Mixing
Determine the value of $α$ for which the $MSE(T)$ is minimal.
$begingroup$
Let $X_1$ be an estimator for the probability $θ$ of unauthorized access.
Let $X_2$ be another estimator for $θ$.
Assume that $X_1$ and $X_2$ are independent, unbiased estimators for $θ$.
Furthermore, let $σ_1^2$ be the variance of $X_1$ and $σ_2$ the variance of $X_2$.
Finally, define a new estimator $T$ for $θ$ by $T = αX_1+(1−α)X_2$.
Determine the value of $α$ for which the $MSE(T)$ is minimal.
$$MSE(T)=Var(T)+(E[T]-Theta)^2$$
$$E[T]=alpha E[X_1]+(1-alpha)E[X_2]=alphacdotTheta+(1-alpha)cdotTheta$$
$$MSE(T)=Var(T)=alpha^2cdot Var(X_1)+(1-alpha)^2cdot Var(X_2)=alpha^2cdot sigma_1^2+(1-alpha)^2cdot sigma_2^2$$
Now from this point how can I calculate the minimum value of $alpha$ that minimize the $MSE(T)$, in my opinion I should put $MSE(T)=0$, then differentiate respect to $alpha$ and solve for $alpha$, but if I do it, then $alpha$ will be equal to $0$, and I think is incorrect and I should find a finite value for it, where I'm wrong?
probability statistics estimation mean-square-error
asked Jan 26 at 22:03
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
Let $X_1$ be an estimator for the probability $θ$ of unauthorized access.
Let $X_2$ be another estimator for $θ$.
Assume that $X_1$ and $X_2$ are independent, unbiased estimators for $θ$.
Furthermore, let $σ_1^2$ be the variance of $X_1$ and $σ_2$ the variance of $X_2$.
Finally, define a new estimator $T$ for $θ$ by $T = αX_1+(1−α)X_2$.
Determine the value of $α$ for which the $MSE(T)$ is minimal.
$$MSE(T)=Var(T)+(E[T]-Theta)^2$$
$$E[T]=alpha E[X_1]+(1-alpha)E[X_2]=alphacdotTheta+(1-alpha)cdotTheta$$
$$MSE(T)=Var(T)=alpha^2cdot Var(X_1)+(1-alpha)^2cdot Var(X_2)=alpha^2cdot sigma_1^2+(1-alpha)^2cdot sigma_2^2$$
Now from this point how can I calculate the minimum value of $alpha$ that minimize the $MSE(T)$, in my opinion I should put $MSE(T)=0$, then differentiate respect to $alpha$ and solve for $alpha$, but if I do it, then $alpha$ will be equal to $0$, and I think is incorrect and I should find a finite value for it, where I'm wrong?
probability statistics estimation mean-square-error
asked Jan 26 at 22:03
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
Let $X_1$ be an estimator for the probability $θ$ of unauthorized access.
Let $X_2$ be another estimator for $θ$.
Assume that $X_1$ and $X_2$ are independent, unbiased estimators for $θ$.
Furthermore, let $σ_1^2$ be the variance of $X_1$ and $σ_2$ the variance of $X_2$.
Finally, define a new estimator $T$ for $θ$ by $T = αX_1+(1−α)X_2$.
Determine the value of $α$ for which the $MSE(T)$ is minimal.
$$MSE(T)=Var(T)+(E[T]-Theta)^2$$
$$E[T]=alpha E[X_1]+(1-alpha)E[X_2]=alphacdotTheta+(1-alpha)cdotTheta$$
$$MSE(T)=Var(T)=alpha^2cdot Var(X_1)+(1-alpha)^2cdot Var(X_2)=alpha^2cdot sigma_1^2+(1-alpha)^2cdot sigma_2^2$$
Now from this point how can I calculate the minimum value of $alpha$ that minimize the $MSE(T)$, in my opinion I should put $MSE(T)=0$, then differentiate respect to $alpha$ and solve for $alpha$, but if I do it, then $alpha$ will be equal to $0$, and I think is incorrect and I should find a finite value for it, where I'm wrong?
probability statistics estimation mean-square-error
asked Jan 26 at 22:03
Mark JaconMark Jacon
1127
$endgroup$
Let $X_1$ be an estimator for the probability $θ$ of unauthorized access.
Let $X_2$ be another estimator for $θ$.
Assume that $X_1$ and $X_2$ are independent, unbiased estimators for $θ$.
Furthermore, let $σ_1^2$ be the variance of $X_1$ and $σ_2$ the variance of $X_2$.
Finally, define a new estimator $T$ for $θ$ by $T = αX_1+(1−α)X_2$.
