Compute integral for expected value in R.
From a previous assignment, I found that the extinction probability of of a branching process is given by solving the equation
$$G(s)=s,$$
where $G(s)=e^{lambda(s-1)}$ is the PGF and noting the smallest root in terms of $s$. The solution to the equation
$$e^{lambda(s-1)}-s=0$$
will be a function of $lambda$, namley $s(lambda)$ which can be computed numerically for a fixed $lambda.$
Now I'm asked to numerically find the expected value of this probability given that $lambdasimGamma(15,9)$. Let's for convention now use the function $s(Lambda)$, where $LambdasimGamma(15,9)$, I know that the expected value of a function of a random variable $X$ is given by
$$mathbb{E}[g(X)]=int_{infty}^{infty}g(x)cdotpi(x)dx,$$
where $pi(x)$ is the pdf of $X$. So in my case I get
$$mathbb{E}[s(Lambda)]=int_{0}^{infty}s(lambda)pi(lambda)dlambda=int_0^{infty}s(lambda)frac{9^{15}}{Gamma(15)}lambda^{15-1}e^{-9lambda}dlambda.$$
Since $9^{15}/Gamma(15)approx 2361.7$ we have that the integral to be numerically computed in R is
$$2361.7int_{0}^{infty}s(lambda)lambda^{14}e^{-9lambda}.$$
I figured, since we don't have a closed form expression for $s(lambda)$, that I need nestled functions, first express the solution of $G(s)-s=0$ in one function depending on $s$, keeping $lambda$ constant and then wrap this with another function only depending on $lambda,$ like this:
fun = function(lambda){
sLambda = function(s){exp(lambda*(s-1))-s}
roots = uniroot.all(sLambda,c(0,1))
probExtinction = roots[2] # since this root is the smallest one.
integrand = probExtinction*exp((-9)*lambda)*lambda^(14)
return(integrand)
}
ExpectedProb = 2361.7*integrate(fun,0,Inf)$value
But here is where trouble starts. I get the error below. I understand what the error means, it's that if the function values at the endpoints are not oposite, then the function never crosses the x-axis, thus no roots will be found. However, plotting the function $f(x)=e^{a(x-1)}-x$ I see that the function dramatically changes behavior when $a$ passes $1$. I don't see how to fix my code in order to deal with this issue.
Error in uniroot(f, lower = xseq[i], upper = xseq[i + 1], ...) :
f() values at end points not of opposite sign
In addition: Warning messages:
1: In lambda * (s - 1) :
longer object length is not a multiple of shorter object length
2: In if (is.na(f.lower)) stop("f.lower = f(lower) is NA") :
the condition has length > 1 and only the first element will be used
3: In if (is.na(f.upper)) stop("f.upper = f(upper) is NA") :
the condition has length > 1 and only the first element will be used
Whats weird is that when I only run this code I don't get the uniroot-error:
lambda = 2
sLambda = function(s){exp(lambda*(s-1))-s}
roots = uniroot.all(sLambda,c(0,1))
probExtinction = roots[2])
Anyone knows how I can fix this?
probability integration probability-theory expected-value
add a comment |
From a previous assignment, I found that the extinction probability of of a branching process is given by solving the equation
$$G(s)=s,$$
where $G(s)=e^{lambda(s-1)}$ is the PGF and noting the smallest root in terms of $s$. The solution to the equation
$$e^{lambda(s-1)}-s=0$$
will be a function of $lambda$, namley $s(lambda)$ which can be computed numerically for a fixed $lambda.$
Now I'm asked to numerically find the expected value of this probability given that $lambdasimGamma(15,9)$. Let's for convention now use the function $s(Lambda)$, where $LambdasimGamma(15,9)$, I know that the expected value of a function of a random variable $X$ is given by
$$mathbb{E}[g(X)]=int_{infty}^{infty}g(x)cdotpi(x)dx,$$
where $pi(x)$ is the pdf of $X$. So in my case I get
$$mathbb{E}[s(Lambda)]=int_{0}^{infty}s(lambda)pi(lambda)dlambda=int_0^{infty}s(lambda)frac{9^{15}}{Gamma(15)}lambda^{15-1}e^{-9lambda}dlambda.$$
Since $9^{15}/Gamma(15)approx 2361.7$ we have that the integral to be numerically computed in R is
$$2361.7int_{0}^{infty}s(lambda)lambda^{14}e^{-9lambda}.$$
I figured, since we don't have a closed form expression for $s(lambda)$, that I need nestled functions, first express the solution of $G(s)-s=0$ in one function depending on $s$, keeping $lambda$ constant and then wrap this with another function only depending on $lambda,$ like this:
fun = function(lambda){
sLambda = function(s){exp(lambda*(s-1))-s}
roots = uniroot.all(sLambda,c(0,1))
probExtinction = roots[2] # since this root is the smallest one.
integrand = probExtinction*exp((-9)*lambda)*lambda^(14)
return(integrand)
}
ExpectedProb = 2361.7*integrate(fun,0,Inf)$value
But here is where trouble starts. I get the error below. I understand what the error means, it's that if the function values at the endpoints are not oposite, then the function never crosses the x-axis, thus no roots will be found. However, plotting the function $f(x)=e^{a(x-1)}-x$ I see that the function dramatically changes behavior when $a$ passes $1$. I don't see how to fix my code in order to deal with this issue.
