How come in these questions we treat the sum of exponential distribution as an exponential dist.? [closed]

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In problems like these, all the solutions I've seen add the lambda value of one device n times (in this case n = 6) for a system. Then they find the probability like they would for exponential distribution. But I thought the sum of exponential distributions was a gamma distribution. How come we are able to solve the probability for the system like we would for one device using exponential distribution formulas?

Here lambda = .1 for one device, and thus for the system, lambda = .1 * 6 = .6

probability

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asked Jul 22 at 2:39

David

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closed as off-topic by Brian Borchers, Taroccoesbrocco, max_zorn, Jose Arnaldo Bebita Dris, Parcly Taxel Jul 22 at 13:32

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." – Brian Borchers, Taroccoesbrocco, max_zorn, Jose Arnaldo Bebita Dris, Parcly Taxel

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

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  You’ve misunderstood the question. It is saying that the system uses six of the devices and all six must continue to function for the system to function. The system fails As soon as any of the six devices fail. Thus you need the minimum of the six times to failure rather than the sum.
  – Brian Borchers
  Jul 22 at 2:50
  

  
  
  
  

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In problems like these, all the solutions I've seen add the lambda value of one device n times (in this case n = 6) for a system. Then they find the probability like they would for exponential distribution. But I thought the sum of exponential distributions was a gamma distribution. How come we are able to solve the probability for the system like we would for one device using exponential distribution formulas?

Here lambda = .1 for one device, and thus for the system, lambda = .1 * 6 = .6

probability

share|cite|improve this question

asked Jul 22 at 2:39

David

603

closed as off-topic by Brian Borchers, Taroccoesbrocco, max_zorn, Jose Arnaldo Bebita Dris, Parcly Taxel Jul 22 at 13:32

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." – Brian Borchers, Taroccoesbrocco, max_zorn, Jose Arnaldo Bebita Dris, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You’ve misunderstood the question. It is saying that the system uses six of the devices and all six must continue to function for the system to function. The system fails As soon as any of the six devices fail. Thus you need the minimum of the six times to failure rather than the sum.
  – Brian Borchers
  Jul 22 at 2:50
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

In problems like these, all the solutions I've seen add the lambda value of one device n times (in this case n = 6) for a system. Then they find the probability like they would for exponential distribution. But I thought the sum of exponential distributions was a gamma distribution. How come we are able to solve the probability for the system like we would for one device using exponential distribution formulas?

Here lambda = .1 for one device, and thus for the system, lambda = .1 * 6 = .6

probability

share|cite|improve this question

asked Jul 22 at 2:39

David

603

In problems like these, all the solutions I've seen add the lambda value of one device n times (in this case n = 6) for a system. Then they find the probability like they would for exponential distribution. But I thought the sum of exponential distributions was a gamma distribution. How come we are able to solve the probability for the system like we would for one device using exponential distribution formulas?

Here lambda = .1 for one device, and thus for the system, lambda = .1 * 6 = .6

probability

share|cite|improve this question

asked Jul 22 at 2:39

David

603

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 2:39

David

603

asked Jul 22 at 2:39

David

603

asked Jul 22 at 2:39

David

603

603

closed as off-topic by Brian Borchers, Taroccoesbrocco, max_zorn, Jose Arnaldo Bebita Dris, Parcly Taxel Jul 22 at 13:32

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." – Brian Borchers, Taroccoesbrocco, max_zorn, Jose Arnaldo Bebita Dris, 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 Brian Borchers, Taroccoesbrocco, max_zorn, Jose Arnaldo Bebita Dris, Parcly Taxel Jul 22 at 13:32

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." – Brian Borchers, Taroccoesbrocco, max_zorn, Jose Arnaldo Bebita Dris, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You’ve misunderstood the question. It is saying that the system uses six of the devices and all six must continue to function for the system to function. The system fails As soon as any of the six devices fail. Thus you need the minimum of the six times to failure rather than the sum.
  – Brian Borchers
  Jul 22 at 2:50
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You’ve misunderstood the question. It is saying that the system uses six of the devices and all six must continue to function for the system to function. The system fails As soon as any of the six devices fail. Thus you need the minimum of the six times to failure rather than the sum.
  – Brian Borchers
  Jul 22 at 2:50
  

  
  
  
  

1

1

You’ve misunderstood the question. It is saying that the system uses six of the devices and all six must continue to function for the system to function. The system fails As soon as any of the six devices fail. Thus you need the minimum of the six times to failure rather than the sum.
– Brian Borchers
Jul 22 at 2:50

You’ve misunderstood the question. It is saying that the system uses six of the devices and all six must continue to function for the system to function. The system fails As soon as any of the six devices fail. Thus you need the minimum of the six times to failure rather than the sum.
– Brian Borchers
Jul 22 at 2:50

add a comment | 

2 Answers
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The system fails as soon as one component fails. Thus the lifetime of the system is the minimum of the six components' lifetimes (not the sum of the lifetimes). Your textbook probably proves that the minimum of $n$ independent $textExponential(lambda)$ random variables is itself a $textExponential(nlambda)$ random variable.

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answered Jul 22 at 2:49

angryavian

34.6k12874

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Note that the minimum of exponential random variables (RV) is also an exponential random variable. In particular, if the PDF of RV $X_i$ is
$$f_X_i(x) = lambda_i exp(-lambda_i x),$$
$Y = minX_1, X_2,cdots, X_n$, and $X_i$ are independent RVs then
beginalign
PrY > y & = PrX_1 >y cdot PrX_2 >y cdots PrX_n >y \
& = exp(-(lambda_1+lambda_2+cdots lambda_n)y)\
therefore f_Y(y) & = lambda exp(-lambda y),
endalign
where $lambda = lambda_1 + lambda_2 + cdots +lambda_n$.

