Minimum of Two Exponential Random Variables

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Elevator $A$ breaks down with an exponential distribution of $4$ per day, while elevator $B$ breaks down with an exponential distribution of $6$ per day. Assume a day begins at midnight. What is the probability that the first breakdown occurs before $2$ AM?

My attempt:

Let $X$ = the time between two consecutive breakdowns for elevator A and $Y$ = the time between two consecutive breakdowns for elevator B. Then $min(X,Y)$ is exponentially distributed with rate $10$. Thus, let $Z = min(X,Y)$, so $P(Z < 2) = 1-e^-20 = 1$ approximately. My only issue with this is that $Z$ is the minimum time between two consecutive breakdowns, not a single breakdown.. Aren't I overcompensating? For example, my answer includes the probability that the elevator breaks down at $5$ AM and then breaks down again at $6$ AM. Also, the probability is $1$, which seems wrong as well.

probability probability-distributions exponential-distribution

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edited Jul 17 at 18:19

Math Lover

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asked Jul 17 at 17:59

Saad

4158

add a comment | 

up vote
1
down vote

favorite

Elevator $A$ breaks down with an exponential distribution of $4$ per day, while elevator $B$ breaks down with an exponential distribution of $6$ per day. Assume a day begins at midnight. What is the probability that the first breakdown occurs before $2$ AM?

My attempt:

Let $X$ = the time between two consecutive breakdowns for elevator A and $Y$ = the time between two consecutive breakdowns for elevator B. Then $min(X,Y)$ is exponentially distributed with rate $10$. Thus, let $Z = min(X,Y)$, so $P(Z < 2) = 1-e^-20 = 1$ approximately. My only issue with this is that $Z$ is the minimum time between two consecutive breakdowns, not a single breakdown.. Aren't I overcompensating? For example, my answer includes the probability that the elevator breaks down at $5$ AM and then breaks down again at $6$ AM. Also, the probability is $1$, which seems wrong as well.

probability probability-distributions exponential-distribution

share|cite|improve this question

edited Jul 17 at 18:19

Math Lover

12.4k21232

asked Jul 17 at 17:59

Saad

4158

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Elevator $A$ breaks down with an exponential distribution of $4$ per day, while elevator $B$ breaks down with an exponential distribution of $6$ per day. Assume a day begins at midnight. What is the probability that the first breakdown occurs before $2$ AM?

My attempt:

Let $X$ = the time between two consecutive breakdowns for elevator A and $Y$ = the time between two consecutive breakdowns for elevator B. Then $min(X,Y)$ is exponentially distributed with rate $10$. Thus, let $Z = min(X,Y)$, so $P(Z < 2) = 1-e^-20 = 1$ approximately. My only issue with this is that $Z$ is the minimum time between two consecutive breakdowns, not a single breakdown.. Aren't I overcompensating? For example, my answer includes the probability that the elevator breaks down at $5$ AM and then breaks down again at $6$ AM. Also, the probability is $1$, which seems wrong as well.

probability probability-distributions exponential-distribution

share|cite|improve this question

edited Jul 17 at 18:19

Math Lover

12.4k21232

asked Jul 17 at 17:59

Saad

4158

Elevator $A$ breaks down with an exponential distribution of $4$ per day, while elevator $B$ breaks down with an exponential distribution of $6$ per day. Assume a day begins at midnight. What is the probability that the first breakdown occurs before $2$ AM?

My attempt:

Let $X$ = the time between two consecutive breakdowns for elevator A and $Y$ = the time between two consecutive breakdowns for elevator B. Then $min(X,Y)$ is exponentially distributed with rate $10$. Thus, let $Z = min(X,Y)$, so $P(Z < 2) = 1-e^-20 = 1$ approximately. My only issue with this is that $Z$ is the minimum time between two consecutive breakdowns, not a single breakdown.. Aren't I overcompensating? For example, my answer includes the probability that the elevator breaks down at $5$ AM and then breaks down again at $6$ AM. Also, the probability is $1$, which seems wrong as well.

