I am playing with the following trinomial (multinomial) distribution which can get values (a,b,c) with the probabilities: $(\theta^2, 2\theta(1-\theta), (1-\theta)^2)$.

Say I have n observations from this distribution, then it is easy to show that the MLE for estimating $\theta$ is: $\hat \theta_{MLE} = {2\#a+ \#b \over 2n}$. However, I am trying to find the variance of this MLE, and keep getting a negative variance (which means I have a mistake somewhere, but I can't find where).

[\begin{array}{l} V\left( {{{\hat \theta }_{MLE}}} \right) = V\left( {\frac{{\left( {2{\#a} + {\#b}} \right)}}{{2n}}} \right) = \frac{1}{{4{n^2}}}V\left( {2{\#a} + {\#b}} \right) = \\ = \frac{1}{{4{n^2}}}\left[ {V\left( {2{\#a}} \right) + V\left( {{\#b}} \right) + 2{\mathop{\rm cov}} \left( {2{\#a},{\#b}} \right)} \right]\\ = \frac{1}{{4{n^2}}}\left[ {4n{\theta ^2}\left( {1 - {\theta ^2}} \right) + 2n\theta \left( {1 - \theta } \right)\left( {1 - 2\theta \left( {1 - \theta } \right)} \right) + 8n\left( {1 - \theta } \right){\theta ^3}\left[ {2\theta \left( {1 - \theta } \right)\left( {1 - n} \right) - 1} \right]} \right]\\ = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}\left[ {2\theta \left( {1 + \theta } \right) + \left( {1 - 2\theta \left( {1 - \theta } \right)} \right) + 4{\theta ^2}\left[ {2\theta \left( {1 - \theta } \right)\left( {1 - n} \right) - 1} \right]} \right]\\ = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}\left[ {4{\theta ^2} + 1 + 4{\theta ^2}\left[ {2\theta \left( {1 - \theta } \right)\left( {1 - n} \right) - 1} \right]} \right]\\ = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}\left[ {1 + 8{\theta ^3}\left( {1 - \theta } \right)\left( {1 - n} \right)} \right] \end{array}]

I am playing with the following trinomial (multinomial) distribution which can get values (a,b,c) with the probabilities: $(\theta^2, 2\theta(1-\theta), (1-\theta)^2)$.

Say I have n observations from this distribution, then it is easy to show that the MLE for estimating $\theta$ is: $\hat \theta_{MLE} = {2\#a+ \#b \over 2n}$. However, I am trying to find the variance of this MLE, and keep getting a negative variance (which means I have a mistake somewhere, but I can't find where).

(EDIT: the correct answer is marked below :) )

I am playing with the following trinomial (multinomial) distribution which can get values (a,b,c) with the probabilities: $(\theta^2, 2\theta(1-\theta), (1-\theta)^2)$.

p.s: does anyone know the name of this distribution? (I remember seeing somewhere that it is a known one, but I can no longer find that source)

Say I have n observations from this distribution, then it is easy to show that the MLE for estimating $\theta$ is: $\hat \theta_{MLE} = {2\#a+ \#b \over 2n}$. However, I am trying to find the variance of this MLE, and keep getting a negative variance (which means I have a mistake somewhere, but I can't find where).

[\begin{array}{l} V\left( {{{\hat \theta }_{MLE}}} \right) = V\left( {\frac{{\left( {2{\#a} + {\#b}} \right)}}{{2n}}} \right) = \frac{1}{{4{n^2}}}V\left( {2{\#a} + {\#b}} \right) = \\ = \frac{1}{{4{n^2}}}\left[ {V\left( {2{\#a}} \right) + V\left( {{\#b}} \right) + 2{\mathop{\rm cov}} \left( {2{\#a},{\#b}} \right)} \right]\\ = \frac{1}{{4{n^2}}}\left[ {4n{\theta ^2}\left( {1 - {\theta ^2}} \right) + 2n\theta \left( {1 - \theta } \right)\left( {1 - 2\theta \left( {1 - \theta } \right)} \right) + 8n\left( {1 - \theta } \right){\theta ^3}\left[ {2\theta \left( {1 - \theta } \right)\left( {1 - n} \right) - 1} \right]} \right]\\ = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}\left[ {2\theta \left( {1 + \theta } \right) + \left( {1 - 2\theta \left( {1 - \theta } \right)} \right) + 4{\theta ^2}\left[ {2\theta \left( {1 - \theta } \right)\left( {1 - n} \right) - 1} \right]} \right]\\ = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}\left[ {4{\theta ^2} + 1 + 4{\theta ^2}\left[ {2\theta \left( {1 - \theta } \right)\left( {1 - n} \right) - 1} \right]} \right]\\ = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}\left[ {1 + 8{\theta ^3}\left( {1 - \theta } \right)\left( {1 - n} \right)} \right] \end{array}]

I am playing with the following trinomial (multinomial) distribution which can get values (a,b,c) with the probabilities: $(\theta^2, 2\theta(1-\theta), (1-\theta)^2)$.

Say I have n observations from this distribution, then it is easy to show that the MLE for estimating $\theta$ is: $\hat \theta_{MLE} = {2\#a+ \#b \over 2n}$. However, I am trying to find the variance of this MLE, and keep getting a negative variance (which means I have a mistake somewhere, but I can't find where).

[\begin{array}{l} V\left( {{{\hat \theta }_{MLE}}} \right) = V\left( {\frac{{\left( {2{\#a} + {\#b}} \right)}}{{2n}}} \right) = \frac{1}{{4{n^2}}}V\left( {2{\#a} + {\#b}} \right) = \\ = \frac{1}{{4{n^2}}}\left[ {V\left( {2{\#a}} \right) + V\left( {{\#b}} \right) + 2{\mathop{\rm cov}} \left( {2{\#a},{\#b}} \right)} \right]\\ = \frac{1}{{4{n^2}}}\left[ {4n{\theta ^2}\left( {1 - {\theta ^2}} \right) + 2n\theta \left( {1 - \theta } \right)\left( {1 - 2\theta \left( {1 - \theta } \right)} \right) + 8n\left( {1 - \theta } \right){\theta ^3}\left[ {2\theta \left( {1 - \theta } \right)\left( {1 - n} \right) - 1} \right]} \right]\\ = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}\left[ {2\theta \left( {1 + \theta } \right) + \left( {1 - 2\theta \left( {1 - \theta } \right)} \right) + 4{\theta ^2}\left[ {2\theta \left( {1 - \theta } \right)\left( {1 - n} \right) - 1} \right]} \right]\\ = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}\left[ {4{\theta ^2} + 1 + 4{\theta ^2}\left[ {2\theta \left( {1 - \theta } \right)\left( {1 - n} \right) - 1} \right]} \right]\\ = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}\left[ {1 + 8{\theta ^3}\left( {1 - \theta } \right)\left( {1 - n} \right)} \right] \end{array}]