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2

1

$begingroup$

Given $Xpropto N(mu,psi^2)$ and $K$ constant, find $E(maxK-X,0)$.

My attempt:
$$
beginalign
E(maxK-X,0)\
&=int_-infty^inftymaxK-x,0f_X(x)dx\
&=int_K^infty(K-x)f_X(x)dx\
&=int_K^infty[(K-mu)-(X-mu)]f_X(x)dx\
&=int_K^infty(K-mu)f_X(x)dx-int_K^infty(x-mu)f_X(x)dx\
&=(K-mu)(1-Phi(K))-int_K^infty(x-mu)f_X(x)dx\
&=(K-mu)(1-Phi(K))-int_K^infty(x-mu)frac1(2pi psi^2)^1/2e^-frac(x-mu)^22psi^2dx\
&=(K-mu)(1-Phi(K))-int_frac(K-mu)^22psi^2^inftypsi^2frac1(2pi psi^2)^1/2e^-zdz\
&=(K-mu)(1-Phi(K))+fracpsisqrt2pie^-z|_frac(K-mu)^22psi^2^infty\
&=(K-mu)(1-Phi(K))-fracpsisqrt2pie^-frac(K-mu)^22psi^2
endalign
$$
where I have used the change of variables $z=frac(x-mu)^22psi^2$.

However, when I check my answer numerically, I see that the integral is not correct. Here is working Python code to check:

import numpy as np
from scipy.stats import norm
from scipy.special import erf

# simulation parameters for checking answers numerically
trials = 100000
K = 0.25
mu = np.pi/2
psi = 3.0

simulated = np.mean([np.max([K-x, 0])
 for x in np.random.normal(mu, psi, trials)])
numerical = (K-mu)*(1-norm.cdf(K, mu, psi)) - np.exp(-((K-mu)**2)/(2*psi**2))*psi/np.sqrt(2*np.pi) 
print(simulated)
print(numerical)

This gives:

simulated = 0.6448298784565597

numerical = -1.9713797586737627

What is the issue with my integration?

statistics numerical-methods normal-distribution expected-value

share|cite|improve this question

edited 14 hours ago

Rodrigo de Azevedo

13.1k41960

asked 16 hours ago

Nick SettjeNick Settje

1112

New contributor

Nick Settje is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  There is a mistake in the second line. $max K-X,0=K-x$ if $x leq K$.
  $endgroup$
  – Kavi Rama Murthy
  16 hours ago
  
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Given $Xpropto N(mu,psi^2)$ and $K$ constant, find $E(maxK-X,0)$.

My attempt:
$$
beginalign
E(maxK-X,0)\
&=int_-infty^inftymaxK-x,0f_X(x)dx\
&=int_K^infty(K-x)f_X(x)dx\
&=int_K^infty[(K-mu)-(X-mu)]f_X(x)dx\
&=int_K^infty(K-mu)f_X(x)dx-int_K^infty(x-mu)f_X(x)dx\
&=(K-mu)(1-Phi(K))-int_K^infty(x-mu)f_X(x)dx\
&=(K-mu)(1-Phi(K))-int_K^infty(x-mu)frac1(2pi psi^2)^1/2e^-frac(x-mu)^22psi^2dx\
&=(K-mu)(1-Phi(K))-int_frac(K-mu)^22psi^2^inftypsi^2frac1(2pi psi^2)^1/2e^-zdz\
&=(K-mu)(1-Phi(K))+fracpsisqrt2pie^-z|_frac(K-mu)^22psi^2^infty\
&=(K-mu)(1-Phi(K))-fracpsisqrt2pie^-frac(K-mu)^22psi^2
endalign
$$
where I have used the change of variables $z=frac(x-mu)^22psi^2$.

