How to find pdf of exponential random variable given the mean [closed]

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Let $X$ be a exponential random variable with mean $beta$.

How we can write the PDF of $X$ given mean $beta$.

Let $Y$ random variable define by
$$ Y=fracPNX$$
were $N$ and $P$ are positive real value. What is the PDF and CDF of Y.

exponential-distribution

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asked Jul 20 at 13:11

BouMokhtar

116

closed as off-topic by Clarinetist, amWhy, Shailesh, Adrian Keister, Parcly Taxel Jul 20 at 13:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clarinetist, amWhy, Shailesh, Adrian Keister, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-2
down vote

favorite

Let $X$ be a exponential random variable with mean $beta$.

How we can write the PDF of $X$ given mean $beta$.

Let $Y$ random variable define by
$$ Y=fracPNX$$
were $N$ and $P$ are positive real value. What is the PDF and CDF of Y.

exponential-distribution

share|cite|improve this question

asked Jul 20 at 13:11

BouMokhtar

116

closed as off-topic by Clarinetist, amWhy, Shailesh, Adrian Keister, Parcly Taxel Jul 20 at 13:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clarinetist, amWhy, Shailesh, Adrian Keister, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

Let $X$ be a exponential random variable with mean $beta$.

How we can write the PDF of $X$ given mean $beta$.

Let $Y$ random variable define by
$$ Y=fracPNX$$
were $N$ and $P$ are positive real value. What is the PDF and CDF of Y.

exponential-distribution

share|cite|improve this question

asked Jul 20 at 13:11

BouMokhtar

116

Let $X$ be a exponential random variable with mean $beta$.

How we can write the PDF of $X$ given mean $beta$.

Let $Y$ random variable define by
$$ Y=fracPNX$$
were $N$ and $P$ are positive real value. What is the PDF and CDF of Y.

exponential-distribution

share|cite|improve this question

asked Jul 20 at 13:11

BouMokhtar

116

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 13:11

BouMokhtar

116

asked Jul 20 at 13:11

BouMokhtar

116

asked Jul 20 at 13:11

BouMokhtar

116

116

closed as off-topic by Clarinetist, amWhy, Shailesh, Adrian Keister, Parcly Taxel Jul 20 at 13:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clarinetist, amWhy, Shailesh, Adrian Keister, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Clarinetist, amWhy, Shailesh, Adrian Keister, Parcly Taxel Jul 20 at 13:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clarinetist, amWhy, Shailesh, Adrian Keister, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

Hints:

• An exponential distribution with rate $lambda$ has pdf $lambda e^-lambda x$ and CDF $1 - e^-lambda x$ for $x ge 0$, and has mean $frac1lambda$




• $Y$ is also exponentially distributed, with mean $fracPNE[X]$

share|cite|improve this answer

answered Jul 20 at 13:17

Henry

93k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK How we drive the CDF and PDF?
  – BouMokhtar
  Jul 20 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You know that the mean is $beta$ and you also know that the mean ist $frac 1 lambda$. Now you have $lambda$ which gives you both the PDF and CDF.
  – Stefan
  Jul 20 at 13:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if mean is $fracPNE[X]$, the pdf is $fracNPE[X]e^-fracNxPE[X]$
  – BouMokhtar
  Jul 20 at 13:32
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Hints:

• An exponential distribution with rate $lambda$ has pdf $lambda e^-lambda x$ and CDF $1 - e^-lambda x$ for $x ge 0$, and has mean $frac1lambda$




• $Y$ is also exponentially distributed, with mean $fracPNE[X]$

share|cite|improve this answer

answered Jul 20 at 13:17

Henry

93k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK How we drive the CDF and PDF?
  – BouMokhtar
  Jul 20 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You know that the mean is $beta$ and you also know that the mean ist $frac 1 lambda$. Now you have $lambda$ which gives you both the PDF and CDF.
  – Stefan
  Jul 20 at 13:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if mean is $fracPNE[X]$, the pdf is $fracNPE[X]e^-fracNxPE[X]$
  – BouMokhtar
  Jul 20 at 13:32
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

Hints:

• An exponential distribution with rate $lambda$ has pdf $lambda e^-lambda x$ and CDF $1 - e^-lambda x$ for $x ge 0$, and has mean $frac1lambda$




• $Y$ is also exponentially distributed, with mean $fracPNE[X]$

share|cite|improve this answer

answered Jul 20 at 13:17

Henry

93k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK How we drive the CDF and PDF?
  – BouMokhtar
  Jul 20 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You know that the mean is $beta$ and you also know that the mean ist $frac 1 lambda$. Now you have $lambda$ which gives you both the PDF and CDF.
  – Stefan
  Jul 20 at 13:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if mean is $fracPNE[X]$, the pdf is $fracNPE[X]e^-fracNxPE[X]$
  – BouMokhtar
  Jul 20 at 13:32
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Hints:

• An exponential distribution with rate $lambda$ has pdf $lambda e^-lambda x$ and CDF $1 - e^-lambda x$ for $x ge 0$, and has mean $frac1lambda$




• $Y$ is also exponentially distributed, with mean $fracPNE[X]$

share|cite|improve this answer

answered Jul 20 at 13:17

Henry

93k469147

Hints:

• An exponential distribution with rate $lambda$ has pdf $lambda e^-lambda x$ and CDF $1 - e^-lambda x$ for $x ge 0$, and has mean $frac1lambda$




• $Y$ is also exponentially distributed, with mean $fracPNE[X]$

share|cite|improve this answer

answered Jul 20 at 13:17

Henry

93k469147

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 13:17

Henry

93k469147

answered Jul 20 at 13:17

Henry

93k469147

answered Jul 20 at 13:17

Henry

93k469147

93k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK How we drive the CDF and PDF?
  – BouMokhtar
  Jul 20 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You know that the mean is $beta$ and you also know that the mean ist $frac 1 lambda$. Now you have $lambda$ which gives you both the PDF and CDF.
  – Stefan
  Jul 20 at 13:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if mean is $fracPNE[X]$, the pdf is $fracNPE[X]e^-fracNxPE[X]$
  – BouMokhtar
  Jul 20 at 13:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK How we drive the CDF and PDF?
  – BouMokhtar
  Jul 20 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You know that the mean is $beta$ and you also know that the mean ist $frac 1 lambda$. Now you have $lambda$ which gives you both the PDF and CDF.
  – Stefan
  Jul 20 at 13:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if mean is $fracPNE[X]$, the pdf is $fracNPE[X]e^-fracNxPE[X]$
  – BouMokhtar
  Jul 20 at 13:32
  

  
  
  
  

OK How we drive the CDF and PDF?
– BouMokhtar
Jul 20 at 13:20

OK How we drive the CDF and PDF?
– BouMokhtar
Jul 20 at 13:20

You know that the mean is $beta$ and you also know that the mean ist $frac 1 lambda$. Now you have $lambda$ which gives you both the PDF and CDF.
– Stefan
Jul 20 at 13:30

You know that the mean is $beta$ and you also know that the mean ist $frac 1 lambda$. Now you have $lambda$ which gives you both the PDF and CDF.
– Stefan
Jul 20 at 13:30

So if mean is $fracPNE[X]$, the pdf is $fracNPE[X]e^-fracNxPE[X]$
– BouMokhtar
Jul 20 at 13:32

So if mean is $fracPNE[X]$, the pdf is $fracNPE[X]e^-fracNxPE[X]$
– BouMokhtar
Jul 20 at 13:32

add a comment |