probability probability-distributions exponential-distribution

share|cite|improve this question

edited Jul 17 at 18:19

Math Lover

12.4k21232

asked Jul 17 at 17:59

Saad

4158

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 18:19

Math Lover

12.4k21232

edited Jul 17 at 18:19

Math Lover

12.4k21232

edited Jul 17 at 18:19

Math Lover

12.4k21232

12.4k21232

asked Jul 17 at 17:59

Saad

4158

asked Jul 17 at 17:59

Saad

4158

asked Jul 17 at 17:59

Saad

4158

4158

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

The number of breakdowns of the first elevator in a day has a Poisson distribution with a mean of $4.$ The thing that has an exponential distribution is the time until the next breakdown, which has an expected value of $1/4text day.$ With the two elevators together the mean waiting time is $1/10text day.$

Since $2text hours = 1/12text day,$ the probability that it happens within that time is $1- e^-(1/12)/(1/10) = 1 - e^-10/12 approx 0.5654.$

One would speak here not of the minimum of two exponential distributions, but of the minimum of two exponentially distributed random variables.

share|cite|improve this answer

answered Jul 17 at 18:14

Michael Hardy

204k23186462

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ahh.. I see.. so I confused the Poisson rate with the exponential rate then? I'm still a bit confused.. If I write the CDF in the form $1-$$e^-λx$, then what does $λ$ represent?
  – Saad
  Jul 17 at 18:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : I'd call it a "rate" either way, but maybe an "average" or "expected value" when talking about the Poisson distribution. The Poisson distribution is the discrete distribution of the number of breakdowns, within the set $0,1,2,3,ldots,$ and the exponential distribution is the continuous distribution of the time until the next breakdown.
  – Michael Hardy
  Jul 17 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in our situation, the mean waiting time is $1/10$ day? But then wikipedia says that if you write the CDF in terms of the mean, it would be $1-$$e^-10x$ as per the alternative parameterization section en.wikipedia.org/wiki/Exponential_distribution where $x$ is the time in days before the next occurence
  – Saad
  Jul 17 at 18:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Saad : Right. The time in days is $1/12.$ so $1-e^-10x = 1 - e^-10/12. qquad$
  – Michael Hardy
  Jul 17 at 18:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

Note that the rates are $4$ breakdowns per $24$ hours and $6$ breakdowns per $24$ hours. Therefore, $lambda_1 = frac424$, and $lambda_2 = frac624$. Using them, you obtain
$$PrZle 2 = 1-exp(-20/24).$$

share|cite|improve this answer

answered Jul 17 at 18:09

Math Lover

12.4k21232

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait.. if $X$~$exp(λ)$, where $λ$ is the rate, then $F(x) = 1-e^-λx$.. isn't the rate 10 in our case?
  – Saad
  Jul 17 at 18:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad It is $(4+6)/24=10/24$.
  – Math Lover
  Jul 17 at 18:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : Proper MathJax code might be $$ Xsim exp(lambda), $$ although I would prefer $XsimoperatornameExp(lambda)$ because often $exp(lambda)$ means $e^-lambda. qquad$
  – Michael Hardy
  Jul 17 at 18:16
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The number of breakdowns of the first elevator in a day has a Poisson distribution with a mean of $4.$ The thing that has an exponential distribution is the time until the next breakdown, which has an expected value of $1/4text day.$ With the two elevators together the mean waiting time is $1/10text day.$

Since $2text hours = 1/12text day,$ the probability that it happens within that time is $1- e^-(1/12)/(1/10) = 1 - e^-10/12 approx 0.5654.$

One would speak here not of the minimum of two exponential distributions, but of the minimum of two exponentially distributed random variables.

share|cite|improve this answer

answered Jul 17 at 18:14

Michael Hardy

204k23186462

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ahh.. I see.. so I confused the Poisson rate with the exponential rate then? I'm still a bit confused.. If I write the CDF in the form $1-$$e^-λx$, then what does $λ$ represent?
  – Saad
  Jul 17 at 18:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : I'd call it a "rate" either way, but maybe an "average" or "expected value" when talking about the Poisson distribution. The Poisson distribution is the discrete distribution of the number of breakdowns, within the set $0,1,2,3,ldots,$ and the exponential distribution is the continuous distribution of the time until the next breakdown.
  – Michael Hardy
  Jul 17 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in our situation, the mean waiting time is $1/10$ day? But then wikipedia says that if you write the CDF in terms of the mean, it would be $1-$$e^-10x$ as per the alternative parameterization section en.wikipedia.org/wiki/Exponential_distribution where $x$ is the time in days before the next occurence
  – Saad
  Jul 17 at 18:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Saad : Right. The time in days is $1/12.$ so $1-e^-10x = 1 - e^-10/12. qquad$
  – Michael Hardy
  Jul 17 at 18:49
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

