Maximum Likelihood Estimation
βνλ₯ κ³Ό ν΅κ³(MATH230)β μμ μμ λ°°μ΄ κ²κ³Ό 곡λΆν κ²μ μ 리ν ν¬μ€νΈμ λλ€. μ 체 ν¬μ€νΈλ Probability and Statisticsμμ νμΈνμ€ μ μμ΅λλ€ π²
Introduction
<MLE; Maximum Likelihood Estimation>μ μ§κΈκΉμ§μ μ κ·Όλ²κ³Όλ μ¬λ λ€λ₯Έ μ κ·Ό λ°©μμ΄λ€. <MLE>λ statistical inferenceλ₯Ό μννλ λ§μ μ κ·Όλ² μ€ νλλ‘, statistical approachμ μλ‘μ΄ μ² νκ³Ό μμΌλ₯Ό μ 곡νλ€ π
Example.
νλ₯ $p$λ₯Ό μ νν μμ§ λͺ»νλ p-coinμ 10λ² λμ§λ€κ³ νμ. 10λ²μ κ²°κ³Όλ μλμ κ°λ€.
μμμ λ°°μ΄ <proportion estimation>μ λ°©λ²μΌλ‘ μ κ·Όνλ©΄, $p$λ Point Estimator $\hat{p} = 6/10$μΌλ‘ μΆμ λλ€.
μ΄λ²μλ <MLE>μ λ°©μμΌλ‘ μ κ·Όν΄λ³΄μ! λ¨Όμ 10μ λμ λμ§κΈ°κ° μμ κ°μ΄ λμ¬ νλ₯ μ μλμ κ°λ€.
\[P(H, H, T, \dots, H, T) = p^6 q^4\]μμ μμ λ€μ μ°λ©΄, $L(p) = p^6 (1-p)^4$λ‘, λμ μ νλ₯ μ΄ $p$μΌ λ βH H β¦ H Tβμ κ²°κ³Όλ₯Ό μ»μ νλ₯ μ μλ―Ένλ€.
μ΄μ , μ΄ ν¨μ $L(p)$λ₯Ό maximize νλ $p$λ₯Ό ꡬν΄λ³΄μ. λ°©λ²μ κ°λ¨νλ€. κ·Έλ₯ $p$μ λν΄ λ―ΈλΆλ°©μ μμ νλ©΄ λλ€. μ΄λ, κ³μ°μ νΈμλ₯Ό μν΄ $\log$λ₯Ό λ¨Όμ μ·¨ν΄μ£Όμ.
\[\ell(p) = \log (L(p)) = 6 \log p + 4 \log (1-p)\] \[\frac{d\ell(p)}{dp} = \frac{6}{p} - \frac{4}{1-p} = 0 \quad \rightarrow \quad p = 6/10\]μ¦, $p=6/10$μ΄ βH H β¦ H TβλΌλ κ²°κ³Όκ° λμ¬ νλ₯ μ Maximizeνλ νλ₯ μ΄λΌλ λ§μ΄λ€!
μ΄μ <MLE>λ₯Ό μνμ μΌλ‘ μ λ¦¬ν΄ λ€μ μ΄ν΄λ³΄μ!
MLE; Maximum Likelihood Estimation
Theorem. MLE for Bernoulli case
Let $X_1, \dots, X_n$ be a $\text{Ber}(p)$ Random Samples, with iid.
Then, the likelihood function $L(p; x_1, \dots, x_n)$ would be
\[\begin{aligned} L(p;\, x_1, \dots, x_n) &= f(x_1;\, p) \cdots f(x_n;\, p) \\ &= p^{x_1} (1-p)^{1-x_1} \cdots p^{x_n} (1-p)^{1-x_n} \\ &= p^{\sum x_i} (1-p)^{n - \sum x_i} \end{aligned}\]Take log on it!
\[\begin{aligned} \ell(p) = \sum x_i \cdot \log p + (n - \sum x_i) \cdot \log (1-p) \end{aligned}\]Take derivative for $p$!
\[\frac{d\ell(p)}{dp} = \frac{\sum x_i}{p} - \frac{n-\sum x_i}{1-p} = 0\]when solve the equation, then
\[p = \frac{\sum x_i}{n} = \bar{x}\]Theorem. MLE for Normal case
Let $X_1, \dots, X_n$ be a $N(\mu, 1)$ Random Samples, with iid.
Find the MLE of $\mu$!
\[\begin{aligned} L(\mu; \, x_1, \dots, x_n) &= f(x_1; \, \mu) \cdots f(x_n; \, \mu) \\ &= \left( \frac{1}{\sqrt{2\pi}}\right)^n \exp \left( - \sum \, (x_i - \mu)^2 / \, 2 \right) \end{aligned}\]Take log on it!
\[\ell(\mu; \, \cdots) = n \cdot \log \left( \frac{1}{\sqrt{2\pi}}\right) - \frac{\sum \, (x_i - \mu)^2}{2}\]Take derivative for $\mu$!
\[\frac{d\ell}{d\mu} = \sum (x_i - \mu) = 0\]when solve the equation, then
\[\mu = \bar{x}\]μ΄μ λ€μ ν¬μ€νΈλΆν° ν΅κ³νμ κ½πΉμ΄λΌκ³ ν μ μλ <κ°μ€ κ²μ ; Hypothesis Tests>μ λν΄ λ€λ£¬λ€!! π μ°λ¦¬κ° μ§κΈκΉμ§ μνν βμΆμ (Estimation)βμ νμ©ν΄ μμ¬κ²°μ μ λ΄λ¦¬λ κ²μ΄ λ°λ‘ <Hypothesis Test>λ€!