간단한 선형 회귀 분석에서 회귀 계수의 편차 도출

간단한 선형 회귀 분석에서 $y = \beta_0 + \beta_1 x + u$ . 여기서 $u \sim iid\;\mathcal N(0,\sigma^2)$ . 추정값을 도출했습니다 :

\hat{β_{1}} = \frac{\sum_{i} (x_{i} - \bar{x}) (y_{i} - \bar{y})}{\sum_{i} (x_{i} - \bar{x})^{2}},

$\hat{\beta_1} = \frac{\sum_i (x_i - \bar{x})(y_i - \bar{y})}{\sum_i (x_i - \bar{x})^2}\ ,$ 여기서

\bar{x}

$\bar{x}$ 및

\bar{y}

$\bar{y}$ 는

x

$x$ 와

의 표본 평균입니다.

y

$y$ .

지금은의 분산 찾으려면 . 나는 다음과 같은 것을 파생시켰다 : $\hat\beta_1$

Var (\hat{β_{1}}) = \frac{σ^{2} (1 - \frac{1}{n})}{\sum_{i} (x_{i} - \bar{x})^{2}} .

$\text{Var}(\hat{\beta_1}) = \frac{\sigma^2(1 - \frac{1}{n})}{\sum_i (x_i - \bar{x})^2}\ .$

도출은 다음과 같습니다.

\begin{aligned} Var (\hat{β_{1}}) \\ = Var (\frac{\sum_{i} (x_{i} - \bar{x}) (y_{i} - \bar{y})}{\sum_{i} (x_{i} - \bar{x})^{2}}) \\ = \frac{1}{(\sum_{i} (x_{i} - \bar{x})^{2})^{2}} Var (\sum_{i} (x_{i} - \bar{x}) (β_{0} + β_{1} x_{i} + u_{i} - \frac{1}{n} \sum_{j} (β_{0} + β_{1} x_{j} + u_{j}))) \\ = \frac{1}{(\sum_{i} (x_{i} - \bar{x})^{2})^{2}} Var (β_{1} \sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2} + \sum_{나는} ({엑스}_{나는} - \bar{엑스}) (유_{나는} - \sum_{제이} \frac{유_{제이}}{엔})) \\ = \frac{1}{(\sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2})^{2}} 바르 (\sum_{나는} ({엑스}_{나는} - \bar{엑스}) (유_{나는} - \sum_{제이} \frac{유_{제이}}{엔})) \\ = \frac{1}{(\sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2})^{2}} \times \\ 이자형 [{(\sum_{나는} ({엑스}_{나는} - \bar{엑스}) (유_{나는} - \sum_{제이} \frac{유_{제이}}{엔}) - \underset{= 0}{\underset{⏟}{이자형 [\sum_{나는} ({엑스}_{나는} - \bar{엑스}) (유_{나는} - \sum_{제이} \frac{유_{제이}}{엔})]}})}^{2}] \\ = \frac{1}{(\sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2})^{2}} 이자형 [{(\sum_{나는} ({엑스}_{나는} - \bar{엑스}) (유_{나는} - \sum_{제이} \frac{유_{제이}}{엔}))}^{2}] \\ = \frac{1}{(\sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2})^{2}} 이자형 [\sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2} (유_{나는} - \sum_{제이} \frac{유_{제이}}{엔})^{2}], 이후 유_{나는} 의 iid입니다 \\ = \frac{1}{(\sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2})^{2}} \sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2} 이자형 {(유_{나는} - \sum_{제이} \frac{유_{제이}}{엔})}^{2} \\ = \frac{1}{(\sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2})^{2}} \sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2} (이자형 (유_{나는}^{2}) - 2 \times 이자형 (유_{나는} \times (\sum_{제이} \frac{유_{제이}}{엔})) + 이자형 {(\sum_{제이} \frac{유_{제이}}{엔})}^{2}) \\ = \frac{1}{(\sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2})^{2}} \sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2} (σ^{2} - \frac{2}{엔} σ^{2} + \frac{σ^{2}}{엔}) \\ = \frac{σ^{2}}{\sum_{나는} ({엑스}_{나는} - \bar{엑스})^{2}} (1 - \frac{1}{엔}) \end{aligned}

