Question

(1 point) A normal distribution with mean 0 and standard deviation Võ is sampled three times, yielding values x, y, z. Find the log-likelihood function In L(O) (type theta for 6): In L(O) = 0 Find the derivative of the log-likelihood with respect to (type theta for €): In L(0)) = Find the maximum likelihood estimator for 2 (note that there is only one positive value): Ô -

Answer 1

Answer:

-0, y, z ~ N(p=0, 0 = 1)

$f(x)=f(y)=f(z)=\frac{1}{\sqrt{2\pi\theta}}e^{-\left ( \frac{(x-\theta)^{2}}{2\theta} \right )}\ \ \ -\infty<\theta<\infty\ ;\ -\infty<x<\infty$

Likelihood function of 2, 4,2

$L(\theta)=f(x).f(y).f(z)$

$=\frac{1}{(\sqrt{2\pi \theta})^{3}}\left \{ e^{-\frac{(x-\theta)^{2}}{2\theta}} \right \}\left \{ e^{-\frac{(y-\theta)^{2}}{2\theta}} \right \}\left \{ e^{-\frac{(z-\theta)^{2}}{2\theta}} \right \}$

  $=\frac{1}{(\sqrt{2\pi \theta})^{3}}e^{-\frac{\left \{ (x^{2}+y^{2}+z^{2})+3\theta^{2}-2\theta(x+y+z) \right \}}{2\theta}}$

$\log L(\theta)=\left [ -\frac{\left \{ (x^{2}+y^{2}+z^{2})+3\theta^{2}-2\theta(x+y+z) \right \}}{2\theta}-\frac{3}{2}(2\pi \theta ) \right ]$

$\frac{\partial \log L(\theta)}{\partial \theta}=\frac{\partial }{\partial \theta}\left [ -\frac{\left \{ (x^{2}+y^{2}+z^{2})+3\theta^{2}-2\theta(x+y+z) \right \}}{2\theta}-\frac{3}{2}(2\pi \theta ) \right ]$

  $=\left [ -\left \{ \frac{-2(x^{2}+y^{2}+z^{2})}{2\theta^{2}}+\frac{3}{2} \right \}-3\pi \right ]$

$For\ MLE\ of\ \theta,we\ have\ to\ equate\ \frac{\partial \log L(\theta)}{\partial \theta} \ with\ 0.$

So

$\frac{\partial \log L(\theta)}{\partial \theta}=0$

$\Rightarrow \left \{ \frac{(x^{2}+y^{2}+z^{2})}{\hat{\theta}^{2}}-\frac{3}{2}-3\pi \right \}=0$

$\Rightarrow \frac{(x^{2}+y^{2}+z^{2})}{\hat{\theta}^{2}}=\left ( 3\pi+\frac{3}{2} \right )$

$\Rightarrow \hat{\theta}^{2}=\frac{(x^{2}+y^{2}+z^{2})}{\left ( 3\pi+\frac{3}{2} \right )}$

$\hat{\theta}_{MLE}=\sqrt{\frac{2.(x^{2}+y^{2}+z^{2})}{3(2\pi+1)}}$