Let $X \square \operatorname { Bin } \left( m , p _ { 1 } \right) \text { and } Y \square \operatorname { Bin } \left( n , p _ { 2 } \right)$ with X and Y independent variables, and let $\hat { p } _ { 1 } = X / m \text { and } \hat { p } _ { 2 } = Y / \mathrm { n } \text {. Then } \mathrm { E } \left( \hat { \mathrm { p } } _ { 1 } - \hat { \mathrm { p } } _ { 2 } \right) =\underline{\quad\quad}$,so $\hat { p } _ { 1 } - \hat { p } _ { 2 }$ is an __________ estimator of $p _ { 1 } - p _ { 2 }$

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The Answer of Let $X \square \operatorname { Bin } \left( m , p _ { 1 } \right) \text { and } Y \square \operatorname { Bin } \left( n , p _ { 2 } \right)$ with X and Y independent variables, and let $\hat { p } _ { 1 } = X / m \text { and } \hat { p } _ { 2 } = Y / \mathrm { n } \text {. Then } \mathrm { E } \left( \hat { \mathrm { p } } _ { 1 } - \hat { \mathrm { p } } _ { 2 } \right) =\underline{\quad\quad}$,so $\hat { p } _ { 1 } - \hat { p } _ { 2 }$ is an __________ estimator of $p _ { 1 } - p _ { 2 }$

Let X□Bin⁡(m,p1) and Y□Bin⁡(n,p2)X \square \operatorname { Bin } \left( m , p _ { 1 } \right) \text { and } Y \square \operatorname { Bin } \left( n , p _ { 2 } \right)X□Bin(m,p1​) and Y□Bin(n,p2​)

Let $X \square \operatorname { Bin } \left( m , p _ { 1 } \right) \text { and } Y \square \operatorname { Bin } \left( n , p _ { 2 } \right)$