โ€œํ™•๋ฅ ๊ณผ ํ†ต๊ณ„(MATH230)โ€ ์ˆ˜์—…์—์„œ ๋ฐฐ์šด ๊ฒƒ๊ณผ ๊ณต๋ถ€ํ•œ ๊ฒƒ์„ ์ •๋ฆฌํ•œ ํฌ์ŠคํŠธ์ž…๋‹ˆ๋‹ค. ์ „์ฒด ํฌ์ŠคํŠธ๋Š” Probability and Statistics์—์„œ ํ™•์ธํ•˜์‹ค ์ˆ˜ ์žˆ์Šต๋‹ˆ๋‹ค ๐ŸŽฒ

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โ€œํ™•๋ฅ ๊ณผ ํ†ต๊ณ„(MATH230)โ€ ์ˆ˜์—…์—์„œ ๋ฐฐ์šด ๊ฒƒ๊ณผ ๊ณต๋ถ€ํ•œ ๊ฒƒ์„ ์ •๋ฆฌํ•œ ํฌ์ŠคํŠธ์ž…๋‹ˆ๋‹ค. ์ „์ฒด ํฌ์ŠคํŠธ๋Š” Probability and Statistics์—์„œ ํ™•์ธํ•˜์‹ค ์ˆ˜ ์žˆ์Šต๋‹ˆ๋‹ค ๐ŸŽฒ

Interval Estimation ํฌ์ŠคํŠธ์—์„œ ๋‹ค๋ฃฌ <Interval Estimation>์„ ํŠน์ • ์ƒํ™ฉ์— ์–ด๋–ป๊ฒŒ ์ ์šฉํ•  ์ˆ˜ ์žˆ๋Š”์ง€๋ฅผ ๋‹ค๋ฃจ๋Š” ํฌ์ŠคํŠธ์ž…๋‹ˆ๋‹ค.

Single Sample Estimation: Variance Estimation

Let $X_1, \dots, X_n$ be a random sample from $N(\mu, \sigma^2)$.

Q. Find $100(1-\alpha)\%$ confidence interval for $\sigma^2$.

1. choose a point estimator for $\sigma^2$.

\[S^2 = \frac{1}{n-1} \cdot \sum^n_{i=1} (X_i - \bar{X})^2\]

and also

\[\frac{(n-1) S^2}{\sigma^2} \; \sim \; \chi^2 (n-1)\]

2. Find confidence interval by using $\dfrac{(n-1) S^2}{\sigma^2} \; \sim \; \chi^2 (n-1)$.

\[1 - \alpha = P \left( \chi^2_{1-\alpha/2} (n-1) \; \le \; \frac{(n-1)s^2}{\sigma^2}\; \le \; \chi^2_{\alpha/2} (n-1)\right)\]

Therefore, the confidence interval for $\sigma^2$ is

\[\left( \frac{(n-1)\cdot s^2}{\chi^2_{\alpha/2}}, \frac{(n-1)\cdot s^2}{\chi^2_{1 - \alpha/2}} \right)\]

where $\chi^2_{\alpha/2}$ and \(\chi^2_{1-\alpha/2}\) are $\chi^2$-values with $n-1$ dof.

๐Ÿ’ฅ NOTE: $X_1, \dots, X_n$ should follow $N(\mu, \sigma^2)$.

๐Ÿ’ฅ NOTE: $\chi^2$ distribution is NOT symmetric!


Two Samples Estimation: The ratio of two variances

Let $X_1, \dots, X_{n_1}$ be random samples from $N(\mu_1, \sigma_1^2)$.

Let $Y_1, \dots, Y_{n_2}$ be random samples from $N(\mu_2, \sigma_2^2)$.

Supp. $X_i$s and $Y_j$s are independent.

Q. Find $100(1-\alpha)\%$ confidence interval for $\sigma_1^2 / \sigma_2^2$.

A. We can use $s_1^2 / s_2^2$ instead!!

\[\frac{s_1^2 / \sigma_1^2}{s_2^2 / \sigma_2^2} \; \sim \; F(n_1-1, n_2-1)\]

๋”ฐ๋ผ์„œ,

\[1 - \alpha = P \left( f_{1-\alpha/2} \, (n_1 - 1, n_2 - 1) \; \le \; \frac{s_1^2 / \sigma_1^2}{s_2^2 / \sigma_2^2} \; \le \; f_{\alpha/2} \, (n_1 - 1, n_2 - 1) \right)\]

Note that $f_{1-\alpha/2} \, (n_1 - 1, n_2 - 1) = \dfrac{1}{f_{\alpha/2} \, (n_2 - 1, n_1 - 1)}$.

Therefore, the confidence interval for $\sigma_1^2 / \sigma_2^2$ is

\[\left( \frac{s_1^2}{s_2^2} \cdot \frac{1}{f_{\alpha/2} \, (n_1 - 1, n_2 - 1)}, \; \frac{s_1^2}{s_2^2} \cdot f_{\alpha/2} \, (n_2 - 1, n_1 - 1) \right)\]

์ง€๊ธˆ๊นŒ์ง€ โ€œ์ถ”์ •(Statistical Estimation)โ€ ๊ณผ์ •์— ๋Œ€ํ•ด ์‚ดํŽด๋ณด์•˜๋‹ค. $\bar{x}$์™€ $s^2$์™€ ๊ฐ™์ด Point Estimator๋ฅผ ๊ตฌํ•˜๋Š” ๊ฒฝ์šฐ๋„ ์žˆ์—ˆ๊ณ , $\bar{x} \pm z_{\alpha/2} \cdot s / \sqrt{n}$ ๊ณผ ๊ฐ™์ด Interval Estimator๋ฅผ ๊ตฌํ•˜๋Š” ๊ฒฝ์šฐ๋„ ์žˆ์—ˆ๋‹ค. ๋˜, Single Sample์—์„œ Estimator๋ฅผ ๊ตฌํ•˜๋Š” ๊ฒƒ๋„ ์žˆ์—ˆ๊ณ , Two Samples์—์„œ Estimator๋ฅผ ๊ตฌํ•˜๋Š” ๊ฒƒ๋„ ์žˆ์—ˆ๋‹ค.

๊ทธ๋Ÿฌ๋‚˜ ์ง€๊ธˆ๊นŒ์ง€ ์‚ดํŽด๋ณธ ๋ฐฉ์‹ ์™ธ์—๋„ ๋˜๋‹ค๋ฅธ Estimation ๋ฐฉ๋ฒ•์ด ์žˆ๋‹ค ๐Ÿ˜ฒ <MLE; Maximum Likelihood Estimation>๋Š” Sample Distribution์„ ์žฌํ˜„ํ•  ํ™•๋ฅ ์ด ๊ฐ€์žฅ ๋†’์€ Parameter $\theta$๋ฅผ ์ฐพ๋Š” ๋ฐฉ์‹์œผ๋กœ Estimator๋ฅผ ์ฐพ๋Š”๋‹ค. <MLE>๋Š” ๋‹น์—ฐํžˆ Point Estimator๋ฅผ ์ œ์‹œํ•˜๋ฉฐ, ๊ทธ ๊ณผ์ •์—์„œ $\theta$์— ๋Œ€ํ•ด ํŽธ๋ฏธ๋ถ„์„ ์ˆ˜ํ–‰ํ•œ๋‹ค.

๐Ÿ‘‰ Maximum Likelihood Estimation