Distribution of sum of independent exponentials with random number of summandsRelationship between poisson...
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Distribution of sum of independent exponentials with random number of summands
Relationship between poisson and exponential distributionHow can I efficiently approximate the sum of Bernoulli random variables for any number of summands in partial sum?The weighted sum of two independent Poisson random variablesRandom number from exponential distribution with a scale parameterSum of two independent random variables (application: inter-arrival times)Which parameter should be considered as “scale” parameter for Gamma distribution?Distribution function of an exponential random variableCompound Poisson Distribution with sum of exponential random variablesSum of linear combination of product of exponentials is exponentialDistribution of sum of exponentialsDistribution of a Random Sum
$begingroup$
Let $tau_isimexpleft(lambdaright)$ be independent and identically distributed exponentials with parameter $lambda$. Then, for given $n$, the sum of these values
$$T_n := sum_{i=0}^n tau_i$$
follows an Erlang-Distribution with probability density function
$$pi(T_n=T| n,lambda)={lambda^n T^{n-1} e^{-lambda T} over (n-1)!}quadmbox{for }T, lambda geq 0.$$
I am interested in the distribution of $T_tilde n$ where $tilde n$ is a random variable such that for $tau_a sim exp(lambda_a)$ exponentially distributed, it holds that
$$T_tilde n leq tau_a \T_{tilde{n}+1} > tau_a.$$
In other words, $T_{tilde n}$ is truncated by an exponential distribution. I fail in deriving the distribution of $tilde n$ but maybe there is an easier way:
$$pileft(tilde n = kright) = pileft(T_n < tau_a|n=kright)\
=1-intlimits_{mathbb{R}^+}sumlimits_{n=0}^{k-1}frac{1}{n!}expleft(-(lambda+lambda_a)tau_aright)(taulambda_a)^nlambda_adtau_a.$$
However, just sampling and eye-balling looks to me like this density isn't that ugly:
iter <- 20000
lambda_a <- 1
lambda <- 2
df <- data.frame(tau=rep(NA, iter), a=rep(NA, iter))
for(i in 1:iter){
set.seed(i)
a <- rexp(1, rate = lambda_a)
s <- cumsum(rexp(500, rate = lambda))
df[i,] <- c(max(s[1], s[s<a]), a)
}
library(tidyverse)
ggplot(df %>% gather(), aes(x = value, fill = key)) +
geom_density(alpha = .3) + theme_bw()
distributions exponential-family truncation
asked 3 hours ago
muffin1974muffin1974
519419
$endgroup$
add a comment |
$begingroup$
Let $tau_isimexpleft(lambdaright)$ be independent and identically distributed exponentials with parameter $lambda$. Then, for given $n$, the sum of these values
$$T_n := sum_{i=0}^n tau_i$$
follows an Erlang-Distribution with probability density function
$$pi(T_n=T| n,lambda)={lambda^n T^{n-1} e^{-lambda T} over (n-1)!}quadmbox{for }T, lambda geq 0.$$
I am interested in the distribution of $T_tilde n$ where $tilde n$ is a random variable such that for $tau_a sim exp(lambda_a)$ exponentially distributed, it holds that
$$T_tilde n leq tau_a \T_{tilde{n}+1} > tau_a.$$
In other words, $T_{tilde n}$ is truncated by an exponential distribution. I fail in deriving the distribution of $tilde n$ but maybe there is an easier way:
$$pileft(tilde n = kright) = pileft(T_n < tau_a|n=kright)\
=1-intlimits_{mathbb{R}^+}sumlimits_{n=0}^{k-1}frac{1}{n!}expleft(-(lambda+lambda_a)tau_aright)(taulambda_a)^nlambda_adtau_a.$$
However, just sampling and eye-balling looks to me like this density isn't that ugly:
iter <- 20000
lambda_a <- 1
lambda <- 2
df <- data.frame(tau=rep(NA, iter), a=rep(NA, iter))
for(i in 1:iter){
set.seed(i)
a <- rexp(1, rate = lambda_a)
s <- cumsum(rexp(500, rate = lambda))
df[i,] <- c(max(s[1], s[s<a]), a)
}
library(tidyverse)
ggplot(df %>% gather(), aes(x = value, fill = key)) +
geom_density(alpha = .3) + theme_bw()
distributions exponential-family truncation
asked 3 hours ago
muffin1974muffin1974
519419
$endgroup$
1
$begingroup$
I would recommend not using the same notation for both $tau_i$ and the sum $tau_n$.
