Mixing
Closed form solution for the following quadratic loss problem?
$begingroup$
I'm trying to find the values of $mu$ and $sigma$ that minimize the following quadratic loss function:
Note that $muin{(-infty, infty}), sigma>0$
$$ f(mu, sigma) = frac{1}{2}big(e^{mu + Asigma} - r_Abig)^2 +frac{1}{2}big(e^{mu + Bsigma} - r_Bbig)^2 $$
Taking the first derivative of both and setting them equal to zero, I get:
$frac{partial f(mu, sigma)}{mu} = e^{mu+Asigma} big(e^{mu + Asigma} - r_Abig) + e^{mu+Bsigma} big(e^{mu + Bsigma} - r_Bbig) = 0$
$frac{partial f(mu, sigma)}{sigma} = Ae^{mu + Asigma} big(e^{mu + Asigma} - r_Abig) + Be^{mu + Bsigma} big(e^{mu + Bsigma} - r_Bbig) = 0$
However I'm getting stuck here on how to combine these two to (try) to get a closed form expression for $mu$ and $sigma$
Any help would be greatly appreciated, or any insight into a closed form solution for the above is even possible
calculus multivariable-calculus optimization
asked Jan 11 at 15:15
measure_theorymeasure_theory
626313
$endgroup$
add a comment |
$begingroup$
I'm trying to find the values of $mu$ and $sigma$ that minimize the following quadratic loss function:
Note that $muin{(-infty, infty}), sigma>0$
$$ f(mu, sigma) = frac{1}{2}big(e^{mu + Asigma} - r_Abig)^2 +frac{1}{2}big(e^{mu + Bsigma} - r_Bbig)^2 $$
Taking the first derivative of both and setting them equal to zero, I get:
$frac{partial f(mu, sigma)}{mu} = e^{mu+Asigma} big(e^{mu + Asigma} - r_Abig) + e^{mu+Bsigma} big(e^{mu + Bsigma} - r_Bbig) = 0$
$frac{partial f(mu, sigma)}{sigma} = Ae^{mu + Asigma} big(e^{mu + Asigma} - r_Abig) + Be^{mu + Bsigma} big(e^{mu + Bsigma} - r_Bbig) = 0$
However I'm getting stuck here on how to combine these two to (try) to get a closed form expression for $mu$ and $sigma$
Any help would be greatly appreciated, or any insight into a closed form solution for the above is even possible
calculus multivariable-calculus optimization
asked Jan 11 at 15:15
measure_theorymeasure_theory
626313
$endgroup$
$begingroup$
You know $de^{mu + Asigma}/dmu = e^{mu + Asigma}$, right?
$endgroup$
– LinAlg
Jan 14 at 15:11
$begingroup$
Right. Edited..
$endgroup$
– measure_theory
Jan 14 at 15:19
$begingroup$
Why did you add a bounty, but then disappear from the website without accepting or even upvoting my answer?
$endgroup$
– LinAlg
Jan 22 at 18:51
add a comment |
$begingroup$
I'm trying to find the values of $mu$ and $sigma$ that minimize the following quadratic loss function:
Note that $muin{(-infty, infty}), sigma>0$
$$ f(mu, sigma) = frac{1}{2}big(e^{mu + Asigma} - r_Abig)^2 +frac{1}{2}big(e^{mu + Bsigma} - r_Bbig)^2 $$
Taking the first derivative of both and setting them equal to zero, I get:
$frac{partial f(mu, sigma)}{mu} = e^{mu+Asigma} big(e^{mu + Asigma} - r_Abig) + e^{mu+Bsigma} big(e^{mu + Bsigma} - r_Bbig) = 0$
$frac{partial f(mu, sigma)}{sigma} = Ae^{mu + Asigma} big(e^{mu + Asigma} - r_Abig) + Be^{mu + Bsigma} big(e^{mu + Bsigma} - r_Bbig) = 0$
However I'm getting stuck here on how to combine these two to (try) to get a closed form expression for $mu$ and $sigma$
Any help would be greatly appreciated, or any insight into a closed form solution for the above is even possible
calculus multivariable-calculus optimization
asked Jan 11 at 15:15
measure_theorymeasure_theory
626313
$endgroup$
