Mean-Variance Criterion Derivation
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Can you help me understand this derivation? This is finance related so $E[X]$ means the expected outcome (of a gamble) and $u(x)$ is the utility function.
With any well-behaved utility function, when the range of possible
outcomes of a gamble is small relative to the risk tolerance of the
decision maker, the certainty equivalent of the gamble can be
approximated by making use of the local curvature of the utility
function.
Consider a gamble $X$ with expected value $E[X]$ and variance $Var(X)$, and
a utility curve $u(x)$. From the first two terms of a Taylor series
expansion of $u(x)$ around the expected value $E[X]$ we obtain:
$E[u(x)]=u(E[X])+1/2 u^{''}(E[X])Var(X)$ i'm good here
We equate this expected utility to the utility of the certainty
equivalent $CE=E[X]+D$, where $D$ is a risk discount.
Again, using a Taylor series expansion, $u(E[X]+D)$ can be approximated
by $u(E[X])+u''(E[X])D$. how did we get to this approximation?
Hence the first order approximation for the risk discount is
$D = 1/2(u''/u') Var(X)$. how did we get this result?
Therefore:
$CE(X)=E[X]+1/2 (u''/u')Var(X)$ how did we get to this conclusion?
The maximum EU (or maximum CE) principle reduces to a linear function
of mean and variance in which the relative weight given the variance
is half the local curvature of the utility function.
Thank you.
taylor-expansion finance utility
asked Dec 14 '18 at 19:46
dreamerboydreamerboy
12
$endgroup$
add a comment |
$begingroup$
Can you help me understand this derivation? This is finance related so $E[X]$ means the expected outcome (of a gamble) and $u(x)$ is the utility function.
With any well-behaved utility function, when the range of possible
outcomes of a gamble is small relative to the risk tolerance of the
decision maker, the certainty equivalent of the gamble can be
approximated by making use of the local curvature of the utility
function.
Consider a gamble $X$ with expected value $E[X]$ and variance $Var(X)$, and
a utility curve $u(x)$. From the first two terms of a Taylor series
expansion of $u(x)$ around the expected value $E[X]$ we obtain:
$E[u(x)]=u(E[X])+1/2 u^{''}(E[X])Var(X)$ i'm good here
We equate this expected utility to the utility of the certainty
equivalent $CE=E[X]+D$, where $D$ is a risk discount.
Again, using a Taylor series expansion, $u(E[X]+D)$ can be approximated
by $u(E[X])+u''(E[X])D$. how did we get to this approximation?
Hence the first order approximation for the risk discount is
$D = 1/2(u''/u') Var(X)$. how did we get this result?
Therefore:
$CE(X)=E[X]+1/2 (u''/u')Var(X)$ how did we get to this conclusion?
The maximum EU (or maximum CE) principle reduces to a linear function
of mean and variance in which the relative weight given the variance
is half the local curvature of the utility function.
Thank you.
taylor-expansion finance utility
asked Dec 14 '18 at 19:46
dreamerboydreamerboy
12
$endgroup$
$begingroup$
It's all Taylor series. Look at how to expand $f(x)$ and $f(x+h).$
$endgroup$
– Sean Roberson
Dec 14 '18 at 20:14
$begingroup$
I think the second approximation should have been $u(E[X]+D) = u(E[X]) + u'(E[X]) D.$
$endgroup$
– angryavian
Dec 14 '18 at 20:35
$begingroup$
I found a more intuitive derivation here: quant.stackexchange.com/questions/20628/…
$endgroup$
– dreamerboy
Dec 14 '18 at 21:21
add a comment |
$begingroup$
Can you help me understand this derivation? This is finance related so $E[X]$ means the expected outcome (of a gamble) and $u(x)$ is the utility function.
With any well-behaved utility function, when the range of possible
outcomes of a gamble is small relative to the risk tolerance of the
decision maker, the certainty equivalent of the gamble can be
approximated by making use of the local curvature of the utility
function.
Consider a gamble $X$ with expected value $E[X]$ and variance $Var(X)$, and
a utility curve $u(x)$. From the first two terms of a Taylor series
expansion of $u(x)$ around the expected value $E[X]$ we obtain:
$E[u(x)]=u(E[X])+1/2 u^{''}(E[X])Var(X)$ i'm good here
We equate this expected utility to the utility of the certainty
equivalent $CE=E[X]+D$, where $D$ is a risk discount.
Again, using a Taylor series expansion, $u(E[X]+D)$ can be approximated
by $u(E[X])+u''(E[X])D$. how did we get to this approximation?
Hence the first order approximation for the risk discount is
$D = 1/2(u''/u') Var(X)$. how did we get this result?
Therefore:
$CE(X)=E[X]+1/2 (u''/u')Var(X)$ how did we get to this conclusion?
