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Edit: I believe this is now correct

Let $X_a$ and $X_b$ be the counts in categories $a$ and $b$. Consider first the case for a single observation. (The variance for $n$ observations is $n$ times the variance of one.$n=1$):

  • $\text{Var}(X_a) = \theta^2(1-\theta^2)$

  • $\text{Var}(X_b) = 2\theta(1-\theta)(1-2\theta(1-\theta))$

  • Since the categories are mutually exclusive, $E(X_aX_b)=0$, so $\text{Cov}(X_a,X_b)=-E(X_a)E(X_b)=-2\theta^3(1-\theta)$.

Hence $Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+4Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-8\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-4\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)$

Hence, (since the variance for $n$ independent observations is $n$ times the variance of one), $Var(2X_a+X_b)=2n\theta(1-\theta)$.

Hence $\text{Var}(\frac{1}{2n}(2X_a+X_b)) = \frac{1}{4n^2}2n\theta(1-\theta)=\frac{1}{2n}\theta(1-\theta)$.

This is non-negative for $0\leq\theta\leq 1$, and simulations agree with this answer.

Edit: I believe this is now correct

Let $X_a$ and $X_b$ be the counts in categories $a$ and $b$. Consider first the case for a single observation. (The variance for $n$ observations is $n$ times the variance of one.)

  • $\text{Var}(X_a) = \theta^2(1-\theta^2)$

  • $\text{Var}(X_b) = 2\theta(1-\theta)(1-2\theta(1-\theta))$

  • Since the categories are mutually exclusive, $E(X_aX_b)=0$, so $\text{Cov}(X_a,X_b)=-E(X_a)E(X_b)=-2\theta^3(1-\theta)$.

Hence $Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+4Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-8\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-4\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)$

Hence, for $n$ independent observations, $Var(2X_a+X_b)=2n\theta(1-\theta)$

Hence $\text{Var}(\frac{1}{2n}(2X_a+X_b)) = \frac{1}{4n^2}2n\theta(1-\theta)=\frac{1}{2n}\theta(1-\theta)$.

This is non-negative for $0\leq\theta\leq 1$, and simulations agree with this answer.

Edit: I believe this is now correct

Let $X_a$ and $X_b$ be the counts in categories $a$ and $b$. Consider first the case for a single observation ($n=1$):

  • $\text{Var}(X_a) = \theta^2(1-\theta^2)$

  • $\text{Var}(X_b) = 2\theta(1-\theta)(1-2\theta(1-\theta))$

  • Since the categories are mutually exclusive, $E(X_aX_b)=0$, so $\text{Cov}(X_a,X_b)=-E(X_a)E(X_b)=-2\theta^3(1-\theta)$.

Hence $Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+4Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-8\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-4\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)$

Hence, (since the variance for $n$ independent observations is $n$ times the variance of one), $Var(2X_a+X_b)=2n\theta(1-\theta)$.

Hence $\text{Var}(\frac{1}{2n}(2X_a+X_b)) = \frac{1}{4n^2}2n\theta(1-\theta)=\frac{1}{2n}\theta(1-\theta)$.

This is non-negative for $0\leq\theta\leq 1$, and simulations agree with this answer.

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Glen_b
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This has an error -- I can't spot it right now, so I'll leave this deleted until I figure out the problem Edit: I believe this is now correct

Let $X_a$ and $X_b$ be the counts in categories $a$ and $b$. Consider first the case for a single observation. (The variance for $n$ observations is $n$ times the variance of one.)

  • $\text{Var}(X_a) = \theta^2(1-\theta^2)$

  • $\text{Var}(X_b) = 2\theta(1-\theta)(1-2\theta(1-\theta))$

  • Since the categories are mutually exclusive, $E(X_aX_b)=0$, so $\text{Cov}(X_a,X_b)=-E(X_a)E(X_b)=-2\theta^3(1-\theta)$.

Hence $Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+4Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-8\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-4\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)$

(Simulations suggest the correct answer Hence, for the variance I was computing$n$ independent observations, $Var(2X_a+X_b)=2n\theta(1-\theta)$

Hence $\text{Var}(\frac{1}{2n}(2X_a+X_b)) = \frac{1}{4n^2}2n\theta(1-\theta)=\frac{1}{2n}\theta(1-\theta)$.

This is actuallynon-negative for $2\theta(1-\theta)$)$0\leq\theta\leq 1$, and simulations agree with this answer.

This has an error -- I can't spot it right now, so I'll leave this deleted until I figure out the problem

Let $X_a$ and $X_b$ be the counts in categories $a$ and $b$. Consider first the case for a single observation. (The variance for $n$ observations is $n$ times the variance of one.)

