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Expectation of pricing algorithm
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I´d like to calculate the expectation of an algorithm. The input to the algorithm are n customers with values $v_1$,..$v_n$(i.i.d) drawn from a normal distribution. Now the algorithm sells an item with a fixed price p and offers this price to the customers sequentially. The first customer i with value $v_i > p$ gets the item.
What is the expectation of the value of the customers we sold to($v$) achieved by the algorithm?
I tried to solve it by calculating the expectation of the normal distribution from p as the lower bound to infinity, but it didn´t match the result I was getting when solving this empirically.
Many thanks in advance!
expectation normal-distribution
asked Jul 27 at 22:39
Dashingo
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up vote
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favorite
I´d like to calculate the expectation of an algorithm. The input to the algorithm are n customers with values $v_1$,..$v_n$(i.i.d) drawn from a normal distribution. Now the algorithm sells an item with a fixed price p and offers this price to the customers sequentially. The first customer i with value $v_i > p$ gets the item.
What is the expectation of the value of the customers we sold to($v$) achieved by the algorithm?
I tried to solve it by calculating the expectation of the normal distribution from p as the lower bound to infinity, but it didn´t match the result I was getting when solving this empirically.
Many thanks in advance!
expectation normal-distribution
asked Jul 27 at 22:39
Dashingo
83
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I´d like to calculate the expectation of an algorithm. The input to the algorithm are n customers with values $v_1$,..$v_n$(i.i.d) drawn from a normal distribution. Now the algorithm sells an item with a fixed price p and offers this price to the customers sequentially. The first customer i with value $v_i > p$ gets the item.
What is the expectation of the value of the customers we sold to($v$) achieved by the algorithm?
I tried to solve it by calculating the expectation of the normal distribution from p as the lower bound to infinity, but it didn´t match the result I was getting when solving this empirically.
Many thanks in advance!
expectation normal-distribution
asked Jul 27 at 22:39
Dashingo
83
I´d like to calculate the expectation of an algorithm. The input to the algorithm are n customers with values $v_1$,..$v_n$(i.i.d) drawn from a normal distribution. Now the algorithm sells an item with a fixed price p and offers this price to the customers sequentially. The first customer i with value $v_i > p$ gets the item.
What is the expectation of the value of the customers we sold to($v$) achieved by the algorithm?
I tried to solve it by calculating the expectation of the normal distribution from p as the lower bound to infinity, but it didn´t match the result I was getting when solving this empirically.
Many thanks in advance!
expectation normal-distribution
asked Jul 27 at 22:39
Dashingo
83
edited Jul 28 at 12:55
asked Jul 27 at 22:39
Dashingo
83
asked Jul 27 at 22:39
Dashingo
83
asked Jul 27 at 22:39
Dashingo
83
83
add a comment |Â
add a comment |Â
1 Answer
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Suppose $v_iin N(mu,sigma^2)$. Let $z_i=(v_i-mu)/sigma$, so the $z_i$ are standard normal, and let $q=(p-mu)/sigma$ be the standardized price. Let $Phi(x)$ be the standard normal cdf, with pdf $phi(x)$.
Conditioned on the existence at least one customer whose value is above the price, then the first customer whose value is larger than $p$ will have a value which is normally distributed, conditional on being greater than $p$. The conditional expected value is
$$
E[vmid v>p]=E[sigma z+mu|z>q]=mu+sigma E[z|z>q] =mu+sigmacdotfracint_q^infty x,phi(x),dxP(z>q)
$$
The normal cdf satisfies $xphi(x)=-phi'(x)$, so using the fundamental theorem of calculus this is
$$
mu+sigmafracphi(q)1-Phi(q)
$$
To make this unconditional, we have to multiply by the probability that at least one customer's value is above the price, resulting in
$$
boxedBig(1-Phi(q)^nBig)left(mu+sigmafracphi(q)1-Phi(q)right)
$$
answered Jul 28 at 4:12
Mike Earnest
14.9k11644
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Suppose $v_iin N(mu,sigma^2)$. Let $z_i=(v_i-mu)/sigma$, so the $z_i$ are standard normal, and let $q=(p-mu)/sigma$ be the standardized price. Let $Phi(x)$ be the standard normal cdf, with pdf $phi(x)$.
