Double integral involving the normal CDFExpected value of minimum order statistic from a normal...
Does the US political system, in principle, allow for a no-party system?
Can a space-faring robot still function over a billion years?
Is being socially reclusive okay for a graduate student?
TikZ rotated text looks washed
Is every open circuit a capacitor?
What can I do if someone tampers with my SSH public key?
Paper published similar to PhD thesis
Insult for someone who "doesn't know anything"
My cassette model name doesn't seem to fit the actual cassette description.. (CS-HG50-9 and CS-HG500-10)
Double integral involving the normal CDF
If nine coins are tossed, what is the probability that the number of heads is even?
Are small insurances worth it
Why do phishing e-mails use faked e-mail addresses instead of the real one?
Has a sovereign Communist government ever run, and conceded loss, on a fair election?
Does restoring a database break external synonyms targeting objects in that database?
How can I portion out frozen cookie dough?
V=IR means causation or equivalence?
Why can't we use freedom of speech and expression to incite people to rebel against government in India?
Under what conditions would I NOT add my Proficiency Bonus to a Spell Attack Roll (or Saving Throw DC)?
Should we avoid writing fiction about historical events without extensive research?
Does unused member variable take up memory?
Should I use HTTPS on a domain that will only be used for redirection?
Was it really inappropriate to write a pull request for the company I interviewed with?
How does a sound wave propagate?
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
$$int_0^infty int_0^infty (1 + alpha u)^{-1}(1 + v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) text{d}u text{d}v,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
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
asked 3 hours ago
user79097user79097
1306
$endgroup$
add a comment |
$begingroup$
I need to compute (or best approximate?) the following integral
$$int_0^infty int_0^infty (1 + alpha u)^{-1}(1 + v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) text{d}u text{d}v,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
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
asked 3 hours ago
user79097user79097
1306
$endgroup$
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
1 hour 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
51 mins ago
add a comment |
$begingroup$
I need to compute (or best approximate?) the following integral
$$int_0^infty int_0^infty (1 + alpha u)^{-1}(1 + v)^{-1} Phileft(frac{beta}{sqrt{gamma + uv}}right) text{d}u text{d}v,qquad alpha, gamma > 0, quad beta in mathbb{R},$$
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
asked 3 hours ago
user79097user79097
1306
$endgroup$
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
asked 3 hours ago
user79097user79097
1306
asked 3 hours ago
user79097user79097
1306
edited 52 mins ago
user79097
asked 3 hours ago
user79097user79097
1306
asked 3 hours ago
user79097user79097
1306
asked 3 hours ago
user79097user79097
1306
1306
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
1 hour 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
51 mins ago
add a comment |
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
1 hour 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
51 mins ago
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
1 hour ago
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
1 hour 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
51 mins 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
51 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Under the conditions that $alpha, gamma, u,$ and $v$ are all positive,
$$beta/sqrt{gamma + uv} ge min(beta/sqrt{gamma}, 0) = delta gt -infty.$$
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 $+infty.$
answered 1 hour ago
whuber♦whuber
205k33448816
$endgroup$
$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
49 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)))
answered 3 hours ago
Christoph HanckChristoph Hanck
17.3k34074
$endgroup$
$begingroup$
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
3 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
3 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
3 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "65"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f396326%2fdouble-integral-involving-the-normal-cdf%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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,
$$beta/sqrt{gamma + uv} ge min(beta/sqrt{gamma}, 0) = delta gt -infty.$$
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 $+infty.$
answered 1 hour ago
whuber♦whuber
205k33448816
$endgroup$
$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
49 mins ago
add a comment |
$begingroup$
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 $+infty.$
answered 1 hour ago
whuber♦whuber
205k33448816
$endgroup$
$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
49 mins ago
add a comment |
$begingroup$
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 $+infty.$
answered 1 hour ago
whuber♦whuber
205k33448816
$endgroup$
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 $+infty.$
answered 1 hour ago
whuber♦whuber
205k33448816
answered 1 hour ago
whuber♦whuber
205k33448816
answered 1 hour ago
whuber♦whuber
205k33448816
answered 1 hour ago
whuber♦whuber
205k33448816
205k33448816
$begingroup$
Thanks, this is a very clear argument. For the "edited" integral, it seems that the argument doesn't apply anymore.
$endgroup$
– user79097
49 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
49 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
49 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
49 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)))
answered 3 hours ago
Christoph HanckChristoph Hanck
17.3k34074
$endgroup$
$begingroup$
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
3 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
3 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
3 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)))
answered 3 hours ago
Christoph HanckChristoph Hanck
17.3k34074
$endgroup$
$begingroup$
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
3 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
3 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
3 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)))
answered 3 hours ago
Christoph HanckChristoph Hanck
17.3k34074
$endgroup$
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)))
answered 3 hours ago
Christoph HanckChristoph Hanck
17.3k34074
edited 3 hours ago
answered 3 hours ago
Christoph HanckChristoph Hanck
17.3k34074
answered 3 hours ago
Christoph HanckChristoph Hanck
17.3k34074
answered 3 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
3 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
3 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
3 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
3 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
3 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
3 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
3 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
3 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
3 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
3 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
3 hours ago
$begingroup$
Yes, sure, the problem is that I really need something analytical unfortunately. But thanks very much for your help!
$endgroup$
– user79097
3 hours ago
add a comment |
Thanks for contributing an answer to Cross Validated!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f396326%2fdouble-integral-involving-the-normal-cdf%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I would advise studying the asymptotic behavior, because this integral diverges.
$endgroup$
– whuber♦
1 hour 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
51 mins ago