I'm trying to find the values of $mu$ and $sigma$ that minimize the following quadratic loss function:
Note that $muin{(-infty, infty}), sigma>0$
$$ f(mu, sigma) = frac{1}{2}big(e^{mu + Asigma} - r_Abig)^2 +frac{1}{2}big(e^{mu + Bsigma} - r_Bbig)^2 $$
Taking the first derivative of both and setting them equal to zero, I get:
$frac{partial f(mu, sigma)}{mu} = e^{mu+Asigma} big(e^{mu + Asigma} - r_Abig) + e^{mu+Bsigma} big(e^{mu + Bsigma} - r_Bbig) = 0$
$frac{partial f(mu, sigma)}{sigma} = Ae^{mu + Asigma} big(e^{mu + Asigma} - r_Abig) + Be^{mu + Bsigma} big(e^{mu + Bsigma} - r_Bbig) = 0$
However I'm getting stuck here on how to combine these two to (try) to get a closed form expression for $mu$ and $sigma$
Any help would be greatly appreciated, or any insight into a closed form solution for the above is even possible
calculus multivariable-calculus optimization
calculus multivariable-calculus optimization
asked Jan 11 at 15:15
measure_theorymeasure_theory
626313
asked Jan 11 at 15:15
measure_theorymeasure_theory
626313
edited Jan 14 at 15:18
measure_theory
asked Jan 11 at 15:15
measure_theorymeasure_theory
626313
asked Jan 11 at 15:15
measure_theorymeasure_theory
626313
asked Jan 11 at 15:15
measure_theorymeasure_theory
626313
626313
$begingroup$
You know $de^{mu + Asigma}/dmu = e^{mu + Asigma}$, right?
$endgroup$
– LinAlg
Jan 14 at 15:11
$begingroup$
Right. Edited..
$endgroup$
– measure_theory
Jan 14 at 15:19
$begingroup$
Why did you add a bounty, but then disappear from the website without accepting or even upvoting my answer?
$endgroup$
– LinAlg
Jan 22 at 18:51
add a comment |
$begingroup$
You know $de^{mu + Asigma}/dmu = e^{mu + Asigma}$, right?
$endgroup$
– LinAlg
Jan 14 at 15:11
$begingroup$
Right. Edited..
$endgroup$
– measure_theory
Jan 14 at 15:19
$begingroup$
Why did you add a bounty, but then disappear from the website without accepting or even upvoting my answer?
$endgroup$
– LinAlg
Jan 22 at 18:51
$begingroup$
You know $de^{mu + Asigma}/dmu = e^{mu + Asigma}$, right?
$endgroup$
– LinAlg
Jan 14 at 15:11
$begingroup$
You know $de^{mu + Asigma}/dmu = e^{mu + Asigma}$, right?
$endgroup$
– LinAlg
Jan 14 at 15:11
$begingroup$
Right. Edited..
$endgroup$
– measure_theory
Jan 14 at 15:19
$begingroup$
Right. Edited..
$endgroup$
– measure_theory
Jan 14 at 15:19
$begingroup$
Why did you add a bounty, but then disappear from the website without accepting or even upvoting my answer?
$endgroup$
– LinAlg
Jan 22 at 18:51
$begingroup$
Why did you add a bounty, but then disappear from the website without accepting or even upvoting my answer?
$endgroup$
– LinAlg
Jan 22 at 18:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $x=e^{mu+Asigma}$ and $y=e^{mu+Bsigma}$, then you have
$$x^2 - r_A x + y^2 - r_B y = 0 text{, and } A (x^2 - r_A x) + B(y^2 - r_B y) = 0.$$
Let us assume that $A neq B$, as otherwise there are infinitely many solutions (which are easy to identify). I will also assume that $r_A$ and $r_B$ are positive. Multiplying the first equation with $A$ and substracting the second equation, you get $(A-B)y^2 - (A-B)r_B y = 0$. Since $Aneq B$, the solutions are $y=0$ and $y=r_B$. The first solution ($y=0$) cannot solve $y=e^{mu+Bsigma}$, so we have $y=r_B$. Similarly, multiplying the first equality with $B$ and subtracting the second equation, you get $x=r_A$.
