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Anti-derivative of Lognormal CDF
solving/approximating integral of lognormal cdfMaxwell-Boltzmann velocity PDF to CDFWhat is $mathbbEleft[fracx1+bxright]$ where $x$ is lognormalNumerical CDF of sum of dependent lognormal variablesHow to prove that given CDF is valid?Expected Value of Maximum of Two Lognormal Random Variables with One Source of RandomnessExpected Value of Maximum of Two Lognormal Random Variablessolving/approximating integral of lognormal cdfExpected value of a lognormal distributionBounds on the two first moments of the absolute value of a random variable?Expectation of maximum of two lognormal random variables (another view)
$begingroup$
I am trying to solve the integral
beginequation
int_0^bPhi_LN(x, mu,sigma)dx
endequation
where beginequation Phi_LN(x, mu,sigma) = int^x_0frac1sqrt2pisigmayexpleft(-frac12left(fraclog y-musigmaright)^2right)dy endequation is the lognormal cdf, evaluated at x.
I found a solution to this question in this thread.
It says the function beginequationI(x) = xPhi_LN(x, mu,sigma) + L(x)endequation
with
beginequationL(x) = frac12e^(mu+sigma^2)/2mboxerfleft(left(sigma - fraclog(x)-musigmaright)/sqrt2right)endequation
is the anti-derivative of the LogNormal CDF.
This makes sense to me as because of $fracddxmboxerf(x) = frac2sqrtpie^-x²$ we get
beginalign
fracddx L(x) &= -frac12e^(mu+sigma^2)/2frac2sqrtpie^-frac12left(sigma - fraclog(x)-musigmaright)^2frac1xsigmasqrt2\
&= -frac1xsigmasqrt2pie^(mu+sigma^2)/2e^-frac12left(sigma^2 + left(fraclog(x)-musigmaright)^2 - 2(log(x)-mu)right)\
&= -frac1xsigmasqrt2pie^(mu+sigma^2)/2e^-frac12sigma^2e^-frac12left(fraclog(x)-musigmaright)^2e^log(x)e^-frac12mu\
&= -frac1sigmasqrt2pie^-frac12left(fraclog(x)-musigmaright)^2\
&= -frac1sigmavarphileft(fraclog(x)-musigmaright).
endalign
which implies that
beginalign
fracddxI(x) &= Phi_LN(x, mu,sigma) + xfracddxPhi_LN(x, mu,sigma) - frac1sigmavarphileft(fraclog(x)-musigmaright)\
&= Phi_LN(x, mu,sigma).
endalign
In order to test this result I wrote the following short R Script.
erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1
L_function<-function(x,mu,sigma)
z<-(sigma^2-log(x)+mu)/sigma
z<-z/sqrt(2)
y<-0.5*exp(0.5*(mu+sigma^2))
return(y*erf(z))
I_function<-function(x,mu,sigma)
return(x*plnorm(x,meanlog = mu,sdlog = sigma)+L_function(x,mu,sigma))
integral_over_LogNormalCDF<-function(lowerBound,upperBound,mu,sigma)
return(I_function(upperBound,mu,sigma)-I_function(lowerBound,mu,sigma))
ded<-2
limit<-3
mu<-1
sigma<-3
n=10000
steplength=0.0001
lognormalCDF<-seq(from=ded,to=limit,by=steplength)
lognormalCDF<-plnorm(lognormalCDF,meanlog = mu,sdlog = sigma)
weights<-replicate(length(lognormalCDF),steplength)
simulated<-lognormalCDF %*% weights
funcValue<-integral_over_LogNormalCDF(ded,limit,mu,sigma)
The results are fine (i.e. simulated equals funcValue) except when $mu neq 0$.
For example:
$mu=10$ -> simulated = 0.00123 and funcValue=0.002579
$mu=1$ -> simulated = 0.488 and funcValue=0.540
$mu=-1$ -> simulated = 0.738 and funcValue=0.667
Does anyone see my mistake?
