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Calculating an infinite integral of log-normal distribution

Expectation of random varible with normal distribution composed with exponentialmoment generating function for folded/absolute normal distributionMoments of the shifted log-normalConditional Expectation of the minimum of two identical log-normal distributionsOn the evaluation of the integral $int_{-frac{b}{a}}^{frac{1-b}{a}}logleft(ax+bright)expleft(-frac{1}{2}x^2right)mathrm{d}x$.Simplifying complicated integral including CDF of normal distributionExpected value of a lognormal distributionClosed form expression for the integralNormal distribution in an intervalCalculating the convolution of an Arcsine law and a Gaussian distribution

0

$begingroup$

The integral is:

$int^infty_0 x exp{Big(frac{-(log{x}-mu)^2}{2sigma^2}Big)}dx $ (it is a second moment of log-normal distribution).

I've tried several subsitutions, such as

$u=log{x}$,

$u=log{x} - mu$,

$u=(log{x} - mu)^2$.

However, all of them lead to more complicated results. I can calculate this integral when there is $x$ instead of $log{x}$, but with the logarithm it gets complicated. Is there a way to calculate it using some trick? (undergraduate level)

Thanks in advance.

probability integration indefinite-integrals

share|cite|improve this question

edited 2 days ago

lkky7

asked Mar 9 at 17:13

lkky7lkky7

295

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a typo? The integral as shown is for the first moment. Which moment are you trying to calculate?
  $endgroup$
  – Lee David Chung Lin
  Mar 9 at 17:28
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The integral in your edited question is the third moment, not the second.
  $endgroup$
  – saz
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hah... Yes, it was all a typo. Thank you.
  $endgroup$
  – lkky7
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

The integral is:

$int^infty_0 x exp{Big(frac{-(log{x}-mu)^2}{2sigma^2}Big)}dx $ (it is a second moment of log-normal distribution).

I've tried several subsitutions, such as

$u=log{x}$,

$u=log{x} - mu$,

$u=(log{x} - mu)^2$.

However, all of them lead to more complicated results. I can calculate this integral when there is $x$ instead of $log{x}$, but with the logarithm it gets complicated. Is there a way to calculate it using some trick? (undergraduate level)

Thanks in advance.

probability integration indefinite-integrals

share|cite|improve this question

edited 2 days ago

lkky7

asked Mar 9 at 17:13

lkky7lkky7

295

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a typo? The integral as shown is for the first moment. Which moment are you trying to calculate?
  $endgroup$
  – Lee David Chung Lin
  Mar 9 at 17:28
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The integral in your edited question is the third moment, not the second.
  $endgroup$
  – saz
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hah... Yes, it was all a typo. Thank you.
  $endgroup$
  – lkky7
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

The integral is:

$int^infty_0 x exp{Big(frac{-(log{x}-mu)^2}{2sigma^2}Big)}dx $ (it is a second moment of log-normal distribution).

I've tried several subsitutions, such as

$u=log{x}$,

$u=log{x} - mu$,

$u=(log{x} - mu)^2$.

However, all of them lead to more complicated results. I can calculate this integral when there is $x$ instead of $log{x}$, but with the logarithm it gets complicated. Is there a way to calculate it using some trick? (undergraduate level)

Thanks in advance.

probability integration indefinite-integrals

share|cite|improve this question

edited 2 days ago

lkky7

asked Mar 9 at 17:13

lkky7lkky7

295

$endgroup$

share|cite|improve this question

edited 2 days ago

lkky7

asked Mar 9 at 17:13

lkky7lkky7

295

share|cite|improve this question

edited 2 days ago

lkky7

asked Mar 9 at 17:13

lkky7lkky7

295

asked Mar 9 at 17:13

lkky7lkky7

295

asked Mar 9 at 17:13

lkky7lkky7

295

1 Answer
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oldest

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0

$begingroup$

Denote by

$$p(x) = frac{1}{x} frac{1}{sqrt{2pi sigma^2}} exp left(- frac{(log x- mu)^2}{2sigma^2} right)$$

the probability density function of the log-normal distribution. For $y := log x- mu$ we have

$$frac{dy}{dx} = frac{1}{x} = exp(-y-mu),$$

i.e.