Conditioned on the existence at least one customer whose value is above the price, then the first customer whose value is larger than $p$ will have a value which is normally distributed, conditional on being greater than $p$. The conditional expected value is
$$
E[vmid v>p]=E[sigma z+mu|z>q]=mu+sigma E[z|z>q] =mu+sigmacdotfracint_q^infty x,phi(x),dxP(z>q)
$$
The normal cdf satisfies $xphi(x)=-phi'(x)$, so using the fundamental theorem of calculus this is
$$
mu+sigmafracphi(q)1-Phi(q)
$$
To make this unconditional, we have to multiply by the probability that at least one customer's value is above the price, resulting in
$$
boxedBig(1-Phi(q)^nBig)left(mu+sigmafracphi(q)1-Phi(q)right)
$$
answered Jul 28 at 4:12
Mike Earnest
14.9k11644
add a comment |Â
up vote
0
down vote
accepted
Suppose $v_iin N(mu,sigma^2)$. Let $z_i=(v_i-mu)/sigma$, so the $z_i$ are standard normal, and let $q=(p-mu)/sigma$ be the standardized price. Let $Phi(x)$ be the standard normal cdf, with pdf $phi(x)$.
Conditioned on the existence at least one customer whose value is above the price, then the first customer whose value is larger than $p$ will have a value which is normally distributed, conditional on being greater than $p$. The conditional expected value is
$$
E[vmid v>p]=E[sigma z+mu|z>q]=mu+sigma E[z|z>q] =mu+sigmacdotfracint_q^infty x,phi(x),dxP(z>q)
$$
The normal cdf satisfies $xphi(x)=-phi'(x)$, so using the fundamental theorem of calculus this is
$$
mu+sigmafracphi(q)1-Phi(q)
$$
To make this unconditional, we have to multiply by the probability that at least one customer's value is above the price, resulting in
$$
boxedBig(1-Phi(q)^nBig)left(mu+sigmafracphi(q)1-Phi(q)right)
$$
answered Jul 28 at 4:12
Mike Earnest
14.9k11644
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Suppose $v_iin N(mu,sigma^2)$. Let $z_i=(v_i-mu)/sigma$, so the $z_i$ are standard normal, and let $q=(p-mu)/sigma$ be the standardized price. Let $Phi(x)$ be the standard normal cdf, with pdf $phi(x)$.
Conditioned on the existence at least one customer whose value is above the price, then the first customer whose value is larger than $p$ will have a value which is normally distributed, conditional on being greater than $p$. The conditional expected value is
$$
E[vmid v>p]=E[sigma z+mu|z>q]=mu+sigma E[z|z>q] =mu+sigmacdotfracint_q^infty x,phi(x),dxP(z>q)
$$
The normal cdf satisfies $xphi(x)=-phi'(x)$, so using the fundamental theorem of calculus this is
$$
mu+sigmafracphi(q)1-Phi(q)
$$
To make this unconditional, we have to multiply by the probability that at least one customer's value is above the price, resulting in
$$
boxedBig(1-Phi(q)^nBig)left(mu+sigmafracphi(q)1-Phi(q)right)
$$
answered Jul 28 at 4:12
Mike Earnest
14.9k11644
Suppose $v_iin N(mu,sigma^2)$. Let $z_i=(v_i-mu)/sigma$, so the $z_i$ are standard normal, and let $q=(p-mu)/sigma$ be the standardized price. Let $Phi(x)$ be the standard normal cdf, with pdf $phi(x)$.
Conditioned on the existence at least one customer whose value is above the price, then the first customer whose value is larger than $p$ will have a value which is normally distributed, conditional on being greater than $p$. The conditional expected value is
$$
E[vmid v>p]=E[sigma z+mu|z>q]=mu+sigma E[z|z>q] =mu+sigmacdotfracint_q^infty x,phi(x),dxP(z>q)
$$
The normal cdf satisfies $xphi(x)=-phi'(x)$, so using the fundamental theorem of calculus this is
$$
mu+sigmafracphi(q)1-Phi(q)
$$
To make this unconditional, we have to multiply by the probability that at least one customer's value is above the price, resulting in
$$
boxedBig(1-Phi(q)^nBig)left(mu+sigmafracphi(q)1-Phi(q)right)
$$
answered Jul 28 at 4:12
Mike Earnest
14.9k11644
answered Jul 28 at 4:12
Mike Earnest
14.9k11644
answered Jul 28 at 4:12
Mike Earnest
14.9k11644
answered Jul 28 at 4:12
Mike Earnest
14.9k11644
14.9k11644
add a comment |Â
add a comment |Â
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