Double integral involving the normal CDF












1












$begingroup$


I need to compute (or best approximate?) the following integral



$$int_0^infty int_0^infty (1 + alpha u)^{-1}(1 + v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) text{d}u text{d}v,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
where $Phi(cdot)$ is the standard normal cumulative distribution function. What strategy would you advise for this?



I don't know if this is helpful but I can also reformulate my problem (up to multiplicative constants) as computing the following expectation:



$$text{E}left{ Phileft(frac{beta}{sqrt{gamma + X^2Y^2}}right)right},qquad gamma > 0, quad beta in mathbb{R},$$
where
$$X sim text{Half-Cauchy}(0,1), qquad Y sim text{Half-Cauchy}(0,alpha^{-1/2}), quad alpha > 0.$$



EDIT: I made a mistake in the first integral, it should be:



$$int_0^infty int_0^infty (1 + alpha x^2)^{-1}(1 + y^2)^{-1} Phileft(frac{beta}{sqrt{gamma + x^2y^2}}right) text{d}x text{d}y,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
which now corresponds to the above expectation.






normal-distribution expected-value approximation integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would advise studying the asymptotic behavior, because this integral diverges.
    $endgroup$
    – whuber♦
    10 hours ago










  • $begingroup$
    Thanks for your point, I actually made a mistake in the first integral, very sorry about this. I have edited the question.
    $endgroup$
    – user79097
    9 hours ago






  • 1




    $begingroup$
    I think it would help to present the original version of this problem. This integral does not seem straightforward to evaluate using any other approach than brute force numerical quadrature, and I suspect that you have made a reformulation which is not necessarily helpful.
    $endgroup$
    – Martin L
    5 hours ago
















1












$begingroup$


I need to compute (or best approximate?) the following integral



$$int_0^infty int_0^infty (1 + alpha u)^{-1}(1 + v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) text{d}u text{d}v,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
where $Phi(cdot)$ is the standard normal cumulative distribution function. What strategy would you advise for this?



I don't know if this is helpful but I can also reformulate my problem (up to multiplicative constants) as computing the following expectation:



$$text{E}left{ Phileft(frac{beta}{sqrt{gamma + X^2Y^2}}right)right},qquad gamma > 0, quad beta in mathbb{R},$$
where
$$X sim text{Half-Cauchy}(0,1), qquad Y sim text{Half-Cauchy}(0,alpha^{-1/2}), quad alpha > 0.$$



EDIT: I made a mistake in the first integral, it should be:



$$int_0^infty int_0^infty (1 + alpha x^2)^{-1}(1 + y^2)^{-1} Phileft(frac{beta}{sqrt{gamma + x^2y^2}}right) text{d}x text{d}y,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
which now corresponds to the above expectation.






normal-distribution expected-value approximation integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would advise studying the asymptotic behavior, because this integral diverges.
    $endgroup$
    – whuber♦
    10 hours ago










  • $begingroup$
    Thanks for your point, I actually made a mistake in the first integral, very sorry about this. I have edited the question.
    $endgroup$
    – user79097
    9 hours ago






  • 1




    $begingroup$
    I think it would help to present the original version of this problem. This integral does not seem straightforward to evaluate using any other approach than brute force numerical quadrature, and I suspect that you have made a reformulation which is not necessarily helpful.
    $endgroup$
    – Martin L
    5 hours ago














1












1








1





$begingroup$


I need to compute (or best approximate?) the following integral



$$int_0^infty int_0^infty (1 + alpha u)^{-1}(1 + v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) text{d}u text{d}v,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
where $Phi(cdot)$ is the standard normal cumulative distribution function. What strategy would you advise for this?



I don't know if this is helpful but I can also reformulate my problem (up to multiplicative constants) as computing the following expectation:



$$text{E}left{ Phileft(frac{beta}{sqrt{gamma + X^2Y^2}}right)right},qquad gamma > 0, quad beta in mathbb{R},$$
where
$$X sim text{Half-Cauchy}(0,1), qquad Y sim text{Half-Cauchy}(0,alpha^{-1/2}), quad alpha > 0.$$



EDIT: I made a mistake in the first integral, it should be:



$$int_0^infty int_0^infty (1 + alpha x^2)^{-1}(1 + y^2)^{-1} Phileft(frac{beta}{sqrt{gamma + x^2y^2}}right) text{d}x text{d}y,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
which now corresponds to the above expectation.






normal-distribution expected-value approximation integral







share|cite|improve this question











$endgroup$




I need to compute (or best approximate?) the following integral



$$int_0^infty int_0^infty (1 + alpha u)^{-1}(1 + v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) text{d}u text{d}v,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
where $Phi(cdot)$ is the standard normal cumulative distribution function. What strategy would you advise for this?



