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Tal Galili
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Thanks to the discussion with Glen, I realized that (assume ${X_a}$ is the number of a's and ${X_b}$ is the count of b's):

$${X_L}|{X_M} \sim B\left( {n - {X_M},{p_L} = \frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}} \right)$$$${X_a}|{X_b} \sim B\left( {n - {X_b},{p_L} = \frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}} \right)$$

Hence:

$$E\left( {{X_L}|{X_M}} \right) = \left( {n - {X_M}} \right)\frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}{\rm{ }}$$$$E\left( {{X_a}|{X_b}} \right) = \left( {n - {X_b}} \right)\frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}{\rm{ }}$$

Making:

$${\mathop{\rm cov}} \left( {{X_L},{X_M}} \right) = - 2n\left( {1 - \theta } \right){\theta ^3}$$$${\mathop{\rm cov}} \left( {{X_a},{X_b}} \right) = - 2n\left( {1 - \theta } \right){\theta ^3}$$

Leading to:

$$V\left( {{{\hat \theta }_{MLE}}} \right) = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}$$

:)

Thanks to the discussion with Glen, I realized that:

$${X_L}|{X_M} \sim B\left( {n - {X_M},{p_L} = \frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}} \right)$$

Hence:

$$E\left( {{X_L}|{X_M}} \right) = \left( {n - {X_M}} \right)\frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}{\rm{ }}$$

Making:

$${\mathop{\rm cov}} \left( {{X_L},{X_M}} \right) = - 2n\left( {1 - \theta } \right){\theta ^3}$$

Leading to:

$$V\left( {{{\hat \theta }_{MLE}}} \right) = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}$$

:)

Thanks to the discussion with Glen, I realized that (assume ${X_a}$ is the number of a's and ${X_b}$ is the count of b's):

$${X_a}|{X_b} \sim B\left( {n - {X_b},{p_L} = \frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}} \right)$$

Hence:

$$E\left( {{X_a}|{X_b}} \right) = \left( {n - {X_b}} \right)\frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}{\rm{ }}$$

Making:

$${\mathop{\rm cov}} \left( {{X_a},{X_b}} \right) = - 2n\left( {1 - \theta } \right){\theta ^3}$$

Leading to:

$$V\left( {{{\hat \theta }_{MLE}}} \right) = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}$$

:)

Source Link
Tal Galili
  • 21.9k
  • 36
  • 147
  • 208

Thanks to the discussion with Glen, I realized that:

$${X_L}|{X_M} \sim B\left( {n - {X_M},{p_L} = \frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}} \right)$$

Hence:

$$E\left( {{X_L}|{X_M}} \right) = \left( {n - {X_M}} \right)\frac{{{\theta ^2}}}{{{\theta ^2} + {{\left( {1 - \theta } \right)}^2}}}{\rm{ }}$$

Making:

$${\mathop{\rm cov}} \left( {{X_L},{X_M}} \right) = - 2n\left( {1 - \theta } \right){\theta ^3}$$

Leading to:

$$V\left( {{{\hat \theta }_{MLE}}} \right) = \frac{{\theta \left( {1 - \theta } \right)}}{{2n}}$$

:)