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5.2 (難) t分布

Last updated at Posted at 2022-05-14

方針は5章後半~7章の内容を含みます.

方針

t分布について.

Z\sim \mathcal{N}(0,1),\ \ Y\sim\chi_m^2

の時,

\frac{Z}{\sqrt{Y/m}}\sim t_m

これが分かっていると,t検定で使う

\frac{\sqrt{n}(\bar{X}-\mu)}{V}\sim t_{n-1}

が分かる.ただし,

X_1,…,X_n:iid\sim\mathcal{N}(\mu,\sigma^2)

として

\bar{X}=\frac{1}{n}\sum_{i=1}^nX_i,\ \ \ V^2=\frac{1}{n-1}(X_i-\bar{X})^2

と置いた.中心極限定理 (CLT) により,

Z=\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}\rightarrow_d\mathcal{N}(0,1)

一方,

Y=\sum_{i=1}^m\left(\frac{X_i-\bar{X}}{\sigma}\right)^2=\frac{(n-1)V^2}{\sigma^2}\sim\chi_{n-1}^2

ゆえ,

\frac{Z}{\sqrt{Y/(n-1)}}\sim t_{n-1}

である.今,

\frac{Z}{\sqrt{Y/(n-1)}}=\frac{\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}}{\sqrt{\frac{V^2}{\sigma^2}}}=\frac{\sqrt{n}(\bar{X}-\mu)}{V}

より,

\frac{\sqrt{n}(\bar{X}-\mu)}{V}\sim t_{n-1}

が得られた.

答案

5.-3.jpg
5.-4.jpg

参考文献

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