I don't know if this is helpful but I can also reformulate my problem (up to multiplicative constants) as computing the following expectation:



$$text{E}left{ Phileft(frac{beta}{sqrt{gamma + X^2Y^2}}right)right},qquad gamma > 0, quad beta in mathbb{R},$$
where
$$X sim text{Half-Cauchy}(0,1), qquad Y sim text{Half-Cauchy}(0,alpha^{-1/2}), quad alpha > 0.$$



EDIT: I made a mistake in the first integral, it should be:



$$int_0^infty int_0^infty (1 + alpha x^2)^{-1}(1 + y^2)^{-1} Phileft(frac{beta}{sqrt{gamma + x^2y^2}}right) text{d}x text{d}y,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
which now corresponds to the above expectation.






normal-distribution expected-value approximation integral




normal-distribution expected-value approximation integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 hours ago







user79097

















asked 11 hours ago









user79097user79097

1306




1306












  • $begingroup$
    I would advise studying the asymptotic behavior, because this integral diverges.
    $endgroup$
    – whuber♦
    10 hours ago










  • $begingroup$
    Thanks for your point, I actually made a mistake in the first integral, very sorry about this. I have edited the question.
    $endgroup$
    – user79097
    9 hours ago






  • 1




    $begingroup$
    I think it would help to present the original version of this problem. This integral does not seem straightforward to evaluate using any other approach than brute force numerical quadrature, and I suspect that you have made a reformulation which is not necessarily helpful.
    $endgroup$
    – Martin L
    5 hours ago


















  • $begingroup$
    I would advise studying the asymptotic behavior, because this integral diverges.
    $endgroup$
    – whuber♦
    10 hours ago










  • $begingroup$
    Thanks for your point, I actually made a mistake in the first integral, very sorry about this. I have edited the question.
    $endgroup$
    – user79097
    9 hours ago






  • 1




    $begingroup$
    I think it would help to present the original version of this problem. This integral does not seem straightforward to evaluate using any other approach than brute force numerical quadrature, and I suspect that you have made a reformulation which is not necessarily helpful.
    $endgroup$
    – Martin L
    5 hours ago
















$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
10 hours ago




$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
10 hours ago












$begingroup$
Thanks for your point, I actually made a mistake in the first integral, very sorry about this. I have edited the question.
$endgroup$
– user79097
9 hours ago




$begingroup$
Thanks for your point, I actually made a mistake in the first integral, very sorry about this. I have edited the question.
$endgroup$
– user79097
9 hours ago




1




1




$begingroup$
I think it would help to present the original version of this problem. This integral does not seem straightforward to evaluate using any other approach than brute force numerical quadrature, and I suspect that you have made a reformulation which is not necessarily helpful.
$endgroup$
– Martin L
5 hours ago




$begingroup$
I think it would help to present the original version of this problem. This integral does not seem straightforward to evaluate using any other approach than brute force numerical quadrature, and I suspect that you have made a reformulation which is not necessarily helpful.
$endgroup$
– Martin L
5 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,



$$beta/sqrt{gamma + uv} ge min(beta/sqrt{gamma}, 0) = delta gt -infty.$$



Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(frac{beta}{sqrt{gamma + uv}}right)ge Phi(delta) = epsilon gt 0.$$



Consequently



$$eqalign{
&int_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) mathrm{d}umathrm{d}v \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} mathrm{d}umathrm{d}v \
&= lim_{Mtoinfty}lim_{Ntoinfty}epsilonint_0^M (1+alpha u)^{-1}mathrm{d}uint_0^N (1+v)^{-1} mathrm{d}v \
&=frac{epsilon}{alpha} lim_{Mtoinfty}lim_{Ntoinfty} log(1 + Malpha)log(1 + N),
}$$



which diverges to $+infty.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
    $endgroup$
    – user79097
    9 hours ago










