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Double integral involving the normal CDF
Expected value of minimum order statistic from a normal sampleExpected value of a random variable involving a standard normal random variableExpected number of events from Poisson distribution with Gamma priorequation involving the cumulative normal distributionFormular Derivation - Expected Value Truncated Normal DistributionExpectation of the softmax transform for Gaussian multivariate variablesSearch solution to an integralMoments of the horseshoe prior?Integral from the Adversarial Spheres paper (maximum of the difference between a constant and a normal random variable)How to calculate the integral of Normal CDF and Normal PDF?
$begingroup$
I need to compute (or best approximate?) the following integral
$$int_0^infty int_0^infty (1 + alpha u)^-1(1 + v)^-1 Phileft(fracbetasqrtgamma + uvright) textdu textdv,qquad alpha, gamma > 0, quad beta in mathbbR,$$
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:
$$textEleft Phileft(fracbetasqrtgamma + X^2Y^2right)right,qquad gamma > 0, quad beta in mathbbR,$$
where
$$X sim textHalf-Cauchy(0,1), qquad Y sim textHalf-Cauchy(0,alpha^-1/2), quad alpha > 0.$$
EDIT: I made a mistake in the first integral, it should be:
$$int_0^infty int_0^infty (1 + alpha x^2)^-1(1 + y^2)^-1 Phileft(fracbetasqrtgamma + x^2y^2right) textdx textdy,qquad alpha, gamma > 0, quad beta in mathbbR,$$
which now corresponds to the above expectation.
EDIT 2: Another way to solve my problem would be to characterise the difference (hopefully small?)
$$textEleft Phileft(fracbetasqrtgamma + X^2Y^2right)right - Phileft(fracbetasqrtgammaright),qquad gamma > 0, quad beta in mathbbR,$$
would it be sensible to approximate the difference
$$Phileft(fracbetasqrtgamma + X^2Y^2right) - Phileft(fracbetasqrtgammaright),$$
with a Taylor series in zero or is this unreasonable given that the half-Cauchy distributions are heavy-tailed?
normal-distribution expected-value approximation integral
$endgroup$
add a comment |
$begingroup$
I need to compute (or best approximate?) the following integral
$$int_0^infty int_0^infty (1 + alpha u)^-1(1 + v)^-1 Phileft(fracbetasqrtgamma + uvright) textdu textdv,qquad alpha, gamma > 0, quad beta in mathbbR,$$
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:
$$textEleft Phileft(fracbetasqrtgamma + X^2Y^2right)right,qquad gamma > 0, quad beta in mathbbR,$$
where
$$X sim textHalf-Cauchy(0,1), qquad Y sim textHalf-Cauchy(0,alpha^-1/2), quad alpha > 0.$$
EDIT: I made a mistake in the first integral, it should be:
$$int_0^infty int_0^infty (1 + alpha x^2)^-1(1 + y^2)^-1 Phileft(fracbetasqrtgamma + x^2y^2right) textdx textdy,qquad alpha, gamma > 0, quad beta in mathbbR,$$
which now corresponds to the above expectation.
EDIT 2: Another way to solve my problem would be to characterise the difference (hopefully small?)
$$textEleft Phileft(fracbetasqrtgamma + X^2Y^2right)right - Phileft(fracbetasqrtgammaright),qquad gamma > 0, quad beta in mathbbR,$$
would it be sensible to approximate the difference
$$Phileft(fracbetasqrtgamma + X^2Y^2right) - Phileft(fracbetasqrtgammaright),$$
with a Taylor series in zero or is this unreasonable given that the half-Cauchy distributions are heavy-tailed?
normal-distribution expected-value approximation integral
$endgroup$
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
19 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
18 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
15 hours ago
$begingroup$
You are probably right, in fact the original problem corresponds to that of the expectation above, with X and Y having their respective half-Cauchy distribution... I should probably have stated the problem directly in this way. Does this make you think of any other strategy?
$endgroup$
– user79097
58 mins ago
add a comment |
$begingroup$
I need to compute (or best approximate?) the following integral
$$int_0^infty int_0^infty (1 + alpha u)^-1(1 + v)^-1 Phileft(fracbetasqrtgamma + uvright) textdu textdv,qquad alpha, gamma > 0, quad beta in mathbbR,$$
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:
$$textEleft Phileft(fracbetasqrtgamma + X^2Y^2right)right,qquad gamma > 0, quad beta in mathbbR,$$
where
$$X sim textHalf-Cauchy(0,1), qquad Y sim textHalf-Cauchy(0,alpha^-1/2), quad alpha > 0.$$
EDIT: I made a mistake in the first integral, it should be:
$$int_0^infty int_0^infty (1 + alpha x^2)^-1(1 + y^2)^-1 Phileft(fracbetasqrtgamma + x^2y^2right) textdx textdy,qquad alpha, gamma > 0, quad beta in mathbbR,$$
which now corresponds to the above expectation.