$endgroup$
– ukemi
3 hours ago
add a comment |
$begingroup$
Let $tau_isimexpleft(lambdaright)$ be independent and identically distributed exponentials with parameter $lambda$. Then, for given $n$, the sum of these values
$$T_n := sum_{i=0}^n tau_i$$
follows an Erlang-Distribution with probability density function
$$pi(T_n=T| n,lambda)={lambda^n T^{n-1} e^{-lambda T} over (n-1)!}quadmbox{for }T, lambda geq 0.$$
I am interested in the distribution of $T_tilde n$ where $tilde n$ is a random variable such that for $tau_a sim exp(lambda_a)$ exponentially distributed, it holds that
$$T_tilde n leq tau_a \T_{tilde{n}+1} > tau_a.$$
In other words, $T_{tilde n}$ is truncated by an exponential distribution. I fail in deriving the distribution of $tilde n$ but maybe there is an easier way:
$$pileft(tilde n = kright) = pileft(T_n < tau_a|n=kright)\
=1-intlimits_{mathbb{R}^+}sumlimits_{n=0}^{k-1}frac{1}{n!}expleft(-(lambda+lambda_a)tau_aright)(taulambda_a)^nlambda_adtau_a.$$
However, just sampling and eye-balling looks to me like this density isn't that ugly:
iter <- 20000
lambda_a <- 1
lambda <- 2
df <- data.frame(tau=rep(NA, iter), a=rep(NA, iter))
for(i in 1:iter){
set.seed(i)
a <- rexp(1, rate = lambda_a)
s <- cumsum(rexp(500, rate = lambda))
df[i,] <- c(max(s[1], s[s<a]), a)
}
library(tidyverse)
ggplot(df %>% gather(), aes(x = value, fill = key)) +
geom_density(alpha = .3) + theme_bw()
distributions exponential-family truncation
asked 3 hours ago
muffin1974muffin1974
519419
$endgroup$
Let $tau_isimexpleft(lambdaright)$ be independent and identically distributed exponentials with parameter $lambda$. Then, for given $n$, the sum of these values
$$T_n := sum_{i=0}^n tau_i$$
follows an Erlang-Distribution with probability density function
$$pi(T_n=T| n,lambda)={lambda^n T^{n-1} e^{-lambda T} over (n-1)!}quadmbox{for }T, lambda geq 0.$$
I am interested in the distribution of $T_tilde n$ where $tilde n$ is a random variable such that for $tau_a sim exp(lambda_a)$ exponentially distributed, it holds that
$$T_tilde n leq tau_a \T_{tilde{n}+1} > tau_a.$$
In other words, $T_{tilde n}$ is truncated by an exponential distribution. I fail in deriving the distribution of $tilde n$ but maybe there is an easier way:
$$pileft(tilde n = kright) = pileft(T_n < tau_a|n=kright)\
=1-intlimits_{mathbb{R}^+}sumlimits_{n=0}^{k-1}frac{1}{n!}expleft(-(lambda+lambda_a)tau_aright)(taulambda_a)^nlambda_adtau_a.$$
However, just sampling and eye-balling looks to me like this density isn't that ugly:
iter <- 20000
lambda_a <- 1
lambda <- 2
df <- data.frame(tau=rep(NA, iter), a=rep(NA, iter))
for(i in 1:iter){
set.seed(i)
a <- rexp(1, rate = lambda_a)
s <- cumsum(rexp(500, rate = lambda))
df[i,] <- c(max(s[1], s[s<a]), a)
}
library(tidyverse)
ggplot(df %>% gather(), aes(x = value, fill = key)) +
geom_density(alpha = .3) + theme_bw()
distributions exponential-family truncation
distributions exponential-family truncation
asked 3 hours ago
muffin1974muffin1974
519419
asked 3 hours ago
muffin1974muffin1974
519419
edited 3 hours ago
muffin1974
asked 3 hours ago
muffin1974muffin1974
519419
asked 3 hours ago
muffin1974muffin1974
519419
asked 3 hours ago
muffin1974muffin1974
519419
519419
1
$begingroup$
I would recommend not using the same notation for both $tau_i$ and the sum $tau_n$.