Thanks in advance
Mathias
probability
asked Mar 21 at 12:29
mathimathi1987mathimathi1987
32
$endgroup$
add a comment |
$begingroup$
I am trying to solve the integral
beginequation
int_0^bPhi_LN(x, mu,sigma)dx
endequation
where beginequation Phi_LN(x, mu,sigma) = int^x_0frac1sqrt2pisigmayexpleft(-frac12left(fraclog y-musigmaright)^2right)dy endequation is the lognormal cdf, evaluated at x.
I found a solution to this question in this thread.
It says the function beginequationI(x) = xPhi_LN(x, mu,sigma) + L(x)endequation
with
beginequationL(x) = frac12e^(mu+sigma^2)/2mboxerfleft(left(sigma - fraclog(x)-musigmaright)/sqrt2right)endequation
is the anti-derivative of the LogNormal CDF.
This makes sense to me as because of $fracddxmboxerf(x) = frac2sqrtpie^-x²$ we get
beginalign
fracddx L(x) &= -frac12e^(mu+sigma^2)/2frac2sqrtpie^-frac12left(sigma - fraclog(x)-musigmaright)^2frac1xsigmasqrt2\
&= -frac1xsigmasqrt2pie^(mu+sigma^2)/2e^-frac12left(sigma^2 + left(fraclog(x)-musigmaright)^2 - 2(log(x)-mu)right)\
&= -frac1xsigmasqrt2pie^(mu+sigma^2)/2e^-frac12sigma^2e^-frac12left(fraclog(x)-musigmaright)^2e^log(x)e^-frac12mu\
&= -frac1sigmasqrt2pie^-frac12left(fraclog(x)-musigmaright)^2\
&= -frac1sigmavarphileft(fraclog(x)-musigmaright).
endalign
which implies that
beginalign
fracddxI(x) &= Phi_LN(x, mu,sigma) + xfracddxPhi_LN(x, mu,sigma) - frac1sigmavarphileft(fraclog(x)-musigmaright)\
&= Phi_LN(x, mu,sigma).
endalign
In order to test this result I wrote the following short R Script.
erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1
L_function<-function(x,mu,sigma)
z<-(sigma^2-log(x)+mu)/sigma
z<-z/sqrt(2)
y<-0.5*exp(0.5*(mu+sigma^2))
return(y*erf(z))
I_function<-function(x,mu,sigma)
return(x*plnorm(x,meanlog = mu,sdlog = sigma)+L_function(x,mu,sigma))
integral_over_LogNormalCDF<-function(lowerBound,upperBound,mu,sigma)
return(I_function(upperBound,mu,sigma)-I_function(lowerBound,mu,sigma))
ded<-2
limit<-3
mu<-1
sigma<-3
n=10000
steplength=0.0001
lognormalCDF<-seq(from=ded,to=limit,by=steplength)
lognormalCDF<-plnorm(lognormalCDF,meanlog = mu,sdlog = sigma)
weights<-replicate(length(lognormalCDF),steplength)
simulated<-lognormalCDF %*% weights
funcValue<-integral_over_LogNormalCDF(ded,limit,mu,sigma)
The results are fine (i.e. simulated equals funcValue) except when $mu neq 0$.
For example:
$mu=10$ -> simulated = 0.00123 and funcValue=0.002579
$mu=1$ -> simulated = 0.488 and funcValue=0.540
$mu=-1$ -> simulated = 0.738 and funcValue=0.667
Does anyone see my mistake?
Thanks in advance
Mathias
probability
asked Mar 21 at 12:29
mathimathi1987mathimathi1987
32
$endgroup$
$begingroup$
How do the values compare when $mu=1$? (Are you sure the differences are large enough to not reflect rounding, or the imprecision of numerical integration?) What about $mu=-1$?
$endgroup$
– J.G.
Mar 21 at 12:43
$begingroup$
You could edit your question to share them.
$endgroup$
– J.G.
Mar 21 at 13:19
$begingroup$
I don't yet see the mistake however using numerical integration in Mathematica agrees completely with the simulated values and not with thefuncValue. So I suspect it has something to do withfuncValue.