EDIT 2: Another way to solve my problem would be to characterise the difference (hopefully small?)
$$textEleft Phileft(fracbetasqrtgamma + X^2Y^2right)right - Phileft(fracbetasqrtgammaright),qquad gamma > 0, quad beta in mathbbR,$$
would it be sensible to approximate the difference
$$Phileft(fracbetasqrtgamma + X^2Y^2right) - Phileft(fracbetasqrtgammaright),$$
with a Taylor series in zero or is this unreasonable given that the half-Cauchy distributions are heavy-tailed?
normal-distribution expected-value approximation integral
$endgroup$
I need to compute (or best approximate?) the following integral
$$int_0^infty int_0^infty (1 + alpha u)^-1(1 + v)^-1 Phileft(fracbetasqrtgamma + uvright) textdu textdv,qquad alpha, gamma > 0, quad beta in mathbbR,$$
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:
$$textEleft Phileft(fracbetasqrtgamma + X^2Y^2right)right,qquad gamma > 0, quad beta in mathbbR,$$
where
$$X sim textHalf-Cauchy(0,1), qquad Y sim textHalf-Cauchy(0,alpha^-1/2), quad alpha > 0.$$
EDIT: I made a mistake in the first integral, it should be:
$$int_0^infty int_0^infty (1 + alpha x^2)^-1(1 + y^2)^-1 Phileft(fracbetasqrtgamma + x^2y^2right) textdx textdy,qquad alpha, gamma > 0, quad beta in mathbbR,$$
which now corresponds to the above expectation.
EDIT 2: Another way to solve my problem would be to characterise the difference (hopefully small?)
$$textEleft Phileft(fracbetasqrtgamma + X^2Y^2right)right - Phileft(fracbetasqrtgammaright),qquad gamma > 0, quad beta in mathbbR,$$
would it be sensible to approximate the difference
$$Phileft(fracbetasqrtgamma + X^2Y^2right) - Phileft(fracbetasqrtgammaright),$$
with a Taylor series in zero or is this unreasonable given that the half-Cauchy distributions are heavy-tailed?
normal-distribution expected-value approximation integral
normal-distribution expected-value approximation integral
edited 33 mins ago
user79097
asked 21 hours ago
user79097user79097
1306
1306
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
19 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
18 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
15 hours ago
$begingroup$
You are probably right, in fact the original problem corresponds to that of the expectation above, with X and Y having their respective half-Cauchy distribution... I should probably have stated the problem directly in this way. Does this make you think of any other strategy?
$endgroup$
– user79097
58 mins ago
add a comment |
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
19 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
18 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
15 hours ago
$begingroup$
You are probably right, in fact the original problem corresponds to that of the expectation above, with X and Y having their respective half-Cauchy distribution... I should probably have stated the problem directly in this way. Does this make you think of any other strategy?
$endgroup$
– user79097
58 mins ago
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
19 hours ago
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
19 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
18 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
18 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
15 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
15 hours ago
$begingroup$
You are probably right, in fact the original problem corresponds to that of the expectation above, with X and Y having their respective half-Cauchy distribution... I should probably have stated the problem directly in this way. Does this make you think of any other strategy?
$endgroup$
– user79097
58 mins ago
$begingroup$
You are probably right, in fact the original problem corresponds to that of the expectation above, with X and Y having their respective half-Cauchy distribution... I should probably have stated the problem directly in this way. Does this make you think of any other strategy?
$endgroup$
– user79097
58 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,
$$beta/sqrtgamma + uv ge min(beta/sqrtgamma, 0) = delta gt -infty.$$
Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(fracbetasqrtgamma + uvright)ge Phi(delta) = epsilon gt 0.$$
Consequently
$$eqalign
&int_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 Phileft(fracbetasqrtgamma + uvright) mathrmdumathrmdv \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 mathrmdumathrmdv \
&= lim_Mtoinftylim_Ntoinftyepsilonint_0^M (1+alpha u)^-1mathrmduint_0^N (1+v)^-1 mathrmdv \
&=fracepsilonalpha lim_Mtoinftylim_Ntoinfty log(1 + Malpha)log(1 + N),
$$
which diverges to $+infty.$
$endgroup$
$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
18 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♦
15 hours ago
$begingroup$
Thanks @whuber, that's helpful, I will try this. In the meantime, I added a second edit, where I give another formulation of my problem.
$endgroup$
– user79097
31 mins ago
add a comment |
$begingroup$
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)))
$endgroup$
$begingroup$
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
21 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
21 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
21 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,
$$beta/sqrtgamma + uv ge min(beta/sqrtgamma, 0) = delta gt -infty.$$
Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(fracbetasqrtgamma + uvright)ge Phi(delta) = epsilon gt 0.$$
Consequently
$$eqalign
&int_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 Phileft(fracbetasqrtgamma + uvright) mathrmdumathrmdv \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 mathrmdumathrmdv \
&= lim_Mtoinftylim_Ntoinftyepsilonint_0^M (1+alpha u)^-1mathrmduint_0^N (1+v)^-1 mathrmdv \
&=fracepsilonalpha lim_Mtoinftylim_Ntoinfty log(1 + Malpha)log(1 + N),
$$
which diverges to $+infty.$
$endgroup$
$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
18 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♦
15 hours ago
$begingroup$
Thanks @whuber, that's helpful, I will try this. In the meantime, I added a second edit, where I give another formulation of my problem.