The remaining task is solving $e^{mu+Asigma} = r_a$ and $y=e^{mu+Bsigma} = r_b$. Taking the natural logarithm you get a linear system: $mu+Asigma=log(r_a)$ and $mu+Bsigma=log(r_b)$. Subtracting the second equation from the first one yields:
$$sigma=frac{log(r_a)-log(r_b)}{A-B}$$
and then you can simply compute $mu=log(r_a)-Asigma$.
answered Jan 14 at 15:35
LinAlgLinAlg
9,4511521
$endgroup$
$begingroup$
Note that $sigma$ could be nonpositive with these formulas. If that is the case, there is an infimum that is not attained ($sigmadownarrow 0$).
$endgroup$
– LinAlg
Jan 14 at 16:14
add a comment |
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1 Answer
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$begingroup$
Let $x=e^{mu+Asigma}$ and $y=e^{mu+Bsigma}$, then you have
$$x^2 - r_A x + y^2 - r_B y = 0 text{, and } A (x^2 - r_A x) + B(y^2 - r_B y) = 0.$$
Let us assume that $A neq B$, as otherwise there are infinitely many solutions (which are easy to identify). I will also assume that $r_A$ and $r_B$ are positive. Multiplying the first equation with $A$ and substracting the second equation, you get $(A-B)y^2 - (A-B)r_B y = 0$. Since $Aneq B$, the solutions are $y=0$ and $y=r_B$. The first solution ($y=0$) cannot solve $y=e^{mu+Bsigma}$, so we have $y=r_B$. Similarly, multiplying the first equality with $B$ and subtracting the second equation, you get $x=r_A$.
The remaining task is solving $e^{mu+Asigma} = r_a$ and $y=e^{mu+Bsigma} = r_b$. Taking the natural logarithm you get a linear system: $mu+Asigma=log(r_a)$ and $mu+Bsigma=log(r_b)$. Subtracting the second equation from the first one yields:
$$sigma=frac{log(r_a)-log(r_b)}{A-B}$$
and then you can simply compute $mu=log(r_a)-Asigma$.
answered Jan 14 at 15:35
LinAlgLinAlg
9,4511521
$endgroup$
$begingroup$
Note that $sigma$ could be nonpositive with these formulas. If that is the case, there is an infimum that is not attained ($sigmadownarrow 0$).
$endgroup$
– LinAlg
Jan 14 at 16:14
add a comment |
$begingroup$
Let $x=e^{mu+Asigma}$ and $y=e^{mu+Bsigma}$, then you have
$$x^2 - r_A x + y^2 - r_B y = 0 text{, and } A (x^2 - r_A x) + B(y^2 - r_B y) = 0.$$
Let us assume that $A neq B$, as otherwise there are infinitely many solutions (which are easy to identify). I will also assume that $r_A$ and $r_B$ are positive. Multiplying the first equation with $A$ and substracting the second equation, you get $(A-B)y^2 - (A-B)r_B y = 0$. Since $Aneq B$, the solutions are $y=0$ and $y=r_B$. The first solution ($y=0$) cannot solve $y=e^{mu+Bsigma}$, so we have $y=r_B$. Similarly, multiplying the first equality with $B$ and subtracting the second equation, you get $x=r_A$.
The remaining task is solving $e^{mu+Asigma} = r_a$ and $y=e^{mu+Bsigma} = r_b$. Taking the natural logarithm you get a linear system: $mu+Asigma=log(r_a)$ and $mu+Bsigma=log(r_b)$. Subtracting the second equation from the first one yields:
$$sigma=frac{log(r_a)-log(r_b)}{A-B}$$
and then you can simply compute $mu=log(r_a)-Asigma$.
answered Jan 14 at 15:35
LinAlgLinAlg
9,4511521
$endgroup$
$begingroup$
Note that $sigma$ could be nonpositive with these formulas. If that is the case, there is an infimum that is not attained ($sigmadownarrow 0$).