$endgroup$
– JimB
Mar 21 at 15:14
add a comment |
$begingroup$
I am trying to solve the integral
beginequation
int_0^bPhi_LN(x, mu,sigma)dx
endequation
where beginequation Phi_LN(x, mu,sigma) = int^x_0frac1sqrt2pisigmayexpleft(-frac12left(fraclog y-musigmaright)^2right)dy endequation is the lognormal cdf, evaluated at x.
I found a solution to this question in this thread.
It says the function beginequationI(x) = xPhi_LN(x, mu,sigma) + L(x)endequation
with
beginequationL(x) = frac12e^(mu+sigma^2)/2mboxerfleft(left(sigma - fraclog(x)-musigmaright)/sqrt2right)endequation
is the anti-derivative of the LogNormal CDF.
This makes sense to me as because of $fracddxmboxerf(x) = frac2sqrtpie^-x²$ we get
beginalign
fracddx L(x) &= -frac12e^(mu+sigma^2)/2frac2sqrtpie^-frac12left(sigma - fraclog(x)-musigmaright)^2frac1xsigmasqrt2\
&= -frac1xsigmasqrt2pie^(mu+sigma^2)/2e^-frac12left(sigma^2 + left(fraclog(x)-musigmaright)^2 - 2(log(x)-mu)right)\
&= -frac1xsigmasqrt2pie^(mu+sigma^2)/2e^-frac12sigma^2e^-frac12left(fraclog(x)-musigmaright)^2e^log(x)e^-frac12mu\
&= -frac1sigmasqrt2pie^-frac12left(fraclog(x)-musigmaright)^2\
&= -frac1sigmavarphileft(fraclog(x)-musigmaright).
endalign
which implies that
beginalign
fracddxI(x) &= Phi_LN(x, mu,sigma) + xfracddxPhi_LN(x, mu,sigma) - frac1sigmavarphileft(fraclog(x)-musigmaright)\
&= Phi_LN(x, mu,sigma).
endalign
In order to test this result I wrote the following short R Script.
erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1
L_function<-function(x,mu,sigma)
z<-(sigma^2-log(x)+mu)/sigma
z<-z/sqrt(2)
y<-0.5*exp(0.5*(mu+sigma^2))
return(y*erf(z))
I_function<-function(x,mu,sigma)
return(x*plnorm(x,meanlog = mu,sdlog = sigma)+L_function(x,mu,sigma))
integral_over_LogNormalCDF<-function(lowerBound,upperBound,mu,sigma)
return(I_function(upperBound,mu,sigma)-I_function(lowerBound,mu,sigma))
ded<-2
limit<-3
mu<-1
sigma<-3
n=10000
steplength=0.0001
lognormalCDF<-seq(from=ded,to=limit,by=steplength)
lognormalCDF<-plnorm(lognormalCDF,meanlog = mu,sdlog = sigma)
weights<-replicate(length(lognormalCDF),steplength)
simulated<-lognormalCDF %*% weights
funcValue<-integral_over_LogNormalCDF(ded,limit,mu,sigma)
The results are fine (i.e. simulated equals funcValue) except when $mu neq 0$.
For example:
$mu=10$ -> simulated = 0.00123 and funcValue=0.002579
$mu=1$ -> simulated = 0.488 and funcValue=0.540
$mu=-1$ -> simulated = 0.738 and funcValue=0.667
Does anyone see my mistake?
Thanks in advance
Mathias
probability
asked Mar 21 at 12:29
mathimathi1987mathimathi1987
32
$endgroup$
I am trying to solve the integral
beginequation
int_0^bPhi_LN(x, mu,sigma)dx
endequation
where beginequation Phi_LN(x, mu,sigma) = int^x_0frac1sqrt2pisigmayexpleft(-frac12left(fraclog y-musigmaright)^2right)dy endequation is the lognormal cdf, evaluated at x.
I found a solution to this question in this thread.
It says the function beginequationI(x) = xPhi_LN(x, mu,sigma) + L(x)endequation
with
beginequationL(x) = frac12e^(mu+sigma^2)/2mboxerfleft(left(sigma - fraclog(x)-musigmaright)/sqrt2right)endequation
is the anti-derivative of the LogNormal CDF.