$$dx = exp(y+mu) , dy,$$

and therefore a change of variables gives

$$int_0^{infty}x p(x) , dx = frac{1}{sqrt{2pi sigma^2}} int_{mathbb{R}} exp(y+mu) exp left(- frac{y^2}{2sigma^2} right) , dy.$$

The right-hand side is the exponential moment of a Gaussian random variable; more precisely, if $Y sim N(0,sigma^2)$ then the right-hand side equals

$$exp(mu) mathbb{E}exp(Y).$$

Since exponential moments of Gaussian random variables can be calculated explicitly, we get

$$int_0^{infty}x p(x) , dx = exp left( mu + frac{1}{2} sigma^2 right).$$

Equivalently,

$$int_{(0,infty)} exp left(- frac{(log x-mu)^2}{2sigma^2} right) , dx = sqrt{2pi sigma^2}exp left( mu + frac{1}{2} sigma^2 right).$$

Remark: The same reasoning works also for higher moments, i.e.

$$int_{(0,infty)} x^k p(x) , dx$$

for $k geq 1$. Following the argumentation from above we get

$$int_{(0,infty)} x^k p(x) , dx = exp(k mu) mathbb{E}exp(kY) = exp left( k mu + frac{1}{2} sigma^2 k^2 right),$$

i.e.

$$int_{(0,infty)} x^{k-1} exp left( - frac{(log x-mu)^2}{2sigma^2} right) , dx = sqrt{2pi sigma^2} exp left( k mu + frac{1}{2} sigma^2 k^2 right).$$

share|cite|improve this answer

edited 2 days ago

answered Mar 9 at 17:43

sazsaz

81.7k861127

$endgroup$

add a comment | 

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0

$begingroup$

Denote by

$$p(x) = frac{1}{x} frac{1}{sqrt{2pi sigma^2}} exp left(- frac{(log x- mu)^2}{2sigma^2} right)$$

the probability density function of the log-normal distribution. For $y := log x- mu$ we have

$$frac{dy}{dx} = frac{1}{x} = exp(-y-mu),$$

i.e.

$$dx = exp(y+mu) , dy,$$

and therefore a change of variables gives

$$int_0^{infty}x p(x) , dx = frac{1}{sqrt{2pi sigma^2}} int_{mathbb{R}} exp(y+mu) exp left(- frac{y^2}{2sigma^2} right) , dy.$$

The right-hand side is the exponential moment of a Gaussian random variable; more precisely, if $Y sim N(0,sigma^2)$ then the right-hand side equals

$$exp(mu) mathbb{E}exp(Y).$$

Since exponential moments of Gaussian random variables can be calculated explicitly, we get

$$int_0^{infty}x p(x) , dx = exp left( mu + frac{1}{2} sigma^2 right).$$

Equivalently,

$$int_{(0,infty)} exp left(- frac{(log x-mu)^2}{2sigma^2} right) , dx = sqrt{2pi sigma^2}exp left( mu + frac{1}{2} sigma^2 right).$$

Remark: The same reasoning works also for higher moments, i.e.

$$int_{(0,infty)} x^k p(x) , dx$$

for $k geq 1$. Following the argumentation from above we get

$$int_{(0,infty)} x^k p(x) , dx = exp(k mu) mathbb{E}exp(kY) = exp left( k mu + frac{1}{2} sigma^2 k^2 right),$$

i.e.

$$int_{(0,infty)} x^{k-1} exp left( - frac{(log x-mu)^2}{2sigma^2} right) , dx = sqrt{2pi sigma^2} exp left( k mu + frac{1}{2} sigma^2 k^2 right).$$

share|cite|improve this answer

edited 2 days ago

answered Mar 9 at 17:43

sazsaz

81.7k861127

$endgroup$

add a comment | 

0

$begingroup$

Denote by

$$p(x) = frac{1}{x} frac{1}{sqrt{2pi sigma^2}} exp left(- frac{(log x- mu)^2}{2sigma^2} right)$$

the probability density function of the log-normal distribution. For $y := log x- mu$ we have

$$frac{dy}{dx} = frac{1}{x} = exp(-y-mu),$$

i.e.