$endgroup$
– ukemi
3 hours ago
add a comment |
1
$begingroup$
I would recommend not using the same notation for both $tau_i$ and the sum $tau_n$.
$endgroup$
– ukemi
3 hours ago
1
1
$begingroup$
I would recommend not using the same notation for both $tau_i$ and the sum $tau_n$.
$endgroup$
– ukemi
3 hours ago
$begingroup$
I would recommend not using the same notation for both $tau_i$ and the sum $tau_n$.
$endgroup$
– ukemi
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As detailed in this X validated answer, waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed one produces an Poisson $mathcal P(lambda)$ variate $N$. Hence waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed $tau_a$ produces an Poisson $mathcal P(tau_alambda)$ variate $N$, conditional on $tau_a$. Therefore
begin{align*}
mathbb P(N=n)&=int_0^infty mathbb P(N=n|tau_a) ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&= int_0^infty dfrac{(lambdatau_a)^n}{n!},e^{-tau_alambda} ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},int_0^infty tau_a^n,e^{-tau_a(lambda+lambda_a)} ,text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},dfrac{Gamma(n+1)}{(lambda_a+lambda)^{n+1}}=dfrac{lambda_alambda^n}{(lambda_a+lambda)^{n+1}}
end{align*}
which is a Geometric $mathcal G(lambda_a/{lambda_a+lambda})$ random variable.
Considering now the distribution of$$zeta=sum_{i=1}^N tau_i$$the moment generating function of $zeta$ is
$$mathbb E[e^{zzeta}]=mathbb E[e^{z{tau_1+cdots+tau_N}}]=mathbb E^N[mathbb E^{tau_1}[e^{ztau_1}]^N]=E^N[{lambda/(lambda-z)}^N]=E^N[e^{N(ln lambda-ln (lambda-z))}]$$and the mgf of a Geometric $mathcal G(p)$ variate is
$$varphi_N(z)=dfrac{pe^z}{1-(1-p)e^z}$$Hence the moment generating function of $zeta$ is $$dfrac{pe^{ln lambda-ln (lambda-z)}}{1-(1-(lambda_a/{lambda_a+lambda}))e^{ln lambda-ln (lambda-z)}}=dfrac{p lambda}{ lambda-z-lambda^2/{lambda_a+lambda}}$$
modulo typing mistakes
answered 3 hours ago
Xi'anXi'an
57.2k895360
$endgroup$
$begingroup$
Thanks a lot @Xi'an! Do I get it right that in your notation $p= lambda_a/(lambda_a+lambda)$ ? Because then the moment generating function in the last line is equivalent to $frac{1}{1-zleft(plambdaright)^{-1}}$ which corresponds to the MGF of an exponential distribution..
$endgroup$
– muffin1974
2 hours ago
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As detailed in this X validated answer, waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed one produces an Poisson $mathcal P(lambda)$ variate $N$. Hence waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed $tau_a$ produces an Poisson $mathcal P(tau_alambda)$ variate $N$, conditional on $tau_a$. Therefore
begin{align*}
mathbb P(N=n)&=int_0^infty mathbb P(N=n|tau_a) ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&= int_0^infty dfrac{(lambdatau_a)^n}{n!},e^{-tau_alambda} ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},int_0^infty tau_a^n,e^{-tau_a(lambda+lambda_a)} ,text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},dfrac{Gamma(n+1)}{(lambda_a+lambda)^{n+1}}=dfrac{lambda_alambda^n}{(lambda_a+lambda)^{n+1}}
end{align*}
which is a Geometric $mathcal G(lambda_a/{lambda_a+lambda})$ random variable.
Considering now the distribution of$$zeta=sum_{i=1}^N tau_i$$the moment generating function of $zeta$ is
$$mathbb E[e^{zzeta}]=mathbb E[e^{z{tau_1+cdots+tau_N}}]=mathbb E^N[mathbb E^{tau_1}[e^{ztau_1}]^N]=E^N[{lambda/(lambda-z)}^N]=E^N[e^{N(ln lambda-ln (lambda-z))}]$$and the mgf of a Geometric $mathcal G(p)$ variate is
$$varphi_N(z)=dfrac{pe^z}{1-(1-p)e^z}$$Hence the moment generating function of $zeta$ is $$dfrac{pe^{ln lambda-ln (lambda-z)}}{1-(1-(lambda_a/{lambda_a+lambda}))e^{ln lambda-ln (lambda-z)}}=dfrac{p lambda}{ lambda-z-lambda^2/{lambda_a+lambda}}$$
modulo typing mistakes
answered 3 hours ago
Xi'anXi'an
57.2k895360
$endgroup$
$begingroup$
Thanks a lot @Xi'an! Do I get it right that in your notation $p= lambda_a/(lambda_a+lambda)$ ? Because then the moment generating function in the last line is equivalent to $frac{1}{1-zleft(plambdaright)^{-1}}$ which corresponds to the MGF of an exponential distribution..
