How to derive the formula for the expected value for maximum of n normal random variables












2












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This is a follow-up question on this one: Expected value for maximum of n normal random variable



@RobertIsrael states the following:




Presumably the $X_i$ are independent. If $Phi$ is the standard normal cdf,
$$P(max_i X_i < mu + t sigma) = prod_i P(X_i < mu + t sigma) = Phi(t)^n$$
so
$$ E[max_i X_i] = mu + sigma int_{-infty}^infty t dfrac{d}{dt} Phi(t)^n dt $$




I can see some of the ideas that lead to this formula (e.g. the multiplication because of independence) but I don't see all of the details. It would be very helpful if somebody could explain the derivation in detail... Thank you!






probability proof-explanation normal-distribution







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    This is a follow-up question on this one: Expected value for maximum of n normal random variable



    @RobertIsrael states the following:




    Presumably the $X_i$ are independent. If $Phi$ is the standard normal cdf,
    $$P(max_i X_i < mu + t sigma) = prod_i P(X_i < mu + t sigma) = Phi(t)^n$$
    so
    $$ E[max_i X_i] = mu + sigma int_{-infty}^infty t dfrac{d}{dt} Phi(t)^n dt $$




    I can see some of the ideas that lead to this formula (e.g. the multiplication because of independence) but I don't see all of the details. It would be very helpful if somebody could explain the derivation in detail... Thank you!






    probability proof-explanation normal-distribution







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      This is a follow-up question on this one: Expected value for maximum of n normal random variable



      @RobertIsrael states the following:




      Presumably the $X_i$ are independent. If $Phi$ is the standard normal cdf,
      $$P(max_i X_i < mu + t sigma) = prod_i P(X_i < mu + t sigma) = Phi(t)^n$$
      so
      $$ E[max_i X_i] = mu + sigma int_{-infty}^infty t dfrac{d}{dt} Phi(t)^n dt $$




      I can see some of the ideas that lead to this formula (e.g. the multiplication because of independence) but I don't see all of the details. It would be very helpful if somebody could explain the derivation in detail... Thank you!






      probability proof-explanation normal-distribution







      share|cite|improve this question









      $endgroup$




      This is a follow-up question on this one: Expected value for maximum of n normal random variable



      @RobertIsrael states the following:




      Presumably the $X_i$ are independent. If $Phi$ is the standard normal cdf,
      $$P(max_i X_i < mu + t sigma) = prod_i P(X_i < mu + t sigma) = Phi(t)^n$$
      so
      $$ E[max_i X_i] = mu + sigma int_{-infty}^infty t dfrac{d}{dt} Phi(t)^n dt $$




      I can see some of the ideas that lead to this formula (e.g. the multiplication because of independence) but I don't see all of the details. It would be very helpful if somebody could explain the derivation in detail... Thank you!






      probability proof-explanation normal-distribution




      probability proof-explanation normal-distribution






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 20 at 9:13









      vonjdvonjd

      4,23574058




      4,23574058






















          1 Answer
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          1












          $begingroup$

          It is basically using the definition of expected value and one simple substitution. The first equation can be rewritten as
          $$
          mathbb{P}(max_i X_i < t) = Phileft(frac{t - mu}{sigma}right)^n.
          $$
          Then, by definition
          $$
          mathbb{E} max_i X_i = int_{-infty}^{+infty} t cdot frac{d}{dt} Phileft(frac{t - mu}{sigma}right)^n dt = int_{-infty}^{+infty} sigma(t + mu) cdot sigma frac{d}{dt} Phi(t)^n frac{dt}{sigma} = mu + sigma int_{-infty}^{+infty} tcdot frac{d}{dt} Phi(t)^n dt,
          $$
          as desired.