$$dx = exp(y+mu) , dy,$$

and therefore a change of variables gives

$$int_0^{infty}x p(x) , dx = frac{1}{sqrt{2pi sigma^2}} int_{mathbb{R}} exp(y+mu) exp left(- frac{y^2}{2sigma^2} right) , dy.$$

The right-hand side is the exponential moment of a Gaussian random variable; more precisely, if $Y sim N(0,sigma^2)$ then the right-hand side equals

$$exp(mu) mathbb{E}exp(Y).$$

Since exponential moments of Gaussian random variables can be calculated explicitly, we get

$$int_0^{infty}x p(x) , dx = exp left( mu + frac{1}{2} sigma^2 right).$$

Equivalently,

$$int_{(0,infty)} exp left(- frac{(log x-mu)^2}{2sigma^2} right) , dx = sqrt{2pi sigma^2}exp left( mu + frac{1}{2} sigma^2 right).$$

Remark: The same reasoning works also for higher moments, i.e.

$$int_{(0,infty)} x^k p(x) , dx$$

for $k geq 1$. Following the argumentation from above we get

$$int_{(0,infty)} x^k p(x) , dx = exp(k mu) mathbb{E}exp(kY) = exp left( k mu + frac{1}{2} sigma^2 k^2 right),$$

i.e.

$$int_{(0,infty)} x^{k-1} exp left( - frac{(log x-mu)^2}{2sigma^2} right) , dx = sqrt{2pi sigma^2} exp left( k mu + frac{1}{2} sigma^2 k^2 right).$$

share|cite|improve this answer

edited 2 days ago

answered Mar 9 at 17:43

sazsaz

81.7k861127

$endgroup$

add a comment | 

$begingroup$

Denote by

$$p(x) = frac{1}{x} frac{1}{sqrt{2pi sigma^2}} exp left(- frac{(log x- mu)^2}{2sigma^2} right)$$

the probability density function of the log-normal distribution. For $y := log x- mu$ we have

$$frac{dy}{dx} = frac{1}{x} = exp(-y-mu),$$

i.e.

$$dx = exp(y+mu) , dy,$$

and therefore a change of variables gives

$$int_0^{infty}x p(x) , dx = frac{1}{sqrt{2pi sigma^2}} int_{mathbb{R}} exp(y+mu) exp left(- frac{y^2}{2sigma^2} right) , dy.$$

The right-hand side is the exponential moment of a Gaussian random variable; more precisely, if $Y sim N(0,sigma^2)$ then the right-hand side equals

$$exp(mu) mathbb{E}exp(Y).$$

Since exponential moments of Gaussian random variables can be calculated explicitly, we get

$$int_0^{infty}x p(x) , dx = exp left( mu + frac{1}{2} sigma^2 right).$$

Equivalently,

$$int_{(0,infty)} exp left(- frac{(log x-mu)^2}{2sigma^2} right) , dx = sqrt{2pi sigma^2}exp left( mu + frac{1}{2} sigma^2 right).$$

Remark: The same reasoning works also for higher moments, i.e.

$$int_{(0,infty)} x^k p(x) , dx$$

for $k geq 1$. Following the argumentation from above we get

$$int_{(0,infty)} x^k p(x) , dx = exp(k mu) mathbb{E}exp(kY) = exp left( k mu + frac{1}{2} sigma^2 k^2 right),$$

i.e.

$$int_{(0,infty)} x^{k-1} exp left( - frac{(log x-mu)^2}{2sigma^2} right) , dx = sqrt{2pi sigma^2} exp left( k mu + frac{1}{2} sigma^2 k^2 right).$$

share|cite|improve this answer

edited 2 days ago

answered Mar 9 at 17:43

sazsaz

81.7k861127

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered Mar 9 at 17:43

sazsaz

81.7k861127

answered Mar 9 at 17:43

sazsaz

81.7k861127

answered Mar 9 at 17:43

sazsaz

81.7k861127