Expectation of max of two normal random variables












1












$begingroup$


I have been reading this paper about the maximum and minimum of two normal distributed variables.



Inside the paper there is the formula for the expectation of this the maximum of the two variables. It is given below:



$$ DeclareMathOperator{E}{mathbb{E}}
E(X) = mu_1 Phileft( frac{mu_1-mu_2}{theta} right) + mu_2 Phileft( frac{mu_2-mu_1}{theta} right) + theta phileft( frac{mu_1-mu_2}{theta} right)
$$



Where:





  • $mu_1$ is the mean of the first normally distributed random variable.


  • $mu_2$ is the mean of the second normally distributed random variable.


  • $Phi$ is the CDF of the standard normal.


  • $phi$ is the PDF of the standard normal.


  • $rho$ is the correlation co-efficient between the variables.


  • $theta$ is $sqrt{sigma_{1}^2 + sigma_{2}^2 + 2rhosigma_{1}sigma_{2}}$


I am trying to use this to get the mean of the new distribution but I run in to what feels like an obvious problem.



If $mu_{1} neq mu_2$ then the argument to the $Phi$ function in either the first or second term of the sum will be negative. The standard CDF does not allow negative values‽



I feel like I have fundamentally misunderstood something here. Where have I gone wrong?






normal-distribution maximum







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    By definition, the CDF evaluated at any real number $x$ gives the chance that the variable (a standard Normal variable in this case) is less than or equal to $x:$ that is a probability and therefore lies between $0$ and $1.$ It cannot be negative.
    $endgroup$
    – whuber♦
    5 hours ago










  • $begingroup$
    Just a hint here Max(X,Y) = 1/2((X+Y)-|X-Y|). So you just have to calculate E[|Z|] where Z is a normal random variable with mean mu and std sigma. Then apply the formula to X-Y and of course E[X+Y]=E[X]+E[Y]
    $endgroup$
    – Daniel Parry
    5 hours ago
















1












$begingroup$


I have been reading this paper about the maximum and minimum of two normal distributed variables.



Inside the paper there is the formula for the expectation of this the maximum of the two variables. It is given below:



$$ DeclareMathOperator{E}{mathbb{E}}
E(X) = mu_1 Phileft( frac{mu_1-mu_2}{theta} right) + mu_2 Phileft( frac{mu_2-mu_1}{theta} right) + theta phileft( frac{mu_1-mu_2}{theta} right)
$$



Where:





  • $mu_1$ is the mean of the first normally distributed random variable.


  • $mu_2$ is the mean of the second normally distributed random variable.


  • $Phi$ is the CDF of the standard normal.


  • $phi$ is the PDF of the standard normal.


  • $rho$ is the correlation co-efficient between the variables.


  • $theta$ is $sqrt{sigma_{1}^2 + sigma_{2}^2 + 2rhosigma_{1}sigma_{2}}$


I am trying to use this to get the mean of the new distribution but I run in to what feels like an obvious problem.



If $mu_{1} neq mu_2$ then the argument to the $Phi$ function in either the first or second term of the sum will be negative. The standard CDF does not allow negative values‽



I feel like I have fundamentally misunderstood something here. Where have I gone wrong?






normal-distribution maximum







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    By definition, the CDF evaluated at any real number $x$ gives the chance that the variable (a standard Normal variable in this case) is less than or equal to $x:$ that is a probability and therefore lies between $0$ and $1.$ It cannot be negative.
    $endgroup$
    – whuber♦
    5 hours ago










  • $begingroup$
    Just a hint here Max(X,Y) = 1/2((X+Y)-|X-Y|). So you just have to calculate E[|Z|] where Z is a normal random variable with mean mu and std sigma. Then apply the formula to X-Y and of course E[X+Y]=E[X]+E[Y]
    $endgroup$
    – Daniel Parry
    5 hours ago














1












1








1





$begingroup$


I have been reading this paper about the maximum and minimum of two normal distributed variables.



Inside the paper there is the formula for the expectation of this the maximum of the two variables. It is given below:



$$ DeclareMathOperator{E}{mathbb{E}}
E(X) = mu_1 Phileft( frac{mu_1-mu_2}{theta} right) + mu_2 Phileft( frac{mu_2-mu_1}{theta} right) + theta phileft( frac{mu_1-mu_2}{theta} right)
$$



Where:





  • $mu_1$ is the mean of the first normally distributed random variable.


  • $mu_2$ is the mean of the second normally distributed random variable.


  • $Phi$ is the CDF of the standard normal.


  • $phi$ is the PDF of the standard normal.


  • $rho$ is the correlation co-efficient between the variables.


  • $theta$ is $sqrt{sigma_{1}^2 + sigma_{2}^2 + 2rhosigma_{1}sigma_{2}}$


I am trying to use this to get the mean of the new distribution but I run in to what feels like an obvious problem.



If $mu_{1} neq mu_2$ then the argument to the $Phi$ function in either the first or second term of the sum will be negative. The standard CDF does not allow negative values‽



I feel like I have fundamentally misunderstood something here. Where have I gone wrong?






normal-distribution maximum







share|cite|improve this question









$endgroup$




I have been reading this paper about the maximum and minimum of two normal distributed variables.



Inside the paper there is the formula for the expectation of this the maximum of the two variables. It is given below:



$$ DeclareMathOperator{E}{mathbb{E}}
E(X) = mu_1 Phileft( frac{mu_1-mu_2}{theta} right) + mu_2 Phileft( frac{mu_2-mu_1}{theta} right) + theta phileft( frac{mu_1-mu_2}{theta} right)
$$



Where:





  • $mu_1$ is the mean of the first normally distributed random variable.


  • $mu_2$ is the mean of the second normally distributed random variable.