The number of breakdowns of the first elevator in a day has a Poisson distribution with a mean of $4.$ The thing that has an exponential distribution is the time until the next breakdown, which has an expected value of $1/4text day.$ With the two elevators together the mean waiting time is $1/10text day.$

Since $2text hours = 1/12text day,$ the probability that it happens within that time is $1- e^-(1/12)/(1/10) = 1 - e^-10/12 approx 0.5654.$

One would speak here not of the minimum of two exponential distributions, but of the minimum of two exponentially distributed random variables.

share|cite|improve this answer

answered Jul 17 at 18:14

Michael Hardy

204k23186462

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ahh.. I see.. so I confused the Poisson rate with the exponential rate then? I'm still a bit confused.. If I write the CDF in the form $1-$$e^-λx$, then what does $λ$ represent?
  – Saad
  Jul 17 at 18:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : I'd call it a "rate" either way, but maybe an "average" or "expected value" when talking about the Poisson distribution. The Poisson distribution is the discrete distribution of the number of breakdowns, within the set $0,1,2,3,ldots,$ and the exponential distribution is the continuous distribution of the time until the next breakdown.
  – Michael Hardy
  Jul 17 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in our situation, the mean waiting time is $1/10$ day? But then wikipedia says that if you write the CDF in terms of the mean, it would be $1-$$e^-10x$ as per the alternative parameterization section en.wikipedia.org/wiki/Exponential_distribution where $x$ is the time in days before the next occurence
  – Saad
  Jul 17 at 18:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Saad : Right. The time in days is $1/12.$ so $1-e^-10x = 1 - e^-10/12. qquad$
  – Michael Hardy
  Jul 17 at 18:49
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The number of breakdowns of the first elevator in a day has a Poisson distribution with a mean of $4.$ The thing that has an exponential distribution is the time until the next breakdown, which has an expected value of $1/4text day.$ With the two elevators together the mean waiting time is $1/10text day.$

Since $2text hours = 1/12text day,$ the probability that it happens within that time is $1- e^-(1/12)/(1/10) = 1 - e^-10/12 approx 0.5654.$

One would speak here not of the minimum of two exponential distributions, but of the minimum of two exponentially distributed random variables.

share|cite|improve this answer

answered Jul 17 at 18:14

Michael Hardy

204k23186462

The number of breakdowns of the first elevator in a day has a Poisson distribution with a mean of $4.$ The thing that has an exponential distribution is the time until the next breakdown, which has an expected value of $1/4text day.$ With the two elevators together the mean waiting time is $1/10text day.$

Since $2text hours = 1/12text day,$ the probability that it happens within that time is $1- e^-(1/12)/(1/10) = 1 - e^-10/12 approx 0.5654.$

One would speak here not of the minimum of two exponential distributions, but of the minimum of two exponentially distributed random variables.

