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Double integral involving the normal CDF

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Double integral involving the normal CDF

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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

edited 52 mins ago

user79097

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 | 

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

edited 52 mins ago

user79097

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

share|cite|improve this question

edited 52 mins ago

user79097

asked 3 hours ago

user79097user79097

1306

$endgroup$

share|cite|improve this question

edited 52 mins ago

user79097

asked 3 hours ago

user79097user79097

1306

share|cite|improve this question

edited 52 mins ago

user79097

asked 3 hours ago

user79097user79097

1306

asked 3 hours ago

user79097user79097

1306

asked 3 hours ago

user79097user79097

1306

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

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 | 

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

edited 3 hours ago

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 | 

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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

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 | 

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

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 $+infty.$

share|cite|improve this answer

answered 1 hour ago

whuber♦whuber

205k33448816

$endgroup$

share|cite|improve this answer

answered 1 hour ago

whuber♦whuber

205k33448816

answered 1 hour ago

whuber♦whuber

205k33448816

answered 1 hour ago

whuber♦whuber

205k33448816

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

edited 3 hours ago

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 | 

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

edited 3 hours ago

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)))

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

Christoph HanckChristoph Hanck

17.3k34074

$endgroup$

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

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