$\begin{align} &\text{Var}(\hat{\beta_1})\\ & = \text{Var} \left(\frac{\sum_i (x_i - \bar{x})(y_i - \bar{y})}{\sum_i (x_i - \bar{x})^2} \right) \\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2} \text{Var}\left( \sum_i (x_i - \bar{x})\left(\beta_0 + \beta_1x_i + u_i - \frac{1}{n}\sum_j(\beta_0 + \beta_1x_j + u_j) \right)\right)\\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2} \text{Var}\left( \beta_1 \sum_i (x_i - \bar{x})^2 + \sum_i(x_i - \bar{x}) \left(u_i - \sum_j \frac{u_j}{n}\right) \right)\\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2}\text{Var}\left( \sum_i(x_i - \bar{x})\left(u_i - \sum_j \frac{u_j}{n}\right)\right)\\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2}\;\times \\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;E\left[\left( \sum_i(x_i - \bar{x})(u_i - \sum_j \frac{u_j}{n}) - \underbrace{E\left[\sum_i(x_i - \bar{x})(u_i - \sum_j \frac{u_j}{n})\right] }_{=0}\right)^2\right]\\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2} E\left[\left( \sum_i(x_i - \bar{x})(u_i - \sum_j \frac{u_j}{n})\right)^2 \right] \\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2} E\left[\sum_i(x_i - \bar{x})^2(u_i - \sum_j \frac{u_j}{n})^2 \right]\;\;\;\;\text{ , since } u_i \text{ 's are iid} \\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2}\sum_i(x_i - \bar{x})^2E\left(u_i - \sum_j \frac{u_j}{n}\right)^2\\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2}\sum_i(x_i - \bar{x})^2 \left(E(u_i^2) - 2 \times E \left(u_i \times (\sum_j \frac{u_j}{n})\right) + E\left(\sum_j \frac{u_j}{n}\right)^2\right)\\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2}\sum_i(x_i - \bar{x})^2 \left(\sigma^2 - \frac{2}{n}\sigma^2 + \frac{\sigma^2}{n}\right)\\ & = \frac{\sigma^2}{\sum_i (x_i - \bar{x})^2}\left(1 - \frac{1}{n}\right) \end{align}$

내가 여기서 뭔가 잘못 했니?

행렬 표기법으로 모든 것을 수행하면 . 그러나 개념을 이해하기 위해 행렬 표기법을 사용하지 않고 답을 도출하려고합니다. ${\rm Var}(\hat{\beta_1}) = \frac{\sigma^2}{\sum_i (x_i - \bar{x})^2}$

— mynameisJEFF
소스

예, 행렬 표기법의 수식이 정확합니다. 문제의 공식을 보면

모집단 표준 편차 대신 샘플 표준 편차를 사용하는 것처럼 보입니까? 파생을 보지 않으면 더 이상 말하기가 어렵습니다.

1 - \frac{1}{n} = \frac{n - 1}{n}

$1-\frac1{n}\,=\,\frac{n-1}{n}$

— TooTone

일반 답변은 stats.stackexchange.com/questions/91750 의 중복 스레드에도 게시되었습니다 .

— 우버

답변:

파생을 시작할 때 와 모두 확장하는 과정에서 괄호 를 곱합니다 . 전자는 합 변수 에 의존 하지만 후자는 그렇지 않습니다. 를 그대로두면 유도가 훨씬 간단합니다. $\sum_i (x_i - \bar{x})(y_i - \bar{y})$ $y_i$ $\bar{y}$ $i$ $\bar{y}$

\begin{aligned} \sum_{i} (x_{i} - \bar{x}) \bar{y} & = \bar{y} \sum_{i} (x_{i} - \bar{x}) \\ = \bar{y} ((\sum_{i} x_{i}) - n \bar{x}) \\ = \bar{y} (n \bar{x} - n \bar{x}) \\ = 0 \end{aligned}

$\begin{align} \sum_i (x_i - \bar{x})\bar{y} &= \bar{y}\sum_i (x_i - \bar{x})\\ &= \bar{y}\left(\left(\sum_i x_i\right) - n\bar{x}\right)\\ &= \bar{y}\left(n\bar{x} - n\bar{x}\right)\\ &= 0 \end{align}$

금후

\begin{aligned} \sum_{i} (x_{i} - \bar{x}) (y_{i} - \bar{y}) & = \sum_{i} (x_{i} - \bar{x}) y_{i} - \sum_{i} (x_{i} - \bar{x}) \bar{y} \\ = \sum_{i} (x_{i} - \bar{x}) y_{i} \\ = \sum_{i} (x_{i} - \bar{x}) (β_{0} + β_{1} x_{i} + u_{i}) \end{aligned}