  • $begingroup$
    I will let this answer stand because (a) the changes you made to the question translate in an obvious way to comparable changes in this analysis and (b) it sheds useful light on issues of numerical integration, which almost surely is going to be the solution. In particular, the convergence of each integral is only $O(1/M)$ and $O(1/N),$ which suggests preliminary manipulations to concentrate the mass of the integral into a smaller region would be useful if you want efficiency or accuracy.
    $endgroup$
    – whuber♦
    5 hours ago



















1












$begingroup$

If approximation is OK, you could simulate the expected value through an average. A law of large numbers implies that the average converges to the expected value - provided the latter exists, which is however not obvious to me. I did simulate a few runs, though, and got very similar results each time, which is not indicative of an issue with heavy tails.



library(LaplacesDemon)



draws <- 1e6

alpha <- 3

gamma <- 1

beta <- 1

x <- rhalfcauchy(draws, scale=1)

y <- rhalfcauchy(draws, scale=1/sqrt(alpha))



mean(pnorm(beta/sqrt(gamma+x^2*y^2)))





share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for your reply. Yes I could do this, but in fact I would need an analytic approximation not a numeric one (in particular, alpha, beta and gamma are generic)...
    $endgroup$
    – user79097
    11 hours ago










  • $begingroup$
    OK, I hope somebody else could weigh in. Of course, you could run the code for any values of the parameters you are interested in.
    $endgroup$
    – Christoph Hanck
    11 hours ago










  • $begingroup$
    Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
    $endgroup$
    – user79097
    11 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,



$$beta/sqrt{gamma + uv} ge min(beta/sqrt{gamma}, 0) = delta gt -infty.$$



Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(frac{beta}{sqrt{gamma + uv}}right)ge Phi(delta) = epsilon gt 0.$$



Consequently



$$eqalign{
&int_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) mathrm{d}umathrm{d}v \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} mathrm{d}umathrm{d}v \
&= lim_{Mtoinfty}lim_{Ntoinfty}epsilonint_0^M (1+alpha u)^{-1}mathrm{d}uint_0^N (1+v)^{-1} mathrm{d}v \
&=frac{epsilon}{alpha} lim_{Mtoinfty}lim_{Ntoinfty} log(1 + Malpha)log(1 + N),
}$$



which diverges to $+infty.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
    $endgroup$
    – user79097
    9 hours ago










  • $begingroup$
    I will let this answer stand because (a) the changes you made to the question translate in an obvious way to comparable changes in this analysis and (b) it sheds useful light on issues of numerical integration, which almost surely is going to be the solution. In particular, the convergence of each integral is only $O(1/M)$ and $O(1/N),$ which suggests preliminary manipulations to concentrate the mass of the integral into a smaller region would be useful if you want efficiency or accuracy.
    $endgroup$
    – whuber♦
    5 hours ago
















2












$begingroup$

Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,



$$beta/sqrt{gamma + uv} ge min(beta/sqrt{gamma}, 0) = delta gt -infty.$$



Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(frac{beta}{sqrt{gamma + uv}}right)ge Phi(delta) = epsilon gt 0.$$



Consequently



$$eqalign{
&int_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) mathrm{d}umathrm{d}v \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} mathrm{d}umathrm{d}v \
&= lim_{Mtoinfty}lim_{Ntoinfty}epsilonint_0^M (1+alpha u)^{-1}mathrm{d}uint_0^N (1+v)^{-1} mathrm{d}v \
&=frac{epsilon}{alpha} lim_{Mtoinfty}lim_{Ntoinfty} log(1 + Malpha)log(1 + N),
}$$



which diverges to $+infty.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
    $endgroup$
    – user79097
    9 hours ago










  • $begingroup$
    I will let this answer stand because (a) the changes you made to the question translate in an obvious way to comparable changes in this analysis and (b) it sheds useful light on issues of numerical integration, which almost surely is going to be the solution. In particular, the convergence of each integral is only $O(1/M)$ and $O(1/N),$ which suggests preliminary manipulations to concentrate the mass of the integral into a smaller region would be useful if you want efficiency or accuracy.
    $endgroup$
    – whuber♦
    5 hours ago














2












2








2





$begingroup$

Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,



$$beta/sqrt{gamma + uv} ge min(beta/sqrt{gamma}, 0) = delta gt -infty.$$



Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(frac{beta}{sqrt{gamma + uv}}right)ge Phi(delta) = epsilon gt 0.$$