$endgroup$
– user79097
31 mins ago
add a comment |
$begingroup$
Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,
$$beta/sqrtgamma + uv ge min(beta/sqrtgamma, 0) = delta gt -infty.$$
Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(fracbetasqrtgamma + uvright)ge Phi(delta) = epsilon gt 0.$$
Consequently
$$eqalign
&int_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 Phileft(fracbetasqrtgamma + uvright) mathrmdumathrmdv \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 mathrmdumathrmdv \
&= lim_Mtoinftylim_Ntoinftyepsilonint_0^M (1+alpha u)^-1mathrmduint_0^N (1+v)^-1 mathrmdv \
&=fracepsilonalpha lim_Mtoinftylim_Ntoinfty log(1 + Malpha)log(1 + N),
$$
which diverges to $+infty.$
$endgroup$
$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
18 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♦
15 hours ago
$begingroup$
Thanks @whuber, that's helpful, I will try this. In the meantime, I added a second edit, where I give another formulation of my problem.
$endgroup$
– user79097
31 mins ago
add a comment |
$begingroup$
Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,
$$beta/sqrtgamma + uv ge min(beta/sqrtgamma, 0) = delta gt -infty.$$
Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(fracbetasqrtgamma + uvright)ge Phi(delta) = epsilon gt 0.$$
Consequently
$$eqalign
&int_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 Phileft(fracbetasqrtgamma + uvright) mathrmdumathrmdv \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 mathrmdumathrmdv \
&= lim_Mtoinftylim_Ntoinftyepsilonint_0^M (1+alpha u)^-1mathrmduint_0^N (1+v)^-1 mathrmdv \
&=fracepsilonalpha lim_Mtoinftylim_Ntoinfty log(1 + Malpha)log(1 + N),
$$
which diverges to $+infty.$
$endgroup$
Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,
$$beta/sqrtgamma + uv ge min(beta/sqrtgamma, 0) = delta gt -infty.$$
Therefore, because $Phi$ is a CDF for a distribution supported on $(-infty,infty),$ $$Phileft(fracbetasqrtgamma + uvright)ge Phi(delta) = epsilon gt 0.$$
Consequently
$$eqalign
&int_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 Phileft(fracbetasqrtgamma + uvright) mathrmdumathrmdv \
&ge
epsilonint_0^infty int_0^infty (1+alpha u)^-1(1+v)^-1 mathrmdumathrmdv \
&= lim_Mtoinftylim_Ntoinftyepsilonint_0^M (1+alpha u)^-1mathrmduint_0^N (1+v)^-1 mathrmdv \
&=fracepsilonalpha lim_Mtoinftylim_Ntoinfty log(1 + Malpha)log(1 + N),
$$
which diverges to $+infty.$
answered 19 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
18 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♦
15 hours ago
$begingroup$
Thanks @whuber, that's helpful, I will try this. In the meantime, I added a second edit, where I give another formulation of my problem.
$endgroup$
– user79097
31 mins ago
add a comment |
$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
18 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♦
15 hours ago
$begingroup$
Thanks @whuber, that's helpful, I will try this. In the meantime, I added a second edit, where I give another formulation of my problem.
$endgroup$
– user79097
31 mins ago
$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
18 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
18 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♦
15 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♦
15 hours ago
$begingroup$
Thanks @whuber, that's helpful, I will try this. In the meantime, I added a second edit, where I give another formulation of my problem.
$endgroup$
– user79097
31 mins ago
$begingroup$
Thanks @whuber, that's helpful, I will try this. In the meantime, I added a second edit, where I give another formulation of my problem.
$endgroup$
– user79097
31 mins ago
add a comment |
$begingroup$
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)))
$endgroup$
$begingroup$
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
21 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
21 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
21 hours ago
add a comment |
$begingroup$
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)))
$endgroup$
$begingroup$
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
21 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
21 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
21 hours ago
add a comment |
$begingroup$
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)))
$endgroup$
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)))
edited 21 hours ago
answered 21 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
21 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
21 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
21 hours ago
add a comment |
$begingroup$
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
21 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
21 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
21 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
21 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
21 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
21 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
21 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
21 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
21 hours ago
add a comment |
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$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
19 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
18 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
15 hours ago
$begingroup$
You are probably right, in fact the original problem corresponds to that of the expectation above, with X and Y having their respective half-Cauchy distribution... I should probably have stated the problem directly in this way. Does this make you think of any other strategy?
$endgroup$
– user79097
58 mins ago