          So, in the first equation we used the substitution $t to sigma t + mu$, in the second one we simplified the expression and used that integral of the density function over the whole domain is $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, which definition are you using for the $d/dt$?
            $endgroup$
            – vonjd
            Jan 20 at 9:54










          • $begingroup$
            $frac{d}{dt}$ is the derivative operator, so $frac{d}{dt} Phi(t)^n$ becomes the density function. is it clear now?
            $endgroup$
            – pointguard0
            Jan 20 at 9:56










          • $begingroup$
            because the probability density function (which is used in the definition of expected value) is the derivative of cumulative distribution function
            $endgroup$
            – pointguard0
            Jan 20 at 10:15






          • 1




            $begingroup$
            Yes, got it - Thank you!
            $endgroup$
            – vonjd
            Jan 20 at 11:02











          Your Answer





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          1 Answer
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          1 Answer
          1






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          active

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          $begingroup$

          It is basically using the definition of expected value and one simple substitution. The first equation can be rewritten as
          $$
          mathbb{P}(max_i X_i < t) = Phileft(frac{t - mu}{sigma}right)^n.
          $$
          Then, by definition
          $$
          mathbb{E} max_i X_i = int_{-infty}^{+infty} t cdot frac{d}{dt} Phileft(frac{t - mu}{sigma}right)^n dt = int_{-infty}^{+infty} sigma(t + mu) cdot sigma frac{d}{dt} Phi(t)^n frac{dt}{sigma} = mu + sigma int_{-infty}^{+infty} tcdot frac{d}{dt} Phi(t)^n dt,
          $$
          as desired.



          So, in the first equation we used the substitution $t to sigma t + mu$, in the second one we simplified the expression and used that integral of the density function over the whole domain is $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, which definition are you using for the $d/dt$?
            $endgroup$
            – vonjd
            Jan 20 at 9:54










          • $begingroup$
            $frac{d}{dt}$ is the derivative operator, so $frac{d}{dt} Phi(t)^n$ becomes the density function. is it clear now?
            $endgroup$
            – pointguard0
            Jan 20 at 9:56










          • $begingroup$
            because the probability density function (which is used in the definition of expected value) is the derivative of cumulative distribution function
            $endgroup$
            – pointguard0
            Jan 20 at 10:15






          • 1




            $begingroup$
            Yes, got it - Thank you!
            $endgroup$
            – vonjd
            Jan 20 at 11:02
















          1












          $begingroup$

          It is basically using the definition of expected value and one simple substitution. The first equation can be rewritten as
          $$
          mathbb{P}(max_i X_i < t) = Phileft(frac{t - mu}{sigma}right)^n.
          $$
          Then, by definition
          $$
          mathbb{E} max_i X_i = int_{-infty}^{+infty} t cdot frac{d}{dt} Phileft(frac{t - mu}{sigma}right)^n dt = int_{-infty}^{+infty} sigma(t + mu) cdot sigma frac{d}{dt} Phi(t)^n frac{dt}{sigma} = mu + sigma int_{-infty}^{+infty} tcdot frac{d}{dt} Phi(t)^n dt,
          $$
          as desired.



          So, in the first equation we used the substitution $t to sigma t + mu$, in the second one we simplified the expression and used that integral of the density function over the whole domain is $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, which definition are you using for the $d/dt$?
            $endgroup$
            – vonjd
            Jan 20 at 9:54










          • $begingroup$
            $frac{d}{dt}$ is the derivative operator, so $frac{d}{dt} Phi(t)^n$ becomes the density function. is it clear now?
            $endgroup$
            – pointguard0
            Jan 20 at 9:56










          • $begingroup$
            because the probability density function (which is used in the definition of expected value) is the derivative of cumulative distribution function
            $endgroup$
            – pointguard0
            Jan 20 at 10:15






          • 1




            $begingroup$
            Yes, got it - Thank you!
            $endgroup$
            – vonjd
            Jan 20 at 11:02














          1












          1








          1





          $begingroup$

          It is basically using the definition of expected value and one simple substitution. The first equation can be rewritten as
          $$
          mathbb{P}(max_i X_i < t) = Phileft(frac{t - mu}{sigma}right)^n.
          $$
          Then, by definition
          $$
          mathbb{E} max_i X_i = int_{-infty}^{+infty} t cdot frac{d}{dt} Phileft(frac{t - mu}{sigma}right)^n dt = int_{-infty}^{+infty} sigma(t + mu) cdot sigma frac{d}{dt} Phi(t)^n frac{dt}{sigma} = mu + sigma int_{-infty}^{+infty} tcdot frac{d}{dt} Phi(t)^n dt,
          $$
          as desired.