share|cite|improve this answer

answered Jul 17 at 18:14

Michael Hardy

204k23186462

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 18:14

Michael Hardy

204k23186462

answered Jul 17 at 18:14

Michael Hardy

204k23186462

answered Jul 17 at 18:14

Michael Hardy

204k23186462

204k23186462

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ahh.. I see.. so I confused the Poisson rate with the exponential rate then? I'm still a bit confused.. If I write the CDF in the form $1-$$e^-λx$, then what does $λ$ represent?
  – Saad
  Jul 17 at 18:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : I'd call it a "rate" either way, but maybe an "average" or "expected value" when talking about the Poisson distribution. The Poisson distribution is the discrete distribution of the number of breakdowns, within the set $0,1,2,3,ldots,$ and the exponential distribution is the continuous distribution of the time until the next breakdown.
  – Michael Hardy
  Jul 17 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in our situation, the mean waiting time is $1/10$ day? But then wikipedia says that if you write the CDF in terms of the mean, it would be $1-$$e^-10x$ as per the alternative parameterization section en.wikipedia.org/wiki/Exponential_distribution where $x$ is the time in days before the next occurence
  – Saad
  Jul 17 at 18:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Saad : Right. The time in days is $1/12.$ so $1-e^-10x = 1 - e^-10/12. qquad$
  – Michael Hardy
  Jul 17 at 18:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ahh.. I see.. so I confused the Poisson rate with the exponential rate then? I'm still a bit confused.. If I write the CDF in the form $1-$$e^-λx$, then what does $λ$ represent?
  – Saad
  Jul 17 at 18:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : I'd call it a "rate" either way, but maybe an "average" or "expected value" when talking about the Poisson distribution. The Poisson distribution is the discrete distribution of the number of breakdowns, within the set $0,1,2,3,ldots,$ and the exponential distribution is the continuous distribution of the time until the next breakdown.
  – Michael Hardy
  Jul 17 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in our situation, the mean waiting time is $1/10$ day? But then wikipedia says that if you write the CDF in terms of the mean, it would be $1-$$e^-10x$ as per the alternative parameterization section en.wikipedia.org/wiki/Exponential_distribution where $x$ is the time in days before the next occurence
  – Saad
  Jul 17 at 18:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Saad : Right. The time in days is $1/12.$ so $1-e^-10x = 1 - e^-10/12. qquad$
  – Michael Hardy
  Jul 17 at 18:49
  

  
  
  
  

Ahh.. I see.. so I confused the Poisson rate with the exponential rate then? I'm still a bit confused.. If I write the CDF in the form $1-$$e^-λx$, then what does $λ$ represent?
– Saad
Jul 17 at 18:23

Ahh.. I see.. so I confused the Poisson rate with the exponential rate then? I'm still a bit confused.. If I write the CDF in the form $1-$$e^-λx$, then what does $λ$ represent?
– Saad
Jul 17 at 18:23

@Saad : I'd call it a "rate" either way, but maybe an "average" or "expected value" when talking about the Poisson distribution. The Poisson distribution is the discrete distribution of the number of breakdowns, within the set $0,1,2,3,ldots,$ and the exponential distribution is the continuous distribution of the time until the next breakdown.
– Michael Hardy
Jul 17 at 18:25

@Saad : I'd call it a "rate" either way, but maybe an "average" or "expected value" when talking about the Poisson distribution. The Poisson distribution is the discrete distribution of the number of breakdowns, within the set $0,1,2,3,ldots,$ and the exponential distribution is the continuous distribution of the time until the next breakdown.
– Michael Hardy
Jul 17 at 18:25

So in our situation, the mean waiting time is $1/10$ day? But then wikipedia says that if you write the CDF in terms of the mean, it would be $1-$$e^-10x$ as per the alternative parameterization section en.wikipedia.org/wiki/Exponential_distribution where $x$ is the time in days before the next occurence
– Saad
Jul 17 at 18:30

So in our situation, the mean waiting time is $1/10$ day? But then wikipedia says that if you write the CDF in terms of the mean, it would be $1-$$e^-10x$ as per the alternative parameterization section en.wikipedia.org/wiki/Exponential_distribution where $x$ is the time in days before the next occurence
– Saad
Jul 17 at 18:30

1

1

@Saad : Right. The time in days is $1/12.$ so $1-e^-10x = 1 - e^-10/12. qquad$
– Michael Hardy
Jul 17 at 18:49

@Saad : Right. The time in days is $1/12.$ so $1-e^-10x = 1 - e^-10/12. qquad$
– Michael Hardy
Jul 17 at 18:49

add a comment | 

up vote
1
down vote

Note that the rates are $4$ breakdowns per $24$ hours and $6$ breakdowns per $24$ hours. Therefore, $lambda_1 = frac424$, and $lambda_2 = frac624$. Using them, you obtain
$$PrZle 2 = 1-exp(-20/24).$$

share|cite|improve this answer

answered Jul 17 at 18:09

Math Lover

12.4k21232

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait.. if $X$~$exp(λ)$, where $λ$ is the rate, then $F(x) = 1-e^-λx$.. isn't the rate 10 in our case?
  – Saad
  Jul 17 at 18:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad It is $(4+6)/24=10/24$.
  – Math Lover
  Jul 17 at 18:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : Proper MathJax code might be $$ Xsim exp(lambda), $$ although I would prefer $XsimoperatornameExp(lambda)$ because often $exp(lambda)$ means $e^-lambda. qquad$
  – Michael Hardy
  Jul 17 at 18:16
  