This makes sense to me as because of $fracddxmboxerf(x) = frac2sqrtpie^-x²$ we get
beginalign
fracddx L(x) &= -frac12e^(mu+sigma^2)/2frac2sqrtpie^-frac12left(sigma - fraclog(x)-musigmaright)^2frac1xsigmasqrt2\
&= -frac1xsigmasqrt2pie^(mu+sigma^2)/2e^-frac12left(sigma^2 + left(fraclog(x)-musigmaright)^2 - 2(log(x)-mu)right)\
&= -frac1xsigmasqrt2pie^(mu+sigma^2)/2e^-frac12sigma^2e^-frac12left(fraclog(x)-musigmaright)^2e^log(x)e^-frac12mu\
&= -frac1sigmasqrt2pie^-frac12left(fraclog(x)-musigmaright)^2\
&= -frac1sigmavarphileft(fraclog(x)-musigmaright).
endalign
which implies that
beginalign
fracddxI(x) &= Phi_LN(x, mu,sigma) + xfracddxPhi_LN(x, mu,sigma) - frac1sigmavarphileft(fraclog(x)-musigmaright)\
&= Phi_LN(x, mu,sigma).
endalign
In order to test this result I wrote the following short R Script.
erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1
L_function<-function(x,mu,sigma)
z<-(sigma^2-log(x)+mu)/sigma
z<-z/sqrt(2)
y<-0.5*exp(0.5*(mu+sigma^2))
return(y*erf(z))
I_function<-function(x,mu,sigma)
return(x*plnorm(x,meanlog = mu,sdlog = sigma)+L_function(x,mu,sigma))
integral_over_LogNormalCDF<-function(lowerBound,upperBound,mu,sigma)
return(I_function(upperBound,mu,sigma)-I_function(lowerBound,mu,sigma))
ded<-2
limit<-3
mu<-1
sigma<-3
n=10000
steplength=0.0001
lognormalCDF<-seq(from=ded,to=limit,by=steplength)
lognormalCDF<-plnorm(lognormalCDF,meanlog = mu,sdlog = sigma)
weights<-replicate(length(lognormalCDF),steplength)
simulated<-lognormalCDF %*% weights
funcValue<-integral_over_LogNormalCDF(ded,limit,mu,sigma)
The results are fine (i.e. simulated equals funcValue) except when $mu neq 0$.
For example:
$mu=10$ -> simulated = 0.00123 and funcValue=0.002579
$mu=1$ -> simulated = 0.488 and funcValue=0.540
$mu=-1$ -> simulated = 0.738 and funcValue=0.667
Does anyone see my mistake?
Thanks in advance
Mathias
probability
probability
asked Mar 21 at 12:29
mathimathi1987mathimathi1987
32
asked Mar 21 at 12:29
mathimathi1987mathimathi1987
32
edited Mar 21 at 13:24
mathimathi1987
asked Mar 21 at 12:29
mathimathi1987mathimathi1987
32
asked Mar 21 at 12:29
mathimathi1987mathimathi1987
32
asked Mar 21 at 12:29
mathimathi1987mathimathi1987
32
32
$begingroup$
How do the values compare when $mu=1$? (Are you sure the differences are large enough to not reflect rounding, or the imprecision of numerical integration?) What about $mu=-1$?
$endgroup$
– J.G.
Mar 21 at 12:43
$begingroup$
You could edit your question to share them.
$endgroup$
– J.G.
Mar 21 at 13:19
$begingroup$
I don't yet see the mistake however using numerical integration in Mathematica agrees completely with the simulated values and not with thefuncValue. So I suspect it has something to do withfuncValue.
$endgroup$
– JimB
Mar 21 at 15:14
add a comment |
$begingroup$
How do the values compare when $mu=1$? (Are you sure the differences are large enough to not reflect rounding, or the imprecision of numerical integration?) What about $mu=-1$?
$endgroup$
– J.G.