$\begin{align} \sum_i (x_i - \bar{x})(y_i - \bar{y}) &= \sum_i (x_i - \bar{x})y_i - \sum_i (x_i - \bar{x})\bar{y}\\ &= \sum_i (x_i - \bar{x})y_i\\ &= \sum_i (x_i - \bar{x})(\beta_0 + \beta_1x_i + u_i )\\ \end{align}$

과

\begin{aligned} Var (\hat{β_{1}}) & = Var (\frac{\sum_{i} (x_{i} - \bar{x}) (y_{i} - \bar{y})}{\sum_{i} (x_{i} - \bar{x})^{2}}) \\ = Var (\frac{\sum_{i} (x_{i} - \bar{x}) (β_{0} + β_{1} x_{i} + u_{i})}{\sum_{i} (x_{i} - \bar{x})^{2}}), substituting in the above \\ = Var (\frac{\sum_{i} (x_{i} - \bar{x}) u_{i}}{\sum_{i} (x_{i} - \bar{x})^{2}}), noting only u_{i} is a random variable \\ = \frac{\sum_{i} (x_{i} - \bar{x})^{2} Var (u_{i})}{{(\sum_{i} (x_{i} - \bar{x})^{2})}^{2}}, independence of u_{i} and, Var (k X) = k^{2} Var (X) \\ = \frac{σ^{2}}{\sum_{i} (x_{i} - \bar{x})^{2}} \end{aligned}

$\begin{align} \text{Var}(\hat{\beta_1}) & = \text{Var} \left(\frac{\sum_i (x_i - \bar{x})(y_i - \bar{y})}{\sum_i (x_i - \bar{x})^2} \right) \\ &= \text{Var} \left(\frac{\sum_i (x_i - \bar{x})(\beta_0 + \beta_1x_i + u_i )}{\sum_i (x_i - \bar{x})^2} \right), \;\;\;\text{substituting in the above} \\ &= \text{Var} \left(\frac{\sum_i (x_i - \bar{x})u_i}{\sum_i (x_i - \bar{x})^2} \right), \;\;\;\text{noting only $u_i$ is a random variable} \\ &= \frac{\sum_i (x_i - \bar{x})^2\text{Var}(u_i)}{\left(\sum_i (x_i - \bar{x})^2\right)^2} , \;\;\;\text{independence of } u_i \text{ and, Var}(kX)=k^2\text{Var}(X) \\ &= \frac{\sigma^2}{\sum_i (x_i - \bar{x})^2} \\ \end{align}$

원하는 결과입니다.

부수적으로, 나는 당신의 파생에서 오류를 찾으려고 오랜 시간을 보냈습니다. 결국 나는 재량이 용기의 더 나은 부분이라고 결정하고 더 간단한 접근법을 시도하는 것이 가장 좋습니다. 그러나 기록을 위해이 단계가 정당화되었는지 확신하지 못했습니다.

\begin{aligned} = . \frac{1}{(\sum_{i} (x_{i} - \bar{x})^{2})^{2}} E [{(\sum_{i} (x_{i} - \bar{x}) (u_{i} - \sum_{j} \frac{u_{j}}{n}))}^{2}] \\ = \frac{1}{(\sum_{i} (x_{i} - \bar{x})^{2})^{2}} E [\sum_{i} (x_{i} - \bar{x})^{2} (u_{i} - \sum_{j} \frac{u_{j}}{n})^{2}], since u_{i}'s are iid \end{aligned}

$\begin{align} & =. \frac{1}{(\sum_i (x_i - \bar{x})^2)^2} E\left[\left( \sum_i(x_i - \bar{x})(u_i - \sum_j \frac{u_j}{n})\right)^2 \right] \\ & = \frac{1}{(\sum_i (x_i - \bar{x})^2)^2} E\left[\sum_i(x_i - \bar{x})^2(u_i - \sum_j \frac{u_j}{n})^2 \right]\;\;\;\;\text{ , since } u_i \text{ 's are iid} \\ \end{align}$ because it misses out the cross terms due to

\sum_{j} \frac{u_{j}}{n}

$\sum_j \frac{u_j}{n}$ .