However, when I check my answer numerically, I see that the integral is not correct. Here is working Python code to check:

import numpy as np
from scipy.stats import norm
from scipy.special import erf

# simulation parameters for checking answers numerically
trials = 100000
K = 0.25
mu = np.pi/2
psi = 3.0

simulated = np.mean([np.max([K-x, 0])
 for x in np.random.normal(mu, psi, trials)])
numerical = (K-mu)*(1-norm.cdf(K, mu, psi)) - np.exp(-((K-mu)**2)/(2*psi**2))*psi/np.sqrt(2*np.pi) 
print(simulated)
print(numerical)

This gives:

simulated = 0.6448298784565597

numerical = -1.9713797586737627

What is the issue with my integration?

statistics numerical-methods normal-distribution expected-value

share|cite|improve this question

edited 14 hours ago

Rodrigo de Azevedo

13.1k41960

asked 16 hours ago

Nick SettjeNick Settje

1112

New contributor

Nick Settje is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  There is a mistake in the second line. $max K-X,0=K-x$ if $x leq K$.
  $endgroup$
  – Kavi Rama Murthy
  16 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Given $Xpropto N(mu,psi^2)$ and $K$ constant, find $E(maxK-X,0)$.

My attempt:
$$
beginalign
E(maxK-X,0)\
&=int_-infty^inftymaxK-x,0f_X(x)dx\
&=int_K^infty(K-x)f_X(x)dx\
&=int_K^infty[(K-mu)-(X-mu)]f_X(x)dx\
&=int_K^infty(K-mu)f_X(x)dx-int_K^infty(x-mu)f_X(x)dx\
&=(K-mu)(1-Phi(K))-int_K^infty(x-mu)f_X(x)dx\
&=(K-mu)(1-Phi(K))-int_K^infty(x-mu)frac1(2pi psi^2)^1/2e^-frac(x-mu)^22psi^2dx\
&=(K-mu)(1-Phi(K))-int_frac(K-mu)^22psi^2^inftypsi^2frac1(2pi psi^2)^1/2e^-zdz\
&=(K-mu)(1-Phi(K))+fracpsisqrt2pie^-z|_frac(K-mu)^22psi^2^infty\
&=(K-mu)(1-Phi(K))-fracpsisqrt2pie^-frac(K-mu)^22psi^2
endalign
$$
where I have used the change of variables $z=frac(x-mu)^22psi^2$.

However, when I check my answer numerically, I see that the integral is not correct. Here is working Python code to check:

import numpy as np
from scipy.stats import norm
from scipy.special import erf

# simulation parameters for checking answers numerically
trials = 100000
K = 0.25
mu = np.pi/2
psi = 3.0

simulated = np.mean([np.max([K-x, 0])
 for x in np.random.normal(mu, psi, trials)])
numerical = (K-mu)*(1-norm.cdf(K, mu, psi)) - np.exp(-((K-mu)**2)/(2*psi**2))*psi/np.sqrt(2*np.pi) 
print(simulated)
print(numerical)

This gives:

simulated = 0.6448298784565597

numerical = -1.9713797586737627

What is the issue with my integration?

statistics numerical-methods normal-distribution expected-value

share|cite|improve this question

edited 14 hours ago

Rodrigo de Azevedo

13.1k41960

asked 16 hours ago

Nick SettjeNick Settje

1112

New contributor

Nick Settje is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

import numpy as np
from scipy.stats import norm
from scipy.special import erf

# simulation parameters for checking answers numerically
trials = 100000
K = 0.25
mu = np.pi/2
psi = 3.0

simulated = np.mean([np.max([K-x, 0])
 for x in np.random.normal(mu, psi, trials)])
numerical = (K-mu)*(1-norm.cdf(K, mu, psi)) - np.exp(-((K-mu)**2)/(2*psi**2))*psi/np.sqrt(2*np.pi) 
print(simulated)
print(numerical)

share|cite|improve this question

edited 14 hours ago

Rodrigo de Azevedo

13.1k41960

asked 16 hours ago

Nick SettjeNick Settje

1112

New contributor

Nick Settje is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 14 hours ago

Rodrigo de Azevedo

13.1k41960

asked 16 hours ago

Nick SettjeNick Settje

1112

New contributor

Nick Settje is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 14 hours ago

Rodrigo de Azevedo

13.1k41960

edited 14 hours ago

Rodrigo de Azevedo

13.1k41960

asked 16 hours ago

Nick SettjeNick Settje

1112

New contributor

Nick Settje is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 16 hours ago

Nick SettjeNick Settje

1112