How to derive the formula for the expected value for maximum of n normal random variables
$begingroup$
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
asked Jan 20 at 9:13
vonjdvonjd
4,23574058
$endgroup$
add a comment |
$begingroup$
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
asked Jan 20 at 9:13
vonjdvonjd
4,23574058
$endgroup$
add a comment |
$begingroup$
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
asked Jan 20 at 9:13
vonjdvonjd
4,23574058
$endgroup$
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
asked Jan 20 at 9:13
vonjdvonjd
4,23574058
asked Jan 20 at 9:13
vonjdvonjd
4,23574058
asked Jan 20 at 9:13
vonjdvonjd
4,23574058
asked Jan 20 at 9:13
vonjdvonjd
4,23574058
asked Jan 20 at 9:13
vonjdvonjd
4,23574058
4,23574058
add a comment |
add a comment |
1 Answer
1
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oldest
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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 $1$.
answered Jan 20 at 9:27
pointguard0pointguard0
1,4801021
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
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 $1$.
answered Jan 20 at 9:27
pointguard0pointguard0
1,4801021
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $1$.
answered Jan 20 at 9:27
pointguard0pointguard0
1,4801021
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $1$.
answered Jan 20 at 9:27
pointguard0pointguard0
1,4801021
$endgroup$
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 $1$.
answered Jan 20 at 9:27
pointguard0pointguard0
1,4801021
answered Jan 20 at 9:27
pointguard0pointguard0
1,4801021
answered Jan 20 at 9:27
pointguard0pointguard0
1,4801021
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
add a comment |
$begingroup$
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
add a comment |
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