Mar 21 at 12:43
$begingroup$
You could edit your question to share them.
$endgroup$
– J.G.
Mar 21 at 13:19
$begingroup$
I don't yet see the mistake however using numerical integration in Mathematica agrees completely with the simulated values and not with thefuncValue. So I suspect it has something to do withfuncValue.
$endgroup$
– JimB
Mar 21 at 15:14
$begingroup$
How do the values compare when $mu=1$? (Are you sure the differences are large enough to not reflect rounding, or the imprecision of numerical integration?) What about $mu=-1$?
$endgroup$
– J.G.
Mar 21 at 12:43
$begingroup$
How do the values compare when $mu=1$? (Are you sure the differences are large enough to not reflect rounding, or the imprecision of numerical integration?) What about $mu=-1$?
$endgroup$
– J.G.
Mar 21 at 12:43
$begingroup$
You could edit your question to share them.
$endgroup$
– J.G.
Mar 21 at 13:19
$begingroup$
You could edit your question to share them.
$endgroup$
– J.G.
Mar 21 at 13:19
$begingroup$
I don't yet see the mistake however using numerical integration in Mathematica agrees completely with the simulated values and not with the
funcValue. So I suspect it has something to do with funcValue.$endgroup$
– JimB
Mar 21 at 15:14
$begingroup$
I don't yet see the mistake however using numerical integration in Mathematica agrees completely with the simulated values and not with the
funcValue. So I suspect it has something to do with funcValue.$endgroup$
– JimB
Mar 21 at 15:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The I_function is not correct. Your R code should be more like the following.
erfc <- function(x) 1 - (2 * pnorm(x * sqrt(2)) - 1)
I_function<-function(x,mu,sigma)
return( (x*erfc((mu - log(x))/(sqrt(2)*sigma)) -
exp(mu + sigma**2/2)*erfc((mu + sigma**2 - log(x))/(sqrt(2)*sigma)))/2)
integral_over_LogNormalCDF<-function(lowerBound,upperBound,mu,sigma)
return(I_function(upperBound,mu,sigma)-I_function(lowerBound,mu,sigma))
The test values you used result in the following:
integral_over_LogNormalCDF(2,3,10,3)
# 0.001231376
integral_over_LogNormalCDF(2,3,1,3)
# 0.4879823
integral_over_LogNormalCDF(2,3,-1,3)
# 0.7376231
answered Mar 21 at 15:46
JimBJimB
61547
$endgroup$
$begingroup$
Hi JimB, thank you for your help! Much appreciated!
$endgroup$
– mathimathi1987
Mar 23 at 14:57
add a comment |
Your Answer
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1 Answer
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$begingroup$
The I_function is not correct. Your R code should be more like the following.
erfc <- function(x) 1 - (2 * pnorm(x * sqrt(2)) - 1)
I_function<-function(x,mu,sigma)
return( (x*erfc((mu - log(x))/(sqrt(2)*sigma)) -
exp(mu + sigma**2/2)*erfc((mu + sigma**2 - log(x))/(sqrt(2)*sigma)))/2)
integral_over_LogNormalCDF<-function(lowerBound,upperBound,mu,sigma)
return(I_function(upperBound,mu,sigma)-I_function(lowerBound,mu,sigma))
The test values you used result in the following:
integral_over_LogNormalCDF(2,3,10,3)
# 0.001231376
integral_over_LogNormalCDF(2,3,1,3)
# 0.4879823
integral_over_LogNormalCDF(2,3,-1,3)
# 0.7376231
answered Mar 21 at 15:46
JimBJimB
61547
$endgroup$
$begingroup$
Hi JimB, thank you for your help! Much appreciated!