Consequently



$$eqalign{
&int_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) mathrm{d}umathrm{d}v \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} mathrm{d}umathrm{d}v \
&= lim_{Mtoinfty}lim_{Ntoinfty}epsilonint_0^M (1+alpha u)^{-1}mathrm{d}uint_0^N (1+v)^{-1} mathrm{d}v \
&=frac{epsilon}{alpha} lim_{Mtoinfty}lim_{Ntoinfty} log(1 + Malpha)log(1 + N),
}$$



which diverges to $+infty.$






share|cite|improve this answer









$endgroup$



Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,



$$beta/sqrt{gamma + uv} ge min(beta/sqrt{gamma}, 0) = delta gt -infty.$$



Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(frac{beta}{sqrt{gamma + uv}}right)ge Phi(delta) = epsilon gt 0.$$



Consequently



$$eqalign{
&int_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) mathrm{d}umathrm{d}v \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^{-1}(1+v)^{-1} mathrm{d}umathrm{d}v \
&= lim_{Mtoinfty}lim_{Ntoinfty}epsilonint_0^M (1+alpha u)^{-1}mathrm{d}uint_0^N (1+v)^{-1} mathrm{d}v \
&=frac{epsilon}{alpha} lim_{Mtoinfty}lim_{Ntoinfty} log(1 + Malpha)log(1 + N),
}$$



which diverges to $+infty.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 9 hours ago









whuber♦whuber

205k33449816




205k33449816












  • $begingroup$
    Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
    $endgroup$
    – user79097
    9 hours ago










  • $begingroup$
    I will let this answer stand because (a) the changes you made to the question translate in an obvious way to comparable changes in this analysis and (b) it sheds useful light on issues of numerical integration, which almost surely is going to be the solution. In particular, the convergence of each integral is only $O(1/M)$ and $O(1/N),$ which suggests preliminary manipulations to concentrate the mass of the integral into a smaller region would be useful if you want efficiency or accuracy.
    $endgroup$
    – whuber♦
    5 hours ago


















  • $begingroup$
    Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
    $endgroup$
    – user79097
    9 hours ago










  • $begingroup$
    I will let this answer stand because (a) the changes you made to the question translate in an obvious way to comparable changes in this analysis and (b) it sheds useful light on issues of numerical integration, which almost surely is going to be the solution. In particular, the convergence of each integral is only $O(1/M)$ and $O(1/N),$ which suggests preliminary manipulations to concentrate the mass of the integral into a smaller region would be useful if you want efficiency or accuracy.
    $endgroup$
    – whuber♦
    5 hours ago
















$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
9 hours ago




$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
9 hours ago












$begingroup$
I will let this answer stand because (a) the changes you made to the question translate in an obvious way to comparable changes in this analysis and (b) it sheds useful light on issues of numerical integration, which almost surely is going to be the solution. In particular, the convergence of each integral is only $O(1/M)$ and $O(1/N),$ which suggests preliminary manipulations to concentrate the mass of the integral into a smaller region would be useful if you want efficiency or accuracy.
$endgroup$
– whuber♦
5 hours ago




$begingroup$
I will let this answer stand because (a) the changes you made to the question translate in an obvious way to comparable changes in this analysis and (b) it sheds useful light on issues of numerical integration, which almost surely is going to be the solution. In particular, the convergence of each integral is only $O(1/M)$ and $O(1/N),$ which suggests preliminary manipulations to concentrate the mass of the integral into a smaller region would be useful if you want efficiency or accuracy.
$endgroup$
– whuber♦
5 hours ago













1












$begingroup$

If approximation is OK, you could simulate the expected value through an average. A law of large numbers implies that the average converges to the expected value - provided the latter exists, which is however not obvious to me. I did simulate a few runs, though, and got very similar results each time, which is not indicative of an issue with heavy tails.



library(LaplacesDemon)



draws <- 1e6

alpha <- 3

gamma <- 1

beta <- 1

x <- rhalfcauchy(draws, scale=1)

y <- rhalfcauchy(draws, scale=1/sqrt(alpha))



mean(pnorm(beta/sqrt(gamma+x^2*y^2)))





share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for your reply. Yes I could do this, but in fact I would need an analytic approximation not a numeric one (in particular, alpha, beta and gamma are generic)...
    $endgroup$
    – user79097
    11 hours ago










  • $begingroup$
    OK, I hope somebody else could weigh in. Of course, you could run the code for any values of the parameters you are interested in.
    $endgroup$
    – Christoph Hanck
    11 hours ago










