum_i=k+1^inftyp(1-p)^i-1
&= p(1-p)^k + p(1-p)^k+1 + p(1-p)^k+2 + dots \
&= p(1-p)^k left1+(1-p)+(1-p)^2+ldotsright\
&= p(1-p)^k frac11-(1-p) = (1-p)^kendalign

---

Added to answer the questions in the comments:

$P(X>k)$ is the probability of $X$ taking values greater than $k$ so:

beginalign
P(X>k) & = P(X=k+1)+P(X=k+2)+P(X=k+3)+dots \
& = p(1-p)^k+1-1 + p(1-p)^k+2-1+ p(1-p)^k+3-1+dots
endalign

This is the same as writing $sum_i=k+1^inftyp(1-p)^i-1$.

$1+(1-p)+(1-p)^2+ldots$ is the geometric series with $a=1$ and $r=1-p$. Plugging them in the formula $fraca1-r$ to get $frac11-(1-p)$.

share|cite|improve this answer

edited Aug 1 at 21:31

answered Jul 31 at 21:28

Just_to_Answer

65317

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 21:31

edited Aug 1 at 21:31

edited Aug 1 at 21:31

answered Jul 31 at 21:28

Just_to_Answer

65317

answered Jul 31 at 21:28

Just_to_Answer

65317

answered Jul 31 at 21:28

Just_to_Answer

65317

65317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tnanks for explanation. Could you maybe also break down how to utilize the geometri series to get to $P(X>k)=sum_i=k+1p(1-p)^i-1$ ?
  – user1607
  Aug 1 at 9:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also how is $1+(1-p)+(1-p)^2+...=frac11-(1-p)$? I appologize if my questions are elementary, by mathematical background is not great.
  – user1607
  Aug 1 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Please see the added explanations.
  – Just_to_Answer
  Aug 1 at 21:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tnanks for explanation. Could you maybe also break down how to utilize the geometri series to get to $P(X>k)=sum_i=k+1p(1-p)^i-1$ ?
  – user1607
  Aug 1 at 9:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also how is $1+(1-p)+(1-p)^2+...=frac11-(1-p)$? I appologize if my questions are elementary, by mathematical background is not great.
  – user1607
  Aug 1 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Please see the added explanations.
  – Just_to_Answer
  Aug 1 at 21:32
  

  
  
  
  

Tnanks for explanation. Could you maybe also break down how to utilize the geometri series to get to $P(X>k)=sum_i=k+1p(1-p)^i-1$ ?
– user1607
Aug 1 at 9:53

Tnanks for explanation. Could you maybe also break down how to utilize the geometri series to get to $P(X>k)=sum_i=k+1p(1-p)^i-1$ ?
– user1607
Aug 1 at 9:53

Also how is $1+(1-p)+(1-p)^2+...=frac11-(1-p)$? I appologize if my questions are elementary, by mathematical background is not great.
– user1607
Aug 1 at 10:03

Also how is $1+(1-p)+(1-p)^2+...=frac11-(1-p)$? I appologize if my questions are elementary, by mathematical background is not great.
– user1607
Aug 1 at 10:03

You are welcome. Please see the added explanations.
– Just_to_Answer
Aug 1 at 21:32

You are welcome. Please see the added explanations.
– Just_to_Answer
Aug 1 at 21:32

add a comment | 

 

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