chre - ahmad alhayek

chre - ahmad alhayek

الرئيسية صورة وعبرة كلام الناس خواطر chre

chre

ahmadalhayk 8:01 ص

Image for 5) Let X1, X2, ...Xn ~ N(theta, theta) 0 < theta < infinity. Find the MLE of theta.

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Since $X_1,X_2,...X_n\sim N(\theta,\theta)$ . Define the likelihood function as
$L=\prod_{i=1}^{n}\left [ \frac{1}{\sqrt{2\pi \theta}}\exp \left \{ -\frac{1}{2\theta}(x_i-\theta)^2 \right \} \right ]\\ ~~~~~~~~=\left ( \frac{1}{\sqrt{2\pi \theta}} \right )^n \exp \left \{ -\sum_{i=1}^{n} \frac{(x_i-\theta)^2}{2\theta} \right \} \\ \Rightarrow \ln L=-\frac{n}{2} \ln (2\pi \theta)-\frac{1}{2\theta}\sum_{i=1}^{n} {(x_i-\theta)^2} \\ ~~~~~~~~~~=-\frac{n\ln 2\pi}{2}-\frac{n\ln \theta}{2}-\frac{1}{2\theta}\sum_{i=1}^{n} {(x_i-\theta)^2}$
Differentiate with respect to $\theta$ gives
$\frac{\partial }{\partial \theta}\ln L=\frac{-n}{2\theta}+\frac{1}{2\theta}\sum_{i=1}^{n} 2(x_i-\theta)+\frac{1}{2\theta}\sum_{i=1}^{n}(x_i-\theta)^2$
Now $\frac{\partial }{\partial \theta}\ln L=0$ gives
$-n+\sum_{i=1}^{n} 2(x_i-\theta)+\frac{1}{\theta}\sum_{i=1}^{n}(x_i-\theta)^2 =0 \\ \Rightarrow \sum_{i=1}^{n}\left ( 2(x_i-\theta)+\frac{(x_i-\theta)^2}{\theta} \right )=n \\\Rightarrow \sum_{i=1}^{n}(x_i^2-\theta^2)=n\theta \\ \Rightarrow n\theta^2+n\theta-\sum_{i=1}^{n}x_i^2=0 \\ \Rightarrow \theta(\theta+1)=\frac{1}{n}\sum_{i=1}^{n}x_i^2$
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