Determine the value of $α$ for which the $MSE(T)$ is minimal.
$$MSE(T)=Var(T)+(E[T]-Theta)^2$$
$$E[T]=alpha E[X_1]+(1-alpha)E[X_2]=alphacdotTheta+(1-alpha)cdotTheta$$
$$MSE(T)=Var(T)=alpha^2cdot Var(X_1)+(1-alpha)^2cdot Var(X_2)=alpha^2cdot sigma_1^2+(1-alpha)^2cdot sigma_2^2$$
Now from this point how can I calculate the minimum value of $alpha$ that minimize the $MSE(T)$, in my opinion I should put $MSE(T)=0$, then differentiate respect to $alpha$ and solve for $alpha$, but if I do it, then $alpha$ will be equal to $0$, and I think is incorrect and I should find a finite value for it, where I'm wrong?
probability statistics estimation mean-square-error
probability statistics estimation mean-square-error
asked Jan 26 at 22:03
Mark JaconMark Jacon
1127
asked Jan 26 at 22:03
Mark JaconMark Jacon
1127
asked Jan 26 at 22:03
Mark JaconMark Jacon
1127
asked Jan 26 at 22:03
Mark JaconMark Jacon
1127
asked Jan 26 at 22:03
Mark JaconMark Jacon
1127
1127
add a comment |
add a comment |
1 Answer
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$begingroup$
First differentiate, then set equal to zero:
begin{align*}
frac{d}{dalpha}text{MSE}(alpha) &= 2alpha sigma_1^2 - 2(1 - alpha)sigma_2^2 = 0\
& alpha (sigma_1^2 +sigma_2^2) - sigma_2^2 = 0\
& alpha = frac{sigma_2^2}{sigma_1^2 + sigma_2^2}
end{align*}
answered Jan 26 at 22:12
inhuretnakhtinhuretnakht
38617
$endgroup$
$begingroup$
Thank you for your answer, thanks to you I found my error! :)
$endgroup$
– Mark Jacon
Jan 26 at 22:19
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First differentiate, then set equal to zero:
begin{align*}
frac{d}{dalpha}text{MSE}(alpha) &= 2alpha sigma_1^2 - 2(1 - alpha)sigma_2^2 = 0\
& alpha (sigma_1^2 +sigma_2^2) - sigma_2^2 = 0\
& alpha = frac{sigma_2^2}{sigma_1^2 + sigma_2^2}
end{align*}
answered Jan 26 at 22:12
inhuretnakhtinhuretnakht
38617
$endgroup$
$begingroup$
Thank you for your answer, thanks to you I found my error! :)
$endgroup$
– Mark Jacon
Jan 26 at 22:19
add a comment |
$begingroup$
First differentiate, then set equal to zero:
begin{align*}
frac{d}{dalpha}text{MSE}(alpha) &= 2alpha sigma_1^2 - 2(1 - alpha)sigma_2^2 = 0\
& alpha (sigma_1^2 +sigma_2^2) - sigma_2^2 = 0\
& alpha = frac{sigma_2^2}{sigma_1^2 + sigma_2^2}
end{align*}
answered Jan 26 at 22:12
inhuretnakhtinhuretnakht
38617
$endgroup$
$begingroup$
Thank you for your answer, thanks to you I found my error! :)
$endgroup$
– Mark Jacon
Jan 26 at 22:19
add a comment |
$begingroup$
First differentiate, then set equal to zero:
begin{align*}
frac{d}{dalpha}text{MSE}(alpha) &= 2alpha sigma_1^2 - 2(1 - alpha)sigma_2^2 = 0\
& alpha (sigma_1^2 +sigma_2^2) - sigma_2^2 = 0\
& alpha = frac{sigma_2^2}{sigma_1^2 + sigma_2^2}
end{align*}
answered Jan 26 at 22:12
inhuretnakhtinhuretnakht
38617
$endgroup$
First differentiate, then set equal to zero:
begin{align*}
frac{d}{dalpha}text{MSE}(alpha) &= 2alpha sigma_1^2 - 2(1 - alpha)sigma_2^2 = 0\
& alpha (sigma_1^2 +sigma_2^2) - sigma_2^2 = 0\
& alpha = frac{sigma_2^2}{sigma_1^2 + sigma_2^2}
end{align*}
answered Jan 26 at 22:12
inhuretnakhtinhuretnakht
38617
answered Jan 26 at 22:12
inhuretnakhtinhuretnakht
38617
answered Jan 26 at 22:12
inhuretnakhtinhuretnakht
38617
answered Jan 26 at 22:12
inhuretnakhtinhuretnakht
38617
38617
$begingroup$
Thank you for your answer, thanks to you I found my error! :)
$endgroup$
– Mark Jacon
Jan 26 at 22:19
add a comment |
$begingroup$
Thank you for your answer, thanks to you I found my error! :)
$endgroup$
– Mark Jacon
Jan 26 at 22:19
$begingroup$
Thank you for your answer, thanks to you I found my error! :)
$endgroup$
– Mark Jacon
Jan 26 at 22:19
$begingroup$
Thank you for your answer, thanks to you I found my error! :)
$endgroup$
– Mark Jacon
Jan 26 at 22:19
add a comment |
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