  • $\text{Var}(X_a) = \theta^2(1-\theta^2)$

  • $\text{Var}(X_b) = 2\theta(1-\theta)(1-2\theta(1-\theta))$

  • Since the categories are mutually exclusive, $E(X_aX_b)=0$, so $\text{Cov}(X_a,X_b)=-E(X_a)E(X_b)=-2\theta^3(1-\theta)$.

Hence $Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+4Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-8\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-4\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)$

(Simulations suggest the correct answer for the variance I was computing is actually $2\theta(1-\theta)$)

Edit: I believe this is now correct

Let $X_a$ and $X_b$ be the counts in categories $a$ and $b$. Consider first the case for a single observation. (The variance for $n$ observations is $n$ times the variance of one.)

  • $\text{Var}(X_a) = \theta^2(1-\theta^2)$

  • $\text{Var}(X_b) = 2\theta(1-\theta)(1-2\theta(1-\theta))$

  • Since the categories are mutually exclusive, $E(X_aX_b)=0$, so $\text{Cov}(X_a,X_b)=-E(X_a)E(X_b)=-2\theta^3(1-\theta)$.

Hence $Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+4Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-8\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-4\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)$

Hence, for $n$ independent observations, $Var(2X_a+X_b)=2n\theta(1-\theta)$

Hence $\text{Var}(\frac{1}{2n}(2X_a+X_b)) = \frac{1}{4n^2}2n\theta(1-\theta)=\frac{1}{2n}\theta(1-\theta)$.

This is non-negative for $0\leq\theta\leq 1$, and simulations agree with this answer.

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Glen_b
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This has an error -- being fixedI can't spot it right now, so I'll leave this deleted until I figure out the problem

Let $X_a$ and $X_b$ be the counts in categories $a$ and $b$. Consider first the case for a single observation. (The variance for $n$ observations is $n$ times the variance of one.)

  • $\text{Var}(X_a) = \theta^2(1-\theta^2)$

  • $\text{Var}(X_b) = 2\theta(1-\theta)(1-2\theta(1-\theta))$

  • Since the categories are mutually exclusive, $E(X_aX_b)=0$, so $\text{Cov}(X_a,X_b)=-E(X_a)E(X_b)=-2\theta^3(1-\theta)$.

Hence $Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+2Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-4\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-2\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)\,[1+2\theta^2]$$Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+4Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-8\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-4\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)$

(Simulations suggest the correct answer for the variance I was computing is actually $2\theta(1-\theta)$)

This has an error -- being fixed

Let $X_a$ and $X_b$ be the counts in categories $a$ and $b$. Consider first the case for a single observation. (The variance for $n$ observations is $n$ times the variance of one.)

  • $\text{Var}(X_a) = \theta^2(1-\theta^2)$

  • $\text{Var}(X_b) = 2\theta(1-\theta)(1-2\theta(1-\theta))$

  • Since the categories are mutually exclusive, $E(X_aX_b)=0$, so $\text{Cov}(X_a,X_b)=-E(X_a)E(X_b)=-2\theta^3(1-\theta)$.

Hence $Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+2Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-4\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-2\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)\,[1+2\theta^2]$

(Simulations suggest the correct answer for the variance I was computing is actually $2\theta(1-\theta)$)

This has an error -- I can't spot it right now, so I'll leave this deleted until I figure out the problem

Let $X_a$ and $X_b$ be the counts in categories $a$ and $b$. Consider first the case for a single observation. (The variance for $n$ observations is $n$ times the variance of one.)

  • $\text{Var}(X_a) = \theta^2(1-\theta^2)$

  • $\text{Var}(X_b) = 2\theta(1-\theta)(1-2\theta(1-\theta))$

  • Since the categories are mutually exclusive, $E(X_aX_b)=0$, so $\text{Cov}(X_a,X_b)=-E(X_a)E(X_b)=-2\theta^3(1-\theta)$.

Hence $Var(2X_a+X_b)=4Var(X_a)+Var(X_b)+4Cov(X_a,X_b)\\ \hspace{2.96cm}=4\theta^2(1-\theta^2)+2\theta(1-\theta)(1-2\theta(1-\theta))-8\theta^3(1-\theta)\\ \hspace{2.96cm}=2\theta(1-\theta)\,[2\theta(1+\theta)+(1-2\theta(1-\theta))-4\theta^2]\\ \hspace{2.96cm}=2\theta(1-\theta)$

(Simulations suggest the correct answer for the variance I was computing is actually $2\theta(1-\theta)$)

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