برآورد درست‌نمایی بیشینه

${\displaystyle {\widehat {\theta }}_{MLE}{\rightarrow }\theta _{0}}$

${\displaystyle \theta \neq \theta _{0}\Leftrightarrow f(.\mid \theta _{0})}$

${\displaystyle Pr[lnf(x\mid \theta )\in C^{0}(\Theta )]=1}$

${\displaystyle \theta \in \Theta }$ for all ${\textstyle |lnf(x\mid \theta )|<D(x)}$

${\displaystyle \sup _{\theta \in \Theta }\left\vert {\widehat {l}}(\theta \mid x)-l(\theta )\right\vert \rightarrow 0}$

${\displaystyle {\sqrt {n}}({\widehat {\theta }}_{mle}-\theta _{0})\longrightarrow \aleph (0,I^{-1})}$.

${\displaystyle {\hat {\alpha }}=g({\hat {\theta }})}$

${\displaystyle {\bar {L}}(\alpha )=\sup _{\theta$ :\alpha =g(\theta )}L(\theta )}

${\displaystyle f_{Y}(y)={\frac {f_{X}(x)}{\mid g'(x)\mid }}}$

${\displaystyle {\sqrt {n}}({\hat {\theta }}_{mle}-\theta _{0})\longrightarrow \aleph (0,I^{-1})}$

${\displaystyle I_{jk}=E_{X}[-{\frac {\delta ^{2}lnf_{\theta _{0}}(X_{t})}{\delta \theta _{j}\delta \theta _{k}}}]}$

${\displaystyle b_{s}\equiv E[({\hat {\theta }}_{mle}-\theta _{0})_{s}]={\frac {1}{n}}.I^{si}I^{jk}({\frac {1}{2}}K_{ijk}+J_{j,ik})}$

${\displaystyle {\frac {1}{2}}K_{ijk}+J_{j,ik}=E_{X}[{\frac {1}{2}}{\frac {\delta ^{3}lnf_{\theta _{0}}(X_{t})}{\delta \theta _{i}\delta \theta _{j}\delta \theta _{k}}}+{\frac {\delta lnf_{\theta _{0}}(x_{t})}{\delta \theta _{j}}}{\frac {\delta ^{2}lnf_{\theta _{0}}(X_{t})}{\delta \theta _{i}\delta \theta _{k}}}}$

${\displaystyle {\hat {\theta }}_{mle}^{*}={\hat {\theta }}_{mle}-{\hat {b}}}$ .

${\displaystyle {\begin{aligned}\Pr(\mathrm {H} =49\mid p=1/3)&={\binom {80}{49}}(1/3)^{49}(1-1/3)^{31}\approx 0.000,\\[6pt]\Pr(\mathrm {H} =49\mid p=1/2)&={\binom {80}{49}}(1/2)^{49}(1-1/2)^{31}\approx 0.012,\\[6pt]\Pr(\mathrm {H} =49\mid p=2/3)&={\binom {80}{49}}(2/3)^{49}(1-2/3)^{31}\approx 0.054.\end{aligned}}}$

${\displaystyle L(p)=f_{D}(\mathrm {H} =49\mid p)={\binom {80}{49}}p^{49}(1-p)^{31},}$

${\displaystyle {\begin{aligned}{0}&{}={\frac {\partial }{\partial p}}\left({\binom {80}{49}}p^{49}(1-p)^{31}\right)\\[8pt]&{}\propto 49p^{48}(1-p)^{31}-31p^{49}(1-p)^{30}\\[8pt]&{}=p^{48}(1-p)^{30}\left[49(1-p)-31p\right]\\[8pt]&{}=p^{48}(1-p)^{30}\left[49-80p\right]\end{aligned}}}$

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${\displaystyle f(x\mid \mu ,\sigma ^{2})={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}exp(-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}),}$

${\displaystyle f(x_{1},...x_{n}\mid \mu ,\sigma ^{2})=\textstyle \prod _{i=1}^{n}f(x_{i}\mid \mu ,\sigma ^{2})\displaystyle =({\frac {1}{2\pi \sigma ^{2}}})^{\frac {n}{2}}exp(-{\frac {\textstyle \sum _{i=1}^{n}(x_{i}-\mu )^{2}\displaystyle }{2\sigma ^{2}}})}$

${\displaystyle f(x_{1},...x_{n}\mid \mu ,\sigma ^{2})=({\frac {1}{2\pi \sigma ^{2}}})^{\frac {n}{2}}exp(-{\frac {\textstyle \sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}+n({\bar {x}}-\mu )^{2}\displaystyle }{2\sigma ^{2}}})}$

${\displaystyle log(L(\mu ,\sigma ))=(-n/2)log(2\pi \sigma ^{2})-{\frac {1}{2\sigma ^{2}}}\sum _{i=1}^{n}(x_{i}-\mu )^{2}}$

${\displaystyle 0={\frac {\partial }{\partial \mu }}log(L(\mu ,\sigma )=0-{\frac {-2n({\bar {x}}-\mu )}{2\sigma ^{2}}}}$

${\displaystyle {\hat {\mu }}={\bar {x}}=\sum _{i=1}^{n}{\frac {x_{i}}{n}}}$

${\displaystyle E[{\hat {\mu }}]=\mu }$

${\displaystyle 0={\frac {\partial }{\partial \sigma }}log(({\frac {1}{2\pi \sigma ^{2}}})^{n/2})exp(-{\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})+n({\bar {x}}-\mu )^{2}}{2\sigma ^{2}}})}$

${\displaystyle ={\frac {\partial }{\partial \sigma }}({\frac {n}{2}}log({\frac {1}{2\pi \sigma ^{2}}})-{\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})+n({\bar {x}}-\mu )^{2}}{2\sigma ^{2}}}}$

${\displaystyle =-{\frac {n}{\sigma }}+{\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})+n({\bar {x}}-\mu )^{2}}{\sigma ^{3}}}}$

${\displaystyle {\hat {\sigma }}^{2}={\frac {1}{n}}\sum _{i=1}^{n}(x_{i}-\mu )^{2}}$

${\displaystyle {\hat {\sigma }}^{2}={\frac {1}{n}}\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}={\frac {1}{n}}\sum _{i=1}^{n}x_{i}^{2}-\sum _{i=1}^{n}\sum _{j=1}^{n}x_{i}x_{j}}$

${\displaystyle {\hat {\sigma }}^{2}={\frac {1}{n}}\sum _{i=1}^{n}(\mu -\delta _{i})^{2}-{\frac {1}{n}}\sum _{i=1}^{n}\sum _{j=1}^{n}(\mu -\delta _{i})(\mu -\delta _{j})}$

${\displaystyle E[{\hat {\sigma }}^{2}]={\frac {n-1}{n}}\sigma ^{2}}$

${\displaystyle {\hat {\theta }}=({\hat {\mu }},{\hat {\sigma }}^{2})}$

${\displaystyle log(L({\hat {\mu }},{\hat {\sigma }}))={\frac {-n}{2}}(log(2\pi \sigma ^{2})+1)}$