The maximum EU (or maximum CE) principle reduces to a linear function
of mean and variance in which the relative weight given the variance
is half the local curvature of the utility function.
Thank you.
taylor-expansion finance utility
asked Dec 14 '18 at 19:46
dreamerboydreamerboy
12
$endgroup$
Can you help me understand this derivation? This is finance related so $E[X]$ means the expected outcome (of a gamble) and $u(x)$ is the utility function.
With any well-behaved utility function, when the range of possible
outcomes of a gamble is small relative to the risk tolerance of the
decision maker, the certainty equivalent of the gamble can be
approximated by making use of the local curvature of the utility
function.
Consider a gamble $X$ with expected value $E[X]$ and variance $Var(X)$, and
a utility curve $u(x)$. From the first two terms of a Taylor series
expansion of $u(x)$ around the expected value $E[X]$ we obtain:
$E[u(x)]=u(E[X])+1/2 u^{''}(E[X])Var(X)$ i'm good here
We equate this expected utility to the utility of the certainty
equivalent $CE=E[X]+D$, where $D$ is a risk discount.
Again, using a Taylor series expansion, $u(E[X]+D)$ can be approximated
by $u(E[X])+u''(E[X])D$. how did we get to this approximation?
Hence the first order approximation for the risk discount is
$D = 1/2(u''/u') Var(X)$. how did we get this result?
Therefore:
$CE(X)=E[X]+1/2 (u''/u')Var(X)$ how did we get to this conclusion?
The maximum EU (or maximum CE) principle reduces to a linear function
of mean and variance in which the relative weight given the variance
is half the local curvature of the utility function.
Thank you.
taylor-expansion finance utility
taylor-expansion finance utility
asked Dec 14 '18 at 19:46
dreamerboydreamerboy
12
asked Dec 14 '18 at 19:46
dreamerboydreamerboy
12
edited Dec 14 '18 at 21:01
dreamerboy
asked Dec 14 '18 at 19:46
dreamerboydreamerboy
12
asked Dec 14 '18 at 19:46
dreamerboydreamerboy
12
asked Dec 14 '18 at 19:46
dreamerboydreamerboy
12
12
$begingroup$
It's all Taylor series. Look at how to expand $f(x)$ and $f(x+h).$
$endgroup$
– Sean Roberson
Dec 14 '18 at 20:14
$begingroup$
I think the second approximation should have been $u(E[X]+D) = u(E[X]) + u'(E[X]) D.$
$endgroup$
– angryavian
Dec 14 '18 at 20:35
$begingroup$
I found a more intuitive derivation here: quant.stackexchange.com/questions/20628/…
$endgroup$
– dreamerboy
Dec 14 '18 at 21:21
add a comment |
$begingroup$
It's all Taylor series. Look at how to expand $f(x)$ and $f(x+h).$
$endgroup$
– Sean Roberson
Dec 14 '18 at 20:14
$begingroup$
I think the second approximation should have been $u(E[X]+D) = u(E[X]) + u'(E[X]) D.$
$endgroup$
– angryavian
Dec 14 '18 at 20:35
$begingroup$
I found a more intuitive derivation here: quant.stackexchange.com/questions/20628/…
$endgroup$
– dreamerboy
Dec 14 '18 at 21:21
$begingroup$
It's all Taylor series. Look at how to expand $f(x)$ and $f(x+h).$
$endgroup$
– Sean Roberson
Dec 14 '18 at 20:14
$begingroup$
It's all Taylor series. Look at how to expand $f(x)$ and $f(x+h).$
$endgroup$
– Sean Roberson
Dec 14 '18 at 20:14
$begingroup$
I think the second approximation should have been $u(E[X]+D) = u(E[X]) + u'(E[X]) D.$
$endgroup$
– angryavian
Dec 14 '18 at 20:35
$begingroup$
I think the second approximation should have been $u(E[X]+D) = u(E[X]) + u'(E[X]) D.$
$endgroup$
– angryavian
Dec 14 '18 at 20:35
$begingroup$
I found a more intuitive derivation here: quant.stackexchange.com/questions/20628/…
$endgroup$
– dreamerboy
Dec 14 '18 at 21:21
$begingroup$
I found a more intuitive derivation here: quant.stackexchange.com/questions/20628/…
$endgroup$
– dreamerboy
Dec 14 '18 at 21:21
add a comment |
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$begingroup$
It's all Taylor series. Look at how to expand $f(x)$ and $f(x+h).$
$endgroup$
– Sean Roberson
Dec 14 '18 at 20:14
$begingroup$
I think the second approximation should have been $u(E[X]+D) = u(E[X]) + u'(E[X]) D.$
$endgroup$
– angryavian
Dec 14 '18 at 20:35
$begingroup$
I found a more intuitive derivation here: quant.stackexchange.com/questions/20628/…
$endgroup$
– dreamerboy
Dec 14 '18 at 21:21