  • $begingroup$
    Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
    $endgroup$
    – user79097
    11 hours ago
















1












$begingroup$

If approximation is OK, you could simulate the expected value through an average. A law of large numbers implies that the average converges to the expected value - provided the latter exists, which is however not obvious to me. I did simulate a few runs, though, and got very similar results each time, which is not indicative of an issue with heavy tails.



library(LaplacesDemon)



draws <- 1e6

alpha <- 3

gamma <- 1

beta <- 1

x <- rhalfcauchy(draws, scale=1)

y <- rhalfcauchy(draws, scale=1/sqrt(alpha))



mean(pnorm(beta/sqrt(gamma+x^2*y^2)))





share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for your reply. Yes I could do this, but in fact I would need an analytic approximation not a numeric one (in particular, alpha, beta and gamma are generic)...
    $endgroup$
    – user79097
    11 hours ago










  • $begingroup$
    OK, I hope somebody else could weigh in. Of course, you could run the code for any values of the parameters you are interested in.
    $endgroup$
    – Christoph Hanck
    11 hours ago










  • $begingroup$
    Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
    $endgroup$
    – user79097
    11 hours ago














1












1








1





$begingroup$

If approximation is OK, you could simulate the expected value through an average. A law of large numbers implies that the average converges to the expected value - provided the latter exists, which is however not obvious to me. I did simulate a few runs, though, and got very similar results each time, which is not indicative of an issue with heavy tails.



library(LaplacesDemon)



draws <- 1e6

alpha <- 3

gamma <- 1

beta <- 1

x <- rhalfcauchy(draws, scale=1)

y <- rhalfcauchy(draws, scale=1/sqrt(alpha))



mean(pnorm(beta/sqrt(gamma+x^2*y^2)))





share|cite|improve this answer











$endgroup$



If approximation is OK, you could simulate the expected value through an average. A law of large numbers implies that the average converges to the expected value - provided the latter exists, which is however not obvious to me. I did simulate a few runs, though, and got very similar results each time, which is not indicative of an issue with heavy tails.



library(LaplacesDemon)



draws <- 1e6

alpha <- 3

gamma <- 1

beta <- 1

x <- rhalfcauchy(draws, scale=1)

y <- rhalfcauchy(draws, scale=1/sqrt(alpha))



mean(pnorm(beta/sqrt(gamma+x^2*y^2)))






share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 11 hours ago

























answered 11 hours ago









Christoph HanckChristoph Hanck

17.3k34074




17.3k34074












  • $begingroup$
    Thanks a lot for your reply. Yes I could do this, but in fact I would need an analytic approximation not a numeric one (in particular, alpha, beta and gamma are generic)...
    $endgroup$
    – user79097
    11 hours ago










  • $begingroup$
    OK, I hope somebody else could weigh in. Of course, you could run the code for any values of the parameters you are interested in.
    $endgroup$
    – Christoph Hanck
    11 hours ago










  • $begingroup$
    Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
    $endgroup$
    – user79097
    11 hours ago


















  • $begingroup$
    Thanks a lot for your reply. Yes I could do this, but in fact I would need an analytic approximation not a numeric one (in particular, alpha, beta and gamma are generic)...
    $endgroup$
    – user79097
    11 hours ago










  • $begingroup$
    OK, I hope somebody else could weigh in. Of course, you could run the code for any values of the parameters you are interested in.
    $endgroup$
    – Christoph Hanck
    11 hours ago










  • $begingroup$
    Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
    $endgroup$
    – user79097
    11 hours ago
















$begingroup$
Thanks a lot for your reply. Yes I could do this, but in fact I would need an analytic approximation not a numeric one (in particular, alpha, beta and gamma are generic)...
$endgroup$
– user79097
11 hours ago




$begingroup$
Thanks a lot for your reply. Yes I could do this, but in fact I would need an analytic approximation not a numeric one (in particular, alpha, beta and gamma are generic)...
$endgroup$
– user79097
11 hours ago












$begingroup$
OK, I hope somebody else could weigh in. Of course, you could run the code for any values of the parameters you are interested in.
$endgroup$
– Christoph Hanck
11 hours ago




$begingroup$
OK, I hope somebody else could weigh in. Of course, you could run the code for any values of the parameters you are interested in.
$endgroup$
– Christoph Hanck
11 hours ago












$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
11 hours ago




$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
11 hours ago


















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