Error in uniroot(f, lower = xseq[i], upper = xseq[i + 1], ...) :
f() values at end points not of opposite sign
In addition: Warning messages:
1: In lambda * (s - 1) :
longer object length is not a multiple of shorter object length
2: In if (is.na(f.lower)) stop("f.lower = f(lower) is NA") :
the condition has length > 1 and only the first element will be used
3: In if (is.na(f.upper)) stop("f.upper = f(upper) is NA") :
the condition has length > 1 and only the first element will be used
Whats weird is that when I only run this code I don't get the uniroot-error:
lambda = 2
sLambda = function(s){exp(lambda*(s-1))-s}
roots = uniroot.all(sLambda,c(0,1))
probExtinction = roots[2])
Anyone knows how I can fix this?
probability integration probability-theory expected-value
In general, if you getthe condition has length > 1 and only the first element will be used
that means your function is not suitable for vector arguments. Either uselapply
,sapply
or just do a loop so the argument is not a vector
– Yuriy S
Dec 4 '18 at 11:27
@YuriyS - Loop with what index? I don't understand where to place the loop. Can you elaborate?
– Parseval
Dec 4 '18 at 11:29
add a comment |
From a previous assignment, I found that the extinction probability of of a branching process is given by solving the equation
$$G(s)=s,$$
where $G(s)=e^{lambda(s-1)}$ is the PGF and noting the smallest root in terms of $s$. The solution to the equation
$$e^{lambda(s-1)}-s=0$$
will be a function of $lambda$, namley $s(lambda)$ which can be computed numerically for a fixed $lambda.$
Now I'm asked to numerically find the expected value of this probability given that $lambdasimGamma(15,9)$. Let's for convention now use the function $s(Lambda)$, where $LambdasimGamma(15,9)$, I know that the expected value of a function of a random variable $X$ is given by
$$mathbb{E}[g(X)]=int_{infty}^{infty}g(x)cdotpi(x)dx,$$
where $pi(x)$ is the pdf of $X$. So in my case I get
$$mathbb{E}[s(Lambda)]=int_{0}^{infty}s(lambda)pi(lambda)dlambda=int_0^{infty}s(lambda)frac{9^{15}}{Gamma(15)}lambda^{15-1}e^{-9lambda}dlambda.$$
Since $9^{15}/Gamma(15)approx 2361.7$ we have that the integral to be numerically computed in R is
$$2361.7int_{0}^{infty}s(lambda)lambda^{14}e^{-9lambda}.$$
I figured, since we don't have a closed form expression for $s(lambda)$, that I need nestled functions, first express the solution of $G(s)-s=0$ in one function depending on $s$, keeping $lambda$ constant and then wrap this with another function only depending on $lambda,$ like this:
fun = function(lambda){
sLambda = function(s){exp(lambda*(s-1))-s}
roots = uniroot.all(sLambda,c(0,1))
probExtinction = roots[2] # since this root is the smallest one.
integrand = probExtinction*exp((-9)*lambda)*lambda^(14)
return(integrand)
}
ExpectedProb = 2361.7*integrate(fun,0,Inf)$value
But here is where trouble starts. I get the error below. I understand what the error means, it's that if the function values at the endpoints are not oposite, then the function never crosses the x-axis, thus no roots will be found. However, plotting the function $f(x)=e^{a(x-1)}-x$ I see that the function dramatically changes behavior when $a$ passes $1$. I don't see how to fix my code in order to deal with this issue.
Error in uniroot(f, lower = xseq[i], upper = xseq[i + 1], ...) :
f() values at end points not of opposite sign
In addition: Warning messages:
1: In lambda * (s - 1) :
longer object length is not a multiple of shorter object length
2: In if (is.na(f.lower)) stop("f.lower = f(lower) is NA") :
the condition has length > 1 and only the first element will be used
3: In if (is.na(f.upper)) stop("f.upper = f(upper) is NA") :
the condition has length > 1 and only the first element will be used
Whats weird is that when I only run this code I don't get the uniroot-error:
lambda = 2
sLambda = function(s){exp(lambda*(s-1))-s}
roots = uniroot.all(sLambda,c(0,1))
probExtinction = roots[2])
Anyone knows how I can fix this?
