Mixing
Problem with deriving weights of a market portfolio in a mean variance framework
$begingroup$
I paste a part of a paper that formulate the MV investment problem.
I don't understand how equation 5 has been derived and, in particular, how the expected return on the zero beta portfolio comes up.
Many thanks in advance for a reply.
finance
asked Jan 5 at 0:16
user279687user279687
112
$endgroup$
add a comment |
$begingroup$
I paste a part of a paper that formulate the MV investment problem.
I don't understand how equation 5 has been derived and, in particular, how the expected return on the zero beta portfolio comes up.
Many thanks in advance for a reply.
finance
asked Jan 5 at 0:16
user279687user279687
112
$endgroup$
add a comment |
$begingroup$
I paste a part of a paper that formulate the MV investment problem.
I don't understand how equation 5 has been derived and, in particular, how the expected return on the zero beta portfolio comes up.
Many thanks in advance for a reply.
finance
asked Jan 5 at 0:16
user279687user279687
112
$endgroup$
I paste a part of a paper that formulate the MV investment problem.
I don't understand how equation 5 has been derived and, in particular, how the expected return on the zero beta portfolio comes up.
Many thanks in advance for a reply.
finance
finance
asked Jan 5 at 0:16
user279687user279687
112
asked Jan 5 at 0:16
user279687user279687
112
asked Jan 5 at 0:16
user279687user279687
112
asked Jan 5 at 0:16
user279687user279687
112
asked Jan 5 at 0:16
user279687user279687
112
112
add a comment |
add a comment |
1 Answer
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$begingroup$
From (3)
$$
pmb Sigmapmb x+lambdapmb iota=spmb muquadLongrightarrowquad pmb x=spmb Sigma^{-1}pmb mu-lambdapmb Sigma^{-1}pmb iotatag a
$$
Substituting in (4)
$$
pmb iota'pmb x=1 quadLongrightarrowquadpmb iota' pmb x=sunderbrace{pmb iota'pmb Sigma^{-1}pmb mu}_a-lambdaunderbrace{pmb iota'pmb Sigma^{-1}pmb iota}_c=sa-lambda c=1tag b
$$
From (b) we have
$$
sa-lambda c=sleft(a-frac{lambda}{s}cright)=sleft(a-bar r_zcright)=1quadLongrightarrowquad s=frac{1}{a-bar r_zc}tag c
$$
where $frac{lambda}{s}=bar r_z$. And then substituting (c) in (a) we have
$$
pmb x_m=sleft(pmb Sigma^{-1}pmb mu-frac{lambda}{s}pmb Sigma^{-1}pmb iotaright)=frac{1}{a-bar r_zc}left(pmb Sigma^{-1}pmb mu-bar r_zpmb Sigma^{-1}pmb iotaright)tag 5
$$
answered Jan 5 at 22:34
alexjoalexjo
12.3k1430
$endgroup$
$begingroup$
First of all many thanks for your reply. Would you be so kind to give me an hint to derive also equation 6?
$endgroup$
– user279687
Jan 9 at 23:15
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
From (3)
$$
pmb Sigmapmb x+lambdapmb iota=spmb muquadLongrightarrowquad pmb x=spmb Sigma^{-1}pmb mu-lambdapmb Sigma^{-1}pmb iotatag a
$$
Substituting in (4)
$$
pmb iota'pmb x=1 quadLongrightarrowquadpmb iota' pmb x=sunderbrace{pmb iota'pmb Sigma^{-1}pmb mu}_a-lambdaunderbrace{pmb iota'pmb Sigma^{-1}pmb iota}_c=sa-lambda c=1tag b
$$
From (b) we have
$$
sa-lambda c=sleft(a-frac{lambda}{s}cright)=sleft(a-bar r_zcright)=1quadLongrightarrowquad s=frac{1}{a-bar r_zc}tag c
$$
where $frac{lambda}{s}=bar r_z$. And then substituting (c) in (a) we have
$$
pmb x_m=sleft(pmb Sigma^{-1}pmb mu-frac{lambda}{s}pmb Sigma^{-1}pmb iotaright)=frac{1}{a-bar r_zc}left(pmb Sigma^{-1}pmb mu-bar r_zpmb Sigma^{-1}pmb iotaright)tag 5
$$
answered Jan 5 at 22:34
alexjoalexjo
12.3k1430
$endgroup$
$begingroup$
First of all many thanks for your reply. Would you be so kind to give me an hint to derive also equation 6?