          So, in the first equation we used the substitution $t to sigma t + mu$, in the second one we simplified the expression and used that integral of the density function over the whole domain is $1$.






          share|cite|improve this answer









          $endgroup$



          It is basically using the definition of expected value and one simple substitution. The first equation can be rewritten as
          $$
          mathbb{P}(max_i X_i < t) = Phileft(frac{t - mu}{sigma}right)^n.
          $$
          Then, by definition
          $$
          mathbb{E} max_i X_i = int_{-infty}^{+infty} t cdot frac{d}{dt} Phileft(frac{t - mu}{sigma}right)^n dt = int_{-infty}^{+infty} sigma(t + mu) cdot sigma frac{d}{dt} Phi(t)^n frac{dt}{sigma} = mu + sigma int_{-infty}^{+infty} tcdot frac{d}{dt} Phi(t)^n dt,
          $$
          as desired.



          So, in the first equation we used the substitution $t to sigma t + mu$, in the second one we simplified the expression and used that integral of the density function over the whole domain is $1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 20 at 9:27









          pointguard0pointguard0

          1,4801021




          1,4801021












          • $begingroup$
            Thank you, which definition are you using for the $d/dt$?
            $endgroup$
            – vonjd
            Jan 20 at 9:54










          • $begingroup$
            $frac{d}{dt}$ is the derivative operator, so $frac{d}{dt} Phi(t)^n$ becomes the density function. is it clear now?
            $endgroup$
            – pointguard0
            Jan 20 at 9:56










          • $begingroup$
            because the probability density function (which is used in the definition of expected value) is the derivative of cumulative distribution function
            $endgroup$
            – pointguard0
            Jan 20 at 10:15






          • 1




            $begingroup$
            Yes, got it - Thank you!
            $endgroup$
            – vonjd
            Jan 20 at 11:02


















          • $begingroup$
            Thank you, which definition are you using for the $d/dt$?
            $endgroup$
            – vonjd
            Jan 20 at 9:54










          • $begingroup$
            $frac{d}{dt}$ is the derivative operator, so $frac{d}{dt} Phi(t)^n$ becomes the density function. is it clear now?
            $endgroup$
            – pointguard0
            Jan 20 at 9:56










          • $begingroup$
            because the probability density function (which is used in the definition of expected value) is the derivative of cumulative distribution function
            $endgroup$
            – pointguard0
            Jan 20 at 10:15






          • 1




            $begingroup$
            Yes, got it - Thank you!
            $endgroup$
            – vonjd
            Jan 20 at 11:02
















          $begingroup$
          Thank you, which definition are you using for the $d/dt$?
          $endgroup$
          – vonjd
          Jan 20 at 9:54




          $begingroup$
          Thank you, which definition are you using for the $d/dt$?
          $endgroup$
          – vonjd
          Jan 20 at 9:54












          $begingroup$
          $frac{d}{dt}$ is the derivative operator, so $frac{d}{dt} Phi(t)^n$ becomes the density function. is it clear now?
          $endgroup$
          – pointguard0
          Jan 20 at 9:56




          $begingroup$
          $frac{d}{dt}$ is the derivative operator, so $frac{d}{dt} Phi(t)^n$ becomes the density function. is it clear now?
          $endgroup$
          – pointguard0
          Jan 20 at 9:56












          $begingroup$
          because the probability density function (which is used in the definition of expected value) is the derivative of cumulative distribution function
          $endgroup$
          – pointguard0
          Jan 20 at 10:15




          $begingroup$
          because the probability density function (which is used in the definition of expected value) is the derivative of cumulative distribution function
          $endgroup$
          – pointguard0
          Jan 20 at 10:15




          1




          1




          $begingroup$
          Yes, got it - Thank you!
          $endgroup$
          – vonjd
          Jan 20 at 11:02




          $begingroup$
          Yes, got it - Thank you!
          $endgroup$
          – vonjd
          Jan 20 at 11:02


















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