— TooTone
소스

I noticed that I could use the simpler approach long ago, but I was determined to dig deep and come up with the same answer using different approaches, in order to ensure that I understand the concepts. I realise that first

\sum_{j} \hat{u_{j}} = 0

$\sum_j \hat{u_j} = 0$ from normal equations (FOC from least square method), so

\bar{\hat{u}} = \frac{\sum_{i} u_{i}}{n} = 0

$\bar{\hat{u}} = \frac{\sum_i u_i}{n}=0$ , plus

\bar{\hat{u}} = \bar{y} - \bar{\hat{y}} = 0

$\bar{\hat{u}} = \bar{y} - \bar{\hat{y}} = 0$ , so

\bar{y} = \bar{\hat{y}}

$\bar{y} = \bar{\hat{y}}$ . So there won't be the term

\sum_{j} \frac{u_{j}}{n}

$\sum_j \frac{u_j}{n}$ in the first place.

— mynameisJEFF

ok, in your question the emphasis was on avoiding matrix notation.

— TooTone

Yes, because I was able to solve it using matrix notation. And notice from my last comment, I did not use any linear algebra. Thanks for your great answer anyway^.^

— mynameisJEFF

sorry are we talking at cross-purposes here? I didn't use any matrix notation in my answer either, and I thought that was what you were asking in your question.

— TooTone

sorry for misunderstanding haha...

— mynameisJEFF

I believe the problem in your proof is the step where you take the expected value of the square of $\sum_i (x_i - \bar{x} )\left( u_i -\sum_j \frac{u_j}{n} \right)$ . This is of the form $E \left[\left(\sum_i a_i b_i \right)^2 \right]$ , where $a_i = x_i -\bar{x}; b_i = u_i -\sum_j \frac{u_j}{n}$ . So, upon squaring, we get $E \left[ \sum_{i,j} a_i a_j b_i b_j \right] = \sum_{i,j} a_i a_j E\left[b_i b_j \right]$ . Now, from explicit computation, $E\left[b_i b_j \right] = \sigma^2 \left( \delta_{ij} -\frac{1}{n} \right)$ , so $E \left[ \sum_{i,j} a_i a_j b_i b_j \right] = \sum_{i,j} a_i a_j \sigma^2 \left( \delta_{ij} -\frac{1}{n} \right) = \sum_i a_i^2 \sigma^2$ as $\sum_i a_i = 0$ .

— Stefano
소스

Begin from "The derivation is as follow:" The 7th "=" is wrong.

Because

$\sum_i (x_i - \bar{x})(u_i - \bar{u})$

$= \sum_i (x_i - \bar{x})u_i - \sum_i (x_i - \bar{x}) \bar{u}$

$= \sum_i (x_i - \bar{x})u_i - \bar{u} \sum_i (x_i - \bar{x})$

$= \sum_i (x_i - \bar{x})u_i - \bar{u} (\sum_i{x_i} -n \bar{x})$

$= \sum_i (x_i - \bar{x})u_i - \bar{u} (\sum_i{x_i} -\sum_i{x_i})$

$= \sum_i (x_i - \bar{x})u_i - \bar{u} 0$

$= \sum_i (x_i - \bar{x})u_i$

So after 7th "=" it should be:

$\frac {1} {(\sum_i(x_i-\bar{x})^2)^2}E\left[\left(\sum_i(x_i-\bar{x})u_i\right)^2\right]$

$=\frac {1} {(\sum_i(x_i-\bar{x})^2)^2}E\left(\sum_i(x_i-\bar{x})^2u_i^2 + 2\sum_{i\ne j}(x_i-\bar{x})(x_j-\bar{x})u_iu_j\right)$

= $\frac {1} {(\sum_i(x_i-\bar{x})^2)^2}E\left(\sum_i(x_i-\bar{x})^2u_i^2\right) + 2E\left(\sum_{i\ne j}(x_i-\bar{x})(x_j-\bar{x})u_iu_j\right)$

= $\frac {1} {(\sum_i(x_i-\bar{x})^2)^2}E\left(\sum_i(x_i-\bar{x})^2u_i^2\right)$ , because $u_i$ and $u_j$ are independent and mean 0, so $E(u_iu_j) =0$

= $\frac {1} {(\sum_i(x_i-\bar{x})^2)^2}\left(\sum_i(x_i-\bar{x})^2E(u_i^2)\right)$

$\frac {\sigma^2} {(\sum_i(x_i-\bar{x})^2)^2}$

— user158565
소스

It might be helpful if you edited your answer to include the correct line.

— mdewey

Your answer is being automatically flagged as low quality because it's very short. Please consider expanding on your answer

— Glen_b