probability integration probability-theory expected-value
From a previous assignment, I found that the extinction probability of of a branching process is given by solving the equation
$$G(s)=s,$$
where $G(s)=e^{lambda(s-1)}$ is the PGF and noting the smallest root in terms of $s$. The solution to the equation
$$e^{lambda(s-1)}-s=0$$
will be a function of $lambda$, namley $s(lambda)$ which can be computed numerically for a fixed $lambda.$
Now I'm asked to numerically find the expected value of this probability given that $lambdasimGamma(15,9)$. Let's for convention now use the function $s(Lambda)$, where $LambdasimGamma(15,9)$, I know that the expected value of a function of a random variable $X$ is given by
$$mathbb{E}[g(X)]=int_{infty}^{infty}g(x)cdotpi(x)dx,$$
where $pi(x)$ is the pdf of $X$. So in my case I get
$$mathbb{E}[s(Lambda)]=int_{0}^{infty}s(lambda)pi(lambda)dlambda=int_0^{infty}s(lambda)frac{9^{15}}{Gamma(15)}lambda^{15-1}e^{-9lambda}dlambda.$$
Since $9^{15}/Gamma(15)approx 2361.7$ we have that the integral to be numerically computed in R is
$$2361.7int_{0}^{infty}s(lambda)lambda^{14}e^{-9lambda}.$$
I figured, since we don't have a closed form expression for $s(lambda)$, that I need nestled functions, first express the solution of $G(s)-s=0$ in one function depending on $s$, keeping $lambda$ constant and then wrap this with another function only depending on $lambda,$ like this:
fun = function(lambda){
sLambda = function(s){exp(lambda*(s-1))-s}
roots = uniroot.all(sLambda,c(0,1))
probExtinction = roots[2] # since this root is the smallest one.
integrand = probExtinction*exp((-9)*lambda)*lambda^(14)
return(integrand)
}
ExpectedProb = 2361.7*integrate(fun,0,Inf)$value
But here is where trouble starts. I get the error below. I understand what the error means, it's that if the function values at the endpoints are not oposite, then the function never crosses the x-axis, thus no roots will be found. However, plotting the function $f(x)=e^{a(x-1)}-x$ I see that the function dramatically changes behavior when $a$ passes $1$. I don't see how to fix my code in order to deal with this issue.
Error in uniroot(f, lower = xseq[i], upper = xseq[i + 1], ...) :
f() values at end points not of opposite sign
In addition: Warning messages:
1: In lambda * (s - 1) :
longer object length is not a multiple of shorter object length
2: In if (is.na(f.lower)) stop("f.lower = f(lower) is NA") :
the condition has length > 1 and only the first element will be used
3: In if (is.na(f.upper)) stop("f.upper = f(upper) is NA") :
the condition has length > 1 and only the first element will be used
Whats weird is that when I only run this code I don't get the uniroot-error:
lambda = 2
sLambda = function(s){exp(lambda*(s-1))-s}
roots = uniroot.all(sLambda,c(0,1))
probExtinction = roots[2])
Anyone knows how I can fix this?
probability integration probability-theory expected-value
probability integration probability-theory expected-value
edited Dec 4 '18 at 11:07
asked Dec 4 '18 at 10:59
Parseval
2,7581718
2,7581718
In general, if you getthe condition has length > 1 and only the first element will be used
that means your function is not suitable for vector arguments. Either uselapply
,sapply
or just do a loop so the argument is not a vector
– Yuriy S
Dec 4 '18 at 11:27
@YuriyS - Loop with what index? I don't understand where to place the loop. Can you elaborate?
– Parseval
Dec 4 '18 at 11:29
add a comment |
In general, if you getthe condition has length > 1 and only the first element will be used
that means your function is not suitable for vector arguments. Either uselapply
,sapply
or just do a loop so the argument is not a vector
– Yuriy S
Dec 4 '18 at 11:27
@YuriyS - Loop with what index? I don't understand where to place the loop. Can you elaborate?
– Parseval
Dec 4 '18 at 11:29
In general, if you get
the condition has length > 1 and only the first element will be used
that means your function is not suitable for vector arguments. Either use lapply
, sapply
or just do a loop so the argument is not a vector– Yuriy S
Dec 4 '18 at 11:27
In general, if you get
the condition has length > 1 and only the first element will be used
that means your function is not suitable for vector arguments. Either use lapply
, sapply
or just do a loop so the argument is not a vector– Yuriy S
Dec 4 '18 at 11:27
@YuriyS - Loop with what index? I don't understand where to place the loop. Can you elaborate?
– Parseval
Dec 4 '18 at 11:29
@YuriyS - Loop with what index? I don't understand where to place the loop. Can you elaborate?