  
  
  
  

add a comment | 

up vote
1
down vote

Note that the rates are $4$ breakdowns per $24$ hours and $6$ breakdowns per $24$ hours. Therefore, $lambda_1 = frac424$, and $lambda_2 = frac624$. Using them, you obtain
$$PrZle 2 = 1-exp(-20/24).$$

share|cite|improve this answer

answered Jul 17 at 18:09

Math Lover

12.4k21232

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait.. if $X$~$exp(λ)$, where $λ$ is the rate, then $F(x) = 1-e^-λx$.. isn't the rate 10 in our case?
  – Saad
  Jul 17 at 18:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad It is $(4+6)/24=10/24$.
  – Math Lover
  Jul 17 at 18:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : Proper MathJax code might be $$ Xsim exp(lambda), $$ although I would prefer $XsimoperatornameExp(lambda)$ because often $exp(lambda)$ means $e^-lambda. qquad$
  – Michael Hardy
  Jul 17 at 18:16
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Note that the rates are $4$ breakdowns per $24$ hours and $6$ breakdowns per $24$ hours. Therefore, $lambda_1 = frac424$, and $lambda_2 = frac624$. Using them, you obtain
$$PrZle 2 = 1-exp(-20/24).$$

share|cite|improve this answer

answered Jul 17 at 18:09

Math Lover

12.4k21232

Note that the rates are $4$ breakdowns per $24$ hours and $6$ breakdowns per $24$ hours. Therefore, $lambda_1 = frac424$, and $lambda_2 = frac624$. Using them, you obtain
$$PrZle 2 = 1-exp(-20/24).$$

share|cite|improve this answer

answered Jul 17 at 18:09

Math Lover

12.4k21232

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 18:09

Math Lover

12.4k21232

answered Jul 17 at 18:09

Math Lover

12.4k21232

answered Jul 17 at 18:09

Math Lover

12.4k21232

12.4k21232

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait.. if $X$~$exp(λ)$, where $λ$ is the rate, then $F(x) = 1-e^-λx$.. isn't the rate 10 in our case?
  – Saad
  Jul 17 at 18:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad It is $(4+6)/24=10/24$.
  – Math Lover
  Jul 17 at 18:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : Proper MathJax code might be $$ Xsim exp(lambda), $$ although I would prefer $XsimoperatornameExp(lambda)$ because often $exp(lambda)$ means $e^-lambda. qquad$
  – Michael Hardy
  Jul 17 at 18:16
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait.. if $X$~$exp(λ)$, where $λ$ is the rate, then $F(x) = 1-e^-λx$.. isn't the rate 10 in our case?
  – Saad
  Jul 17 at 18:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad It is $(4+6)/24=10/24$.
  – Math Lover
  Jul 17 at 18:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Saad : Proper MathJax code might be $$ Xsim exp(lambda), $$ although I would prefer $XsimoperatornameExp(lambda)$ because often $exp(lambda)$ means $e^-lambda. qquad$
  – Michael Hardy
  Jul 17 at 18:16
  

  
  
  
  

Wait.. if $X$~$exp(λ)$, where $λ$ is the rate, then $F(x) = 1-e^-λx$.. isn't the rate 10 in our case?
– Saad
Jul 17 at 18:13

Wait.. if $X$~$exp(λ)$, where $λ$ is the rate, then $F(x) = 1-e^-λx$.. isn't the rate 10 in our case?
– Saad
Jul 17 at 18:13

@Saad It is $(4+6)/24=10/24$.
– Math Lover
Jul 17 at 18:14

@Saad It is $(4+6)/24=10/24$.
– Math Lover
Jul 17 at 18:14

@Saad : Proper MathJax code might be $$ Xsim exp(lambda), $$ although I would prefer $XsimoperatornameExp(lambda)$ because often $exp(lambda)$ means $e^-lambda. qquad$
– Michael Hardy
Jul 17 at 18:16

@Saad : Proper MathJax code might be $$ Xsim exp(lambda), $$ although I would prefer $XsimoperatornameExp(lambda)$ because often $exp(lambda)$ means $e^-lambda. qquad$
– Michael Hardy
Jul 17 at 18:16

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