$endgroup$
– user279687
Jan 9 at 23:15
add a comment |
$begingroup$
From (3)
$$
pmb Sigmapmb x+lambdapmb iota=spmb muquadLongrightarrowquad pmb x=spmb Sigma^{-1}pmb mu-lambdapmb Sigma^{-1}pmb iotatag a
$$
Substituting in (4)
$$
pmb iota'pmb x=1 quadLongrightarrowquadpmb iota' pmb x=sunderbrace{pmb iota'pmb Sigma^{-1}pmb mu}_a-lambdaunderbrace{pmb iota'pmb Sigma^{-1}pmb iota}_c=sa-lambda c=1tag b
$$
From (b) we have
$$
sa-lambda c=sleft(a-frac{lambda}{s}cright)=sleft(a-bar r_zcright)=1quadLongrightarrowquad s=frac{1}{a-bar r_zc}tag c
$$
where $frac{lambda}{s}=bar r_z$. And then substituting (c) in (a) we have
$$
pmb x_m=sleft(pmb Sigma^{-1}pmb mu-frac{lambda}{s}pmb Sigma^{-1}pmb iotaright)=frac{1}{a-bar r_zc}left(pmb Sigma^{-1}pmb mu-bar r_zpmb Sigma^{-1}pmb iotaright)tag 5
$$
answered Jan 5 at 22:34
alexjoalexjo
12.3k1430
$endgroup$
$begingroup$
First of all many thanks for your reply. Would you be so kind to give me an hint to derive also equation 6?
$endgroup$
– user279687
Jan 9 at 23:15
add a comment |
$begingroup$
From (3)
$$
pmb Sigmapmb x+lambdapmb iota=spmb muquadLongrightarrowquad pmb x=spmb Sigma^{-1}pmb mu-lambdapmb Sigma^{-1}pmb iotatag a
$$
Substituting in (4)
$$
pmb iota'pmb x=1 quadLongrightarrowquadpmb iota' pmb x=sunderbrace{pmb iota'pmb Sigma^{-1}pmb mu}_a-lambdaunderbrace{pmb iota'pmb Sigma^{-1}pmb iota}_c=sa-lambda c=1tag b
$$
From (b) we have
$$
sa-lambda c=sleft(a-frac{lambda}{s}cright)=sleft(a-bar r_zcright)=1quadLongrightarrowquad s=frac{1}{a-bar r_zc}tag c
$$
where $frac{lambda}{s}=bar r_z$. And then substituting (c) in (a) we have
$$
pmb x_m=sleft(pmb Sigma^{-1}pmb mu-frac{lambda}{s}pmb Sigma^{-1}pmb iotaright)=frac{1}{a-bar r_zc}left(pmb Sigma^{-1}pmb mu-bar r_zpmb Sigma^{-1}pmb iotaright)tag 5
$$
answered Jan 5 at 22:34
alexjoalexjo
12.3k1430
$endgroup$
From (3)
$$
pmb Sigmapmb x+lambdapmb iota=spmb muquadLongrightarrowquad pmb x=spmb Sigma^{-1}pmb mu-lambdapmb Sigma^{-1}pmb iotatag a
$$
Substituting in (4)
$$
pmb iota'pmb x=1 quadLongrightarrowquadpmb iota' pmb x=sunderbrace{pmb iota'pmb Sigma^{-1}pmb mu}_a-lambdaunderbrace{pmb iota'pmb Sigma^{-1}pmb iota}_c=sa-lambda c=1tag b
$$
From (b) we have
$$
sa-lambda c=sleft(a-frac{lambda}{s}cright)=sleft(a-bar r_zcright)=1quadLongrightarrowquad s=frac{1}{a-bar r_zc}tag c
$$
where $frac{lambda}{s}=bar r_z$. And then substituting (c) in (a) we have
$$
pmb x_m=sleft(pmb Sigma^{-1}pmb mu-frac{lambda}{s}pmb Sigma^{-1}pmb iotaright)=frac{1}{a-bar r_zc}left(pmb Sigma^{-1}pmb mu-bar r_zpmb Sigma^{-1}pmb iotaright)tag 5
$$
answered Jan 5 at 22:34
alexjoalexjo
12.3k1430
answered Jan 5 at 22:34
alexjoalexjo
12.3k1430
answered Jan 5 at 22:34
alexjoalexjo
12.3k1430
answered Jan 5 at 22:34
alexjoalexjo
12.3k1430
12.3k1430
$begingroup$
First of all many thanks for your reply. Would you be so kind to give me an hint to derive also equation 6?
$endgroup$
– user279687
Jan 9 at 23:15
add a comment |
$begingroup$
First of all many thanks for your reply. Would you be so kind to give me an hint to derive also equation 6?
$endgroup$
– user279687
Jan 9 at 23:15
$begingroup$
First of all many thanks for your reply. Would you be so kind to give me an hint to derive also equation 6?
$endgroup$
– user279687
Jan 9 at 23:15
$begingroup$
First of all many thanks for your reply. Would you be so kind to give me an hint to derive also equation 6?
$endgroup$
– user279687
Jan 9 at 23:15
add a comment |
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