Erlang Case of a Gamma Distribution
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For part a) I get $ E(X)=alphabeta=frac{n}{lambda}. $ Thus the answer is $frac{10}{0.5}=20$ minutes. I am not sure how to do b). Any help?
The special case of the gamma distribution in
which $alpha$ is a positive integer $n$ is called an Erlang
distribution. If we replace $beta$ by $1/lambda$ in Expression
(4.7), the Erlang pdf is
$ f(x;lambda,n)=frac{lambda(lambda{x})^{n-1}e^{-lambda{x}}}{(n-1!)} $ for $ xgeq0$ and $ 0 $ otherwise.
It can be shown that if the times between successive
events are independent, each with an exponential
distribution with parameter $lambda$, then the
total time $X$ that elapses before all of the next $n$
events occur has pdf $f(x; l, n)$.
a. What is the expected value of $X$? If the time (in
minutes) between arrivals of successive customers
is exponentially distributed with $lambda = .5$,
how much time can be expected to elapse
before the tenth customer arrives?
b. If customer interarrival time is exponentially
distributed with $lambda = .5$, what is the probability
that the tenth customer (after the one who has
just arrived) will arrive within the next
30 min?
c. The event ${X leq t}$ occurs if and only if at least $n$
events occur in the next t units of time. Use the
fact that the number of events occurring in an
interval of length $t$ has a Poisson distribution
with parameter $lambda{t}$ to write an expression (involving
Poisson probabilities) for the Erlang
cumulative distribution function $F(t;lambda,n)=O(Xleq t) $.
statistics gamma-distribution
asked Nov 12 at 17:33
David Wang
134
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up vote
1
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For part a) I get $ E(X)=alphabeta=frac{n}{lambda}. $ Thus the answer is $frac{10}{0.5}=20$ minutes. I am not sure how to do b). Any help?
The special case of the gamma distribution in
which $alpha$ is a positive integer $n$ is called an Erlang
distribution. If we replace $beta$ by $1/lambda$ in Expression
(4.7), the Erlang pdf is
$ f(x;lambda,n)=frac{lambda(lambda{x})^{n-1}e^{-lambda{x}}}{(n-1!)} $ for $ xgeq0$ and $ 0 $ otherwise.
It can be shown that if the times between successive
events are independent, each with an exponential
distribution with parameter $lambda$, then the
total time $X$ that elapses before all of the next $n$
events occur has pdf $f(x; l, n)$.
a. What is the expected value of $X$? If the time (in
minutes) between arrivals of successive customers
is exponentially distributed with $lambda = .5$,
how much time can be expected to elapse
before the tenth customer arrives?
b. If customer interarrival time is exponentially
distributed with $lambda = .5$, what is the probability
that the tenth customer (after the one who has
just arrived) will arrive within the next
30 min?
c. The event ${X leq t}$ occurs if and only if at least $n$
events occur in the next t units of time. Use the
fact that the number of events occurring in an
interval of length $t$ has a Poisson distribution
with parameter $lambda{t}$ to write an expression (involving
Poisson probabilities) for the Erlang
cumulative distribution function $F(t;lambda,n)=O(Xleq t) $.
statistics gamma-distribution
asked Nov 12 at 17:33
David Wang
134
New contributor
David Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
(b) does not seem to have a simple closed form. So what methods do you allow? For example in R you might use something likepgamma(30, shape=10, rate=0.5)
– Henry
Nov 12 at 20:28
Sorry, where should I input those numbers? I tried integrating to get a cdf but it seemed to complicated. Thanks!
– David Wang
Nov 12 at 20:35
Sorry, didn't seem to read your question right, I am not sure what methods can be used to solve this problem, for example I am not sure if finding a cdf will work or if I should find std. deviation and try to use that.
– David Wang
Nov 12 at 20:44
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For part a) I get $ E(X)=alphabeta=frac{n}{lambda}. $ Thus the answer is $frac{10}{0.5}=20$ minutes. I am not sure how to do b). Any help?
The special case of the gamma distribution in
which $alpha$ is a positive integer $n$ is called an Erlang
distribution. If we replace $beta$ by $1/lambda$ in Expression
(4.7), the Erlang pdf is
$ f(x;lambda,n)=frac{lambda(lambda{x})^{n-1}e^{-lambda{x}}}{(n-1!)} $ for $ xgeq0$ and $ 0 $ otherwise.
It can be shown that if the times between successive
events are independent, each with an exponential
distribution with parameter $lambda$, then the
total time $X$ that elapses before all of the next $n$
events occur has pdf $f(x; l, n)$.
a. What is the expected value of $X$? If the time (in
minutes) between arrivals of successive customers
is exponentially distributed with $lambda = .5$,
how much time can be expected to elapse
before the tenth customer arrives?
b. If customer interarrival time is exponentially
distributed with $lambda = .5$, what is the probability
that the tenth customer (after the one who has
just arrived) will arrive within the next
30 min?
c. The event ${X leq t}$ occurs if and only if at least $n$
events occur in the next t units of time. Use the
fact that the number of events occurring in an
interval of length $t$ has a Poisson distribution
with parameter $lambda{t}$ to write an expression (involving
Poisson probabilities) for the Erlang
cumulative distribution function $F(t;lambda,n)=O(Xleq t) $.
statistics gamma-distribution
asked Nov 12 at 17:33
David Wang
134
New contributor
David Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For part a) I get $ E(X)=alphabeta=frac{n}{lambda}. $ Thus the answer is $frac{10}{0.5}=20$ minutes. I am not sure how to do b). Any help?