In your question, you need to consider the minimum of the six random variables as the system fails if at least one of them fails.

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answered Jul 22 at 2:52

Math Lover

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

The system fails as soon as one component fails. Thus the lifetime of the system is the minimum of the six components' lifetimes (not the sum of the lifetimes). Your textbook probably proves that the minimum of $n$ independent $textExponential(lambda)$ random variables is itself a $textExponential(nlambda)$ random variable.

share|cite|improve this answer

answered Jul 22 at 2:49

angryavian

34.6k12874

add a comment | 

up vote
2
down vote

The system fails as soon as one component fails. Thus the lifetime of the system is the minimum of the six components' lifetimes (not the sum of the lifetimes). Your textbook probably proves that the minimum of $n$ independent $textExponential(lambda)$ random variables is itself a $textExponential(nlambda)$ random variable.

share|cite|improve this answer

answered Jul 22 at 2:49

angryavian

34.6k12874

add a comment | 

up vote
2
down vote

up vote
2
down vote

The system fails as soon as one component fails. Thus the lifetime of the system is the minimum of the six components' lifetimes (not the sum of the lifetimes). Your textbook probably proves that the minimum of $n$ independent $textExponential(lambda)$ random variables is itself a $textExponential(nlambda)$ random variable.

share|cite|improve this answer

answered Jul 22 at 2:49

angryavian

34.6k12874

The system fails as soon as one component fails. Thus the lifetime of the system is the minimum of the six components' lifetimes (not the sum of the lifetimes). Your textbook probably proves that the minimum of $n$ independent $textExponential(lambda)$ random variables is itself a $textExponential(nlambda)$ random variable.

share|cite|improve this answer

answered Jul 22 at 2:49

angryavian

34.6k12874

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 2:49

angryavian

34.6k12874

answered Jul 22 at 2:49

angryavian

34.6k12874

answered Jul 22 at 2:49

angryavian

34.6k12874

34.6k12874

add a comment | 

add a comment | 

up vote
0
down vote

Note that the minimum of exponential random variables (RV) is also an exponential random variable. In particular, if the PDF of RV $X_i$ is
$$f_X_i(x) = lambda_i exp(-lambda_i x),$$
$Y = minX_1, X_2,cdots, X_n$, and $X_i$ are independent RVs then
beginalign
PrY > y & = PrX_1 >y cdot PrX_2 >y cdots PrX_n >y \
& = exp(-(lambda_1+lambda_2+cdots lambda_n)y)\
therefore f_Y(y) & = lambda exp(-lambda y),
endalign
where $lambda = lambda_1 + lambda_2 + cdots +lambda_n$.

In your question, you need to consider the minimum of the six random variables as the system fails if at least one of them fails.

share|cite|improve this answer

answered Jul 22 at 2:52

Math Lover

12.4k21232

add a comment | 

up vote
0
down vote

Note that the minimum of exponential random variables (RV) is also an exponential random variable. In particular, if the PDF of RV $X_i$ is
$$f_X_i(x) = lambda_i exp(-lambda_i x),$$
$Y = minX_1, X_2,cdots, X_n$, and $X_i$ are independent RVs then
beginalign
PrY > y & = PrX_1 >y cdot PrX_2 >y cdots PrX_n >y \
& = exp(-(lambda_1+lambda_2+cdots lambda_n)y)\
therefore f_Y(y) & = lambda exp(-lambda y),
endalign
where $lambda = lambda_1 + lambda_2 + cdots +lambda_n$.

In your question, you need to consider the minimum of the six random variables as the system fails if at least one of them fails.

share|cite|improve this answer

answered Jul 22 at 2:52

Math Lover

12.4k21232

add a comment | 

up vote
0
down vote

up vote
0
down vote

Note that the minimum of exponential random variables (RV) is also an exponential random variable. In particular, if the PDF of RV $X_i$ is
$$f_X_i(x) = lambda_i exp(-lambda_i x),$$
$Y = minX_1, X_2,cdots, X_n$, and $X_i$ are independent RVs then
beginalign
PrY > y & = PrX_1 >y cdot PrX_2 >y cdots PrX_n >y \
& = exp(-(lambda_1+lambda_2+cdots lambda_n)y)\
therefore f_Y(y) & = lambda exp(-lambda y),
endalign
where $lambda = lambda_1 + lambda_2 + cdots +lambda_n$.

In your question, you need to consider the minimum of the six random variables as the system fails if at least one of them fails.

share|cite|improve this answer

answered Jul 22 at 2:52

Math Lover

12.4k21232

Note that the minimum of exponential random variables (RV) is also an exponential random variable. In particular, if the PDF of RV $X_i$ is
$$f_X_i(x) = lambda_i exp(-lambda_i x),$$
$Y = minX_1, X_2,cdots, X_n$, and $X_i$ are independent RVs then
beginalign
PrY > y & = PrX_1 >y cdot PrX_2 >y cdots PrX_n >y \
& = exp(-(lambda_1+lambda_2+cdots lambda_n)y)\
therefore f_Y(y) & = lambda exp(-lambda y),
endalign
where $lambda = lambda_1 + lambda_2 + cdots +lambda_n$.

In your question, you need to consider the minimum of the six random variables as the system fails if at least one of them fails.

share|cite|improve this answer

answered Jul 22 at 2:52

Math Lover

12.4k21232

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 2:52

Math Lover

12.4k21232

answered Jul 22 at 2:52

Math Lover

12.4k21232

answered Jul 22 at 2:52

Math Lover

12.4k21232

12.4k21232

add a comment | 

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