$endgroup$
– muffin1974
2 hours ago
add a comment |
$begingroup$
As detailed in this X validated answer, waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed one produces an Poisson $mathcal P(lambda)$ variate $N$. Hence waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed $tau_a$ produces an Poisson $mathcal P(tau_alambda)$ variate $N$, conditional on $tau_a$. Therefore
begin{align*}
mathbb P(N=n)&=int_0^infty mathbb P(N=n|tau_a) ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&= int_0^infty dfrac{(lambdatau_a)^n}{n!},e^{-tau_alambda} ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},int_0^infty tau_a^n,e^{-tau_a(lambda+lambda_a)} ,text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},dfrac{Gamma(n+1)}{(lambda_a+lambda)^{n+1}}=dfrac{lambda_alambda^n}{(lambda_a+lambda)^{n+1}}
end{align*}
which is a Geometric $mathcal G(lambda_a/{lambda_a+lambda})$ random variable.
Considering now the distribution of$$zeta=sum_{i=1}^N tau_i$$the moment generating function of $zeta$ is
$$mathbb E[e^{zzeta}]=mathbb E[e^{z{tau_1+cdots+tau_N}}]=mathbb E^N[mathbb E^{tau_1}[e^{ztau_1}]^N]=E^N[{lambda/(lambda-z)}^N]=E^N[e^{N(ln lambda-ln (lambda-z))}]$$and the mgf of a Geometric $mathcal G(p)$ variate is
$$varphi_N(z)=dfrac{pe^z}{1-(1-p)e^z}$$Hence the moment generating function of $zeta$ is $$dfrac{pe^{ln lambda-ln (lambda-z)}}{1-(1-(lambda_a/{lambda_a+lambda}))e^{ln lambda-ln (lambda-z)}}=dfrac{p lambda}{ lambda-z-lambda^2/{lambda_a+lambda}}$$
modulo typing mistakes
answered 3 hours ago
Xi'anXi'an
57.2k895360
$endgroup$
$begingroup$
Thanks a lot @Xi'an! Do I get it right that in your notation $p= lambda_a/(lambda_a+lambda)$ ? Because then the moment generating function in the last line is equivalent to $frac{1}{1-zleft(plambdaright)^{-1}}$ which corresponds to the MGF of an exponential distribution..
$endgroup$
– muffin1974
2 hours ago
add a comment |
$begingroup$
As detailed in this X validated answer, waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed one produces an Poisson $mathcal P(lambda)$ variate $N$. Hence waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed $tau_a$ produces an Poisson $mathcal P(tau_alambda)$ variate $N$, conditional on $tau_a$. Therefore
begin{align*}
mathbb P(N=n)&=int_0^infty mathbb P(N=n|tau_a) ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&= int_0^infty dfrac{(lambdatau_a)^n}{n!},e^{-tau_alambda} ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},int_0^infty tau_a^n,e^{-tau_a(lambda+lambda_a)} ,text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},dfrac{Gamma(n+1)}{(lambda_a+lambda)^{n+1}}=dfrac{lambda_alambda^n}{(lambda_a+lambda)^{n+1}}
end{align*}
which is a Geometric $mathcal G(lambda_a/{lambda_a+lambda})$ random variable.