$endgroup$
– LinAlg
Jan 14 at 16:14
add a comment |
$begingroup$
Let $x=e^{mu+Asigma}$ and $y=e^{mu+Bsigma}$, then you have
$$x^2 - r_A x + y^2 - r_B y = 0 text{, and } A (x^2 - r_A x) + B(y^2 - r_B y) = 0.$$
Let us assume that $A neq B$, as otherwise there are infinitely many solutions (which are easy to identify). I will also assume that $r_A$ and $r_B$ are positive. Multiplying the first equation with $A$ and substracting the second equation, you get $(A-B)y^2 - (A-B)r_B y = 0$. Since $Aneq B$, the solutions are $y=0$ and $y=r_B$. The first solution ($y=0$) cannot solve $y=e^{mu+Bsigma}$, so we have $y=r_B$. Similarly, multiplying the first equality with $B$ and subtracting the second equation, you get $x=r_A$.
The remaining task is solving $e^{mu+Asigma} = r_a$ and $y=e^{mu+Bsigma} = r_b$. Taking the natural logarithm you get a linear system: $mu+Asigma=log(r_a)$ and $mu+Bsigma=log(r_b)$. Subtracting the second equation from the first one yields:
$$sigma=frac{log(r_a)-log(r_b)}{A-B}$$
and then you can simply compute $mu=log(r_a)-Asigma$.
answered Jan 14 at 15:35
LinAlgLinAlg
9,4511521
$endgroup$
Let $x=e^{mu+Asigma}$ and $y=e^{mu+Bsigma}$, then you have
$$x^2 - r_A x + y^2 - r_B y = 0 text{, and } A (x^2 - r_A x) + B(y^2 - r_B y) = 0.$$
Let us assume that $A neq B$, as otherwise there are infinitely many solutions (which are easy to identify). I will also assume that $r_A$ and $r_B$ are positive. Multiplying the first equation with $A$ and substracting the second equation, you get $(A-B)y^2 - (A-B)r_B y = 0$. Since $Aneq B$, the solutions are $y=0$ and $y=r_B$. The first solution ($y=0$) cannot solve $y=e^{mu+Bsigma}$, so we have $y=r_B$. Similarly, multiplying the first equality with $B$ and subtracting the second equation, you get $x=r_A$.
The remaining task is solving $e^{mu+Asigma} = r_a$ and $y=e^{mu+Bsigma} = r_b$. Taking the natural logarithm you get a linear system: $mu+Asigma=log(r_a)$ and $mu+Bsigma=log(r_b)$. Subtracting the second equation from the first one yields:
$$sigma=frac{log(r_a)-log(r_b)}{A-B}$$
and then you can simply compute $mu=log(r_a)-Asigma$.
answered Jan 14 at 15:35
LinAlgLinAlg
9,4511521
answered Jan 14 at 15:35
LinAlgLinAlg
9,4511521
answered Jan 14 at 15:35
LinAlgLinAlg
9,4511521
answered Jan 14 at 15:35
LinAlgLinAlg
9,4511521
9,4511521
$begingroup$
Note that $sigma$ could be nonpositive with these formulas. If that is the case, there is an infimum that is not attained ($sigmadownarrow 0$).
$endgroup$
– LinAlg
Jan 14 at 16:14
add a comment |
$begingroup$
Note that $sigma$ could be nonpositive with these formulas. If that is the case, there is an infimum that is not attained ($sigmadownarrow 0$).
$endgroup$
– LinAlg
Jan 14 at 16:14
$begingroup$
Note that $sigma$ could be nonpositive with these formulas. If that is the case, there is an infimum that is not attained ($sigmadownarrow 0$).
$endgroup$
– LinAlg
Jan 14 at 16:14
$begingroup$
Note that $sigma$ could be nonpositive with these formulas. If that is the case, there is an infimum that is not attained ($sigmadownarrow 0$).
$endgroup$
– LinAlg
Jan 14 at 16:14
add a comment |
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$begingroup$
You know $de^{mu + Asigma}/dmu = e^{mu + Asigma}$, right?
$endgroup$
– LinAlg
Jan 14 at 15:11
$begingroup$
Right. Edited..
$endgroup$
– measure_theory
Jan 14 at 15:19
$begingroup$
Why did you add a bounty, but then disappear from the website without accepting or even upvoting my answer?
$endgroup$
– LinAlg
Jan 22 at 18:51