– Parseval
Dec 4 '18 at 11:29
add a comment |
1 Answer
1
active
oldest
votes
The solution of
$$e^{lambda(s-1)}-s=0$$ is given in terms of Lambert function and it is
$$s=-frac{Wleft(-lambda, e^{-lambda } right)}{lambda }$$ This function is available in R (have a look here).
When plotted, the integrand looks very nice but you still need numerical integration. If I did it properly, the result is $approx 0.4002$.
Thanks, I know that it can be expressed in the Lambert function, I did that too and got the same answer but since we have not covered this Lambert function, we are not supposed to ue it. Is there a way to do this without using that function?
– Parseval
Dec 4 '18 at 11:27
1
@Parseval. No idea ! Sorry for that. Cheers :-(
– Claude Leibovici
Dec 4 '18 at 11:30
add a comment |
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1 Answer
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The solution of
$$e^{lambda(s-1)}-s=0$$ is given in terms of Lambert function and it is
$$s=-frac{Wleft(-lambda, e^{-lambda } right)}{lambda }$$ This function is available in R (have a look here).
When plotted, the integrand looks very nice but you still need numerical integration. If I did it properly, the result is $approx 0.4002$.
Thanks, I know that it can be expressed in the Lambert function, I did that too and got the same answer but since we have not covered this Lambert function, we are not supposed to ue it. Is there a way to do this without using that function?
– Parseval
Dec 4 '18 at 11:27
1
@Parseval. No idea ! Sorry for that. Cheers :-(
– Claude Leibovici
Dec 4 '18 at 11:30
add a comment |
The solution of
$$e^{lambda(s-1)}-s=0$$ is given in terms of Lambert function and it is
$$s=-frac{Wleft(-lambda, e^{-lambda } right)}{lambda }$$ This function is available in R (have a look here).
When plotted, the integrand looks very nice but you still need numerical integration. If I did it properly, the result is $approx 0.4002$.
Thanks, I know that it can be expressed in the Lambert function, I did that too and got the same answer but since we have not covered this Lambert function, we are not supposed to ue it. Is there a way to do this without using that function?
– Parseval
Dec 4 '18 at 11:27
1
@Parseval. No idea ! Sorry for that. Cheers :-(
– Claude Leibovici
Dec 4 '18 at 11:30
add a comment |
The solution of
$$e^{lambda(s-1)}-s=0$$ is given in terms of Lambert function and it is
$$s=-frac{Wleft(-lambda, e^{-lambda } right)}{lambda }$$ This function is available in R (have a look here).
When plotted, the integrand looks very nice but you still need numerical integration. If I did it properly, the result is $approx 0.4002$.
The solution of
$$e^{lambda(s-1)}-s=0$$ is given in terms of Lambert function and it is
$$s=-frac{Wleft(-lambda, e^{-lambda } right)}{lambda }$$ This function is available in R (have a look here).
When plotted, the integrand looks very nice but you still need numerical integration. If I did it properly, the result is $approx 0.4002$.
answered Dec 4 '18 at 11:23
Claude Leibovici
119k1157132
119k1157132
Thanks, I know that it can be expressed in the Lambert function, I did that too and got the same answer but since we have not covered this Lambert function, we are not supposed to ue it. Is there a way to do this without using that function?
– Parseval
Dec 4 '18 at 11:27
1
@Parseval. No idea ! Sorry for that. Cheers :-(
– Claude Leibovici
Dec 4 '18 at 11:30
add a comment |
Thanks, I know that it can be expressed in the Lambert function, I did that too and got the same answer but since we have not covered this Lambert function, we are not supposed to ue it. Is there a way to do this without using that function?
– Parseval
Dec 4 '18 at 11:27
1
@Parseval. No idea ! Sorry for that. Cheers :-(
– Claude Leibovici
Dec 4 '18 at 11:30
Thanks, I know that it can be expressed in the Lambert function, I did that too and got the same answer but since we have not covered this Lambert function, we are not supposed to ue it. Is there a way to do this without using that function?
– Parseval
Dec 4 '18 at 11:27
Thanks, I know that it can be expressed in the Lambert function, I did that too and got the same answer but since we have not covered this Lambert function, we are not supposed to ue it. Is there a way to do this without using that function?
– Parseval
Dec 4 '18 at 11:27
1
1
@Parseval. No idea ! Sorry for that. Cheers :-(
– Claude Leibovici
Dec 4 '18 at 11:30
@Parseval. No idea ! Sorry for that. Cheers :-(
– Claude Leibovici
Dec 4 '18 at 11:30
add a comment |
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In general, if you get
the condition has length > 1 and only the first element will be used
that means your function is not suitable for vector arguments. Either uselapply
,sapply
or just do a loop so the argument is not a vector– Yuriy S
Dec 4 '18 at 11:27
@YuriyS - Loop with what index? I don't understand where to place the loop. Can you elaborate?
– Parseval
Dec 4 '18 at 11:29