Considering now the distribution of$$zeta=sum_{i=1}^N tau_i$$the moment generating function of $zeta$ is
$$mathbb E[e^{zzeta}]=mathbb E[e^{z{tau_1+cdots+tau_N}}]=mathbb E^N[mathbb E^{tau_1}[e^{ztau_1}]^N]=E^N[{lambda/(lambda-z)}^N]=E^N[e^{N(ln lambda-ln (lambda-z))}]$$and the mgf of a Geometric $mathcal G(p)$ variate is
$$varphi_N(z)=dfrac{pe^z}{1-(1-p)e^z}$$Hence the moment generating function of $zeta$ is $$dfrac{pe^{ln lambda-ln (lambda-z)}}{1-(1-(lambda_a/{lambda_a+lambda}))e^{ln lambda-ln (lambda-z)}}=dfrac{p lambda}{ lambda-z-lambda^2/{lambda_a+lambda}}$$
modulo typing mistakes
answered 3 hours ago
Xi'anXi'an
57.2k895360
$endgroup$
As detailed in this X validated answer, waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed one produces an Poisson $mathcal P(lambda)$ variate $N$. Hence waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed $tau_a$ produces an Poisson $mathcal P(tau_alambda)$ variate $N$, conditional on $tau_a$. Therefore
begin{align*}
mathbb P(N=n)&=int_0^infty mathbb P(N=n|tau_a) ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&= int_0^infty dfrac{(lambdatau_a)^n}{n!},e^{-tau_alambda} ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},int_0^infty tau_a^n,e^{-tau_a(lambda+lambda_a)} ,text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},dfrac{Gamma(n+1)}{(lambda_a+lambda)^{n+1}}=dfrac{lambda_alambda^n}{(lambda_a+lambda)^{n+1}}
end{align*}
which is a Geometric $mathcal G(lambda_a/{lambda_a+lambda})$ random variable.
Considering now the distribution of$$zeta=sum_{i=1}^N tau_i$$the moment generating function of $zeta$ is
$$mathbb E[e^{zzeta}]=mathbb E[e^{z{tau_1+cdots+tau_N}}]=mathbb E^N[mathbb E^{tau_1}[e^{ztau_1}]^N]=E^N[{lambda/(lambda-z)}^N]=E^N[e^{N(ln lambda-ln (lambda-z))}]$$and the mgf of a Geometric $mathcal G(p)$ variate is
$$varphi_N(z)=dfrac{pe^z}{1-(1-p)e^z}$$Hence the moment generating function of $zeta$ is $$dfrac{pe^{ln lambda-ln (lambda-z)}}{1-(1-(lambda_a/{lambda_a+lambda}))e^{ln lambda-ln (lambda-z)}}=dfrac{p lambda}{ lambda-z-lambda^2/{lambda_a+lambda}}$$
modulo typing mistakes
answered 3 hours ago
Xi'anXi'an
57.2k895360
answered 3 hours ago
Xi'anXi'an
57.2k895360
answered 3 hours ago
Xi'anXi'an
57.2k895360
answered 3 hours ago
Xi'anXi'an
57.2k895360
57.2k895360
$begingroup$
Thanks a lot @Xi'an! Do I get it right that in your notation $p= lambda_a/(lambda_a+lambda)$ ? Because then the moment generating function in the last line is equivalent to $frac{1}{1-zleft(plambdaright)^{-1}}$ which corresponds to the MGF of an exponential distribution..
$endgroup$
– muffin1974
2 hours ago
add a comment |
$begingroup$
Thanks a lot @Xi'an! Do I get it right that in your notation $p= lambda_a/(lambda_a+lambda)$ ? Because then the moment generating function in the last line is equivalent to $frac{1}{1-zleft(plambdaright)^{-1}}$ which corresponds to the MGF of an exponential distribution..
$endgroup$
– muffin1974
2 hours ago
$begingroup$
Thanks a lot @Xi'an! Do I get it right that in your notation $p= lambda_a/(lambda_a+lambda)$ ? Because then the moment generating function in the last line is equivalent to $frac{1}{1-zleft(plambdaright)^{-1}}$ which corresponds to the MGF of an exponential distribution..
$endgroup$
– muffin1974
2 hours ago
$begingroup$
Thanks a lot @Xi'an! Do I get it right that in your notation $p= lambda_a/(lambda_a+lambda)$ ? Because then the moment generating function in the last line is equivalent to $frac{1}{1-zleft(plambdaright)^{-1}}$ which corresponds to the MGF of an exponential distribution..
$endgroup$
– muffin1974
2 hours ago
add a comment |
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$begingroup$
I would recommend not using the same notation for both $tau_i$ and the sum $tau_n$.
$endgroup$
– ukemi
3 hours ago