The special case of the gamma distribution in
which $alpha$ is a positive integer $n$ is called an Erlang
distribution. If we replace $beta$ by $1/lambda$ in Expression
(4.7), the Erlang pdf is
$ f(x;lambda,n)=frac{lambda(lambda{x})^{n-1}e^{-lambda{x}}}{(n-1!)} $ for $ xgeq0$ and $ 0 $ otherwise.
It can be shown that if the times between successive
events are independent, each with an exponential
distribution with parameter $lambda$, then the
total time $X$ that elapses before all of the next $n$
events occur has pdf $f(x; l, n)$.
a. What is the expected value of $X$? If the time (in
minutes) between arrivals of successive customers
is exponentially distributed with $lambda = .5$,
how much time can be expected to elapse
before the tenth customer arrives?
b. If customer interarrival time is exponentially
distributed with $lambda = .5$, what is the probability
that the tenth customer (after the one who has
just arrived) will arrive within the next
30 min?
c. The event ${X leq t}$ occurs if and only if at least $n$
events occur in the next t units of time. Use the
fact that the number of events occurring in an
interval of length $t$ has a Poisson distribution
with parameter $lambda{t}$ to write an expression (involving
Poisson probabilities) for the Erlang
cumulative distribution function $F(t;lambda,n)=O(Xleq t) $.
statistics gamma-distribution
statistics gamma-distribution
asked Nov 12 at 17:33
David Wang
134
New contributor
David Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 17:33
David Wang
134
New contributor
David Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 12 at 20:38
asked Nov 12 at 17:33
David Wang
134
New contributor
David Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 17:33
David Wang
134
asked Nov 12 at 17:33
David Wang
134
134
New contributor
David Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
David Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
David Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
(b) does not seem to have a simple closed form. So what methods do you allow? For example in R you might use something likepgamma(30, shape=10, rate=0.5)
– Henry
Nov 12 at 20:28
Sorry, where should I input those numbers? I tried integrating to get a cdf but it seemed to complicated. Thanks!
– David Wang
Nov 12 at 20:35
Sorry, didn't seem to read your question right, I am not sure what methods can be used to solve this problem, for example I am not sure if finding a cdf will work or if I should find std. deviation and try to use that.
– David Wang
Nov 12 at 20:44
add a comment |
(b) does not seem to have a simple closed form. So what methods do you allow? For example in R you might use something likepgamma(30, shape=10, rate=0.5)
– Henry
Nov 12 at 20:28
Sorry, where should I input those numbers? I tried integrating to get a cdf but it seemed to complicated. Thanks!
– David Wang
Nov 12 at 20:35
Sorry, didn't seem to read your question right, I am not sure what methods can be used to solve this problem, for example I am not sure if finding a cdf will work or if I should find std. deviation and try to use that.
– David Wang
Nov 12 at 20:44
(b) does not seem to have a simple closed form. So what methods do you allow? For example in R you might use something like
pgamma(30, shape=10, rate=0.5)– Henry
Nov 12 at 20:28
(b) does not seem to have a simple closed form. So what methods do you allow? For example in R you might use something like
pgamma(30, shape=10, rate=0.5)– Henry
Nov 12 at 20:28
Sorry, where should I input those numbers? I tried integrating to get a cdf but it seemed to complicated. Thanks!
– David Wang
Nov 12 at 20:35
Sorry, where should I input those numbers? I tried integrating to get a cdf but it seemed to complicated. Thanks!
– David Wang
Nov 12 at 20:35
Sorry, didn't seem to read your question right, I am not sure what methods can be used to solve this problem, for example I am not sure if finding a cdf will work or if I should find std. deviation and try to use that.
– David Wang
Nov 12 at 20:44
Sorry, didn't seem to read your question right, I am not sure what methods can be used to solve this problem, for example I am not sure if finding a cdf will work or if I should find std. deviation and try to use that.
– David Wang
Nov 12 at 20:44
add a comment |
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David Wang is a new contributor. Be nice, and check out our Code of Conduct.
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(b) does not seem to have a simple closed form. So what methods do you allow? For example in R you might use something like
pgamma(30, shape=10, rate=0.5)– Henry
Nov 12 at 20:28
Sorry, where should I input those numbers? I tried integrating to get a cdf but it seemed to complicated. Thanks!
– David Wang
Nov 12 at 20:35
Sorry, didn't seem to read your question right, I am not sure what methods can be used to solve this problem, for example I am not sure if finding a cdf will work or if I should find std. deviation and try to use that.
– David Wang
Nov 12 at 20:44