$endgroup$
– mathimathi1987
Mar 23 at 14:57
add a comment |
$begingroup$
The I_function is not correct. Your R code should be more like the following.
erfc <- function(x) 1 - (2 * pnorm(x * sqrt(2)) - 1)
I_function<-function(x,mu,sigma)
return( (x*erfc((mu - log(x))/(sqrt(2)*sigma)) -
exp(mu + sigma**2/2)*erfc((mu + sigma**2 - log(x))/(sqrt(2)*sigma)))/2)
integral_over_LogNormalCDF<-function(lowerBound,upperBound,mu,sigma)
return(I_function(upperBound,mu,sigma)-I_function(lowerBound,mu,sigma))
The test values you used result in the following:
integral_over_LogNormalCDF(2,3,10,3)
# 0.001231376
integral_over_LogNormalCDF(2,3,1,3)
# 0.4879823
integral_over_LogNormalCDF(2,3,-1,3)
# 0.7376231
answered Mar 21 at 15:46
JimBJimB
61547
$endgroup$
$begingroup$
Hi JimB, thank you for your help! Much appreciated!
$endgroup$
– mathimathi1987
Mar 23 at 14:57
add a comment |
$begingroup$
The I_function is not correct. Your R code should be more like the following.
erfc <- function(x) 1 - (2 * pnorm(x * sqrt(2)) - 1)
I_function<-function(x,mu,sigma)
return( (x*erfc((mu - log(x))/(sqrt(2)*sigma)) -
exp(mu + sigma**2/2)*erfc((mu + sigma**2 - log(x))/(sqrt(2)*sigma)))/2)
integral_over_LogNormalCDF<-function(lowerBound,upperBound,mu,sigma)
return(I_function(upperBound,mu,sigma)-I_function(lowerBound,mu,sigma))
The test values you used result in the following:
integral_over_LogNormalCDF(2,3,10,3)
# 0.001231376
integral_over_LogNormalCDF(2,3,1,3)
# 0.4879823
integral_over_LogNormalCDF(2,3,-1,3)
# 0.7376231
answered Mar 21 at 15:46
JimBJimB
61547
$endgroup$
The I_function is not correct. Your R code should be more like the following.
erfc <- function(x) 1 - (2 * pnorm(x * sqrt(2)) - 1)
I_function<-function(x,mu,sigma)
return( (x*erfc((mu - log(x))/(sqrt(2)*sigma)) -
exp(mu + sigma**2/2)*erfc((mu + sigma**2 - log(x))/(sqrt(2)*sigma)))/2)
integral_over_LogNormalCDF<-function(lowerBound,upperBound,mu,sigma)
return(I_function(upperBound,mu,sigma)-I_function(lowerBound,mu,sigma))
The test values you used result in the following:
integral_over_LogNormalCDF(2,3,10,3)
# 0.001231376
integral_over_LogNormalCDF(2,3,1,3)
# 0.4879823
integral_over_LogNormalCDF(2,3,-1,3)
# 0.7376231
answered Mar 21 at 15:46
JimBJimB
61547
answered Mar 21 at 15:46
JimBJimB
61547
answered Mar 21 at 15:46
JimBJimB
61547
answered Mar 21 at 15:46
JimBJimB
61547
61547
$begingroup$
Hi JimB, thank you for your help! Much appreciated!
$endgroup$
– mathimathi1987
Mar 23 at 14:57
add a comment |
$begingroup$
Hi JimB, thank you for your help! Much appreciated!
$endgroup$
– mathimathi1987
Mar 23 at 14:57
$begingroup$
Hi JimB, thank you for your help! Much appreciated!
$endgroup$
– mathimathi1987
Mar 23 at 14:57
$begingroup$
Hi JimB, thank you for your help! Much appreciated!
$endgroup$
– mathimathi1987
Mar 23 at 14:57
add a comment |
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$begingroup$
How do the values compare when $mu=1$? (Are you sure the differences are large enough to not reflect rounding, or the imprecision of numerical integration?) What about $mu=-1$?
$endgroup$
– J.G.
Mar 21 at 12:43
$begingroup$
You could edit your question to share them.
$endgroup$
– J.G.
Mar 21 at 13:19
$begingroup$
I don't yet see the mistake however using numerical integration in Mathematica agrees completely with the simulated values and not with the
funcValue. So I suspect it has something to do withfuncValue.$endgroup$
– JimB
Mar 21 at 15:14