  • $Phi$ is the CDF of the standard normal.


  • $phi$ is the PDF of the standard normal.


  • $rho$ is the correlation co-efficient between the variables.


  • $theta$ is $sqrt{sigma_{1}^2 + sigma_{2}^2 + 2rhosigma_{1}sigma_{2}}$


I am trying to use this to get the mean of the new distribution but I run in to what feels like an obvious problem.



If $mu_{1} neq mu_2$ then the argument to the $Phi$ function in either the first or second term of the sum will be negative. The standard CDF does not allow negative values‽



I feel like I have fundamentally misunderstood something here. Where have I gone wrong?






normal-distribution maximum




normal-distribution maximum






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









Simon JohnsonSimon Johnson

1084




1084








  • 2




    $begingroup$
    By definition, the CDF evaluated at any real number $x$ gives the chance that the variable (a standard Normal variable in this case) is less than or equal to $x:$ that is a probability and therefore lies between $0$ and $1.$ It cannot be negative.
    $endgroup$
    – whuber♦
    5 hours ago










  • $begingroup$
    Just a hint here Max(X,Y) = 1/2((X+Y)-|X-Y|). So you just have to calculate E[|Z|] where Z is a normal random variable with mean mu and std sigma. Then apply the formula to X-Y and of course E[X+Y]=E[X]+E[Y]
    $endgroup$
    – Daniel Parry
    5 hours ago














  • 2




    $begingroup$
    By definition, the CDF evaluated at any real number $x$ gives the chance that the variable (a standard Normal variable in this case) is less than or equal to $x:$ that is a probability and therefore lies between $0$ and $1.$ It cannot be negative.
    $endgroup$
    – whuber♦
    5 hours ago










  • $begingroup$
    Just a hint here Max(X,Y) = 1/2((X+Y)-|X-Y|). So you just have to calculate E[|Z|] where Z is a normal random variable with mean mu and std sigma. Then apply the formula to X-Y and of course E[X+Y]=E[X]+E[Y]
    $endgroup$
    – Daniel Parry
    5 hours ago








2




2




$begingroup$
By definition, the CDF evaluated at any real number $x$ gives the chance that the variable (a standard Normal variable in this case) is less than or equal to $x:$ that is a probability and therefore lies between $0$ and $1.$ It cannot be negative.
$endgroup$
– whuber♦
5 hours ago




$begingroup$
By definition, the CDF evaluated at any real number $x$ gives the chance that the variable (a standard Normal variable in this case) is less than or equal to $x:$ that is a probability and therefore lies between $0$ and $1.$ It cannot be negative.
$endgroup$
– whuber♦
5 hours ago












$begingroup$
Just a hint here Max(X,Y) = 1/2((X+Y)-|X-Y|). So you just have to calculate E[|Z|] where Z is a normal random variable with mean mu and std sigma. Then apply the formula to X-Y and of course E[X+Y]=E[X]+E[Y]
$endgroup$
– Daniel Parry
5 hours ago




$begingroup$
Just a hint here Max(X,Y) = 1/2((X+Y)-|X-Y|). So you just have to calculate E[|Z|] where Z is a normal random variable with mean mu and std sigma. Then apply the formula to X-Y and of course E[X+Y]=E[X]+E[Y]
$endgroup$
– Daniel Parry
5 hours ago










1 Answer
1






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oldest

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5












$begingroup$

The CDF does not yield negative results (it only yields numbers between 0 and 1), but it can certainly take negative arguments.



> pnorm(-1,0,1)

[1] 0.1586553





share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Stupid mistake on my part. In Excel I was using the Norm.Inv function and then slapped my forehead when I saw this answer.
    $endgroup$
    – Simon Johnson
    5 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

The CDF does not yield negative results (it only yields numbers between 0 and 1), but it can certainly take negative arguments.



> pnorm(-1,0,1)

[1] 0.1586553





share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Stupid mistake on my part. In Excel I was using the Norm.Inv function and then slapped my forehead when I saw this answer.
    $endgroup$
    – Simon Johnson
    5 hours ago
















5












$begingroup$

The CDF does not yield negative results (it only yields numbers between 0 and 1), but it can certainly take negative arguments.



> pnorm(-1,0,1)

[1] 0.1586553





share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Stupid mistake on my part. In Excel I was using the Norm.Inv function and then slapped my forehead when I saw this answer.
    $endgroup$
    – Simon Johnson
    5 hours ago














5












5








5





$begingroup$

The CDF does not yield negative results (it only yields numbers between 0 and 1), but it can certainly take negative arguments.



> pnorm(-1,0,1)

[1] 0.1586553





share|cite|improve this answer









$endgroup$



The CDF does not yield negative results (it only yields numbers between 0 and 1), but it can certainly take negative arguments.



> pnorm(-1,0,1)

[1] 0.1586553






share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









Stephan KolassaStephan Kolassa

45.7k695167




45.7k695167












  • $begingroup$
    Thanks! Stupid mistake on my part. In Excel I was using the Norm.Inv function and then slapped my forehead when I saw this answer.
    $endgroup$
    – Simon Johnson
    5 hours ago


















  • $begingroup$
    Thanks! Stupid mistake on my part. In Excel I was using the Norm.Inv function and then slapped my forehead when I saw this answer.
    $endgroup$
    – Simon Johnson
    5 hours ago
















$begingroup$
Thanks! Stupid mistake on my part. In Excel I was using the Norm.Inv function and then slapped my forehead when I saw this answer.
$endgroup$
– Simon Johnson
5 hours ago




$begingroup$
Thanks! Stupid mistake on my part. In Excel I was using the Norm.Inv function and then slapped my forehead when I saw this answer.
$endgroup$
– Simon Johnson
5 hours ago


















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