2021-08-16

Consider the probability density function $f\left(x\right)=c\left(1+\theta x\right),-1\le x\le 1$ a. Find the value of the constant c b. Find both the moment and maximum likelihood estimator for eslimators for $\theta$

2k1enyvp

Given: $f\left(x\right)=C\left(1+\theta x\right)-1\le x\le 1$
a)A valid PDF satisfies ${\int }_{x}f\left(x\right)dx=1$
${\int }_{-1}^{1}c\left(1+\theta x\right)dx=1$
$C{\left[x+\theta \frac{{x}^{2}}{2}\right]}_{-1}^{1}=1$
$c=\frac{1}{2}$
b) Moment estimator for $\theta ,{\mu }_{1}^{1}={\int }_{-1}^{1}xf\left(x\right)dx$
${\mu }_{1}^{1}=s\frac{1}{2}{\int }_{1}^{-1}\left(x+\theta {x}^{2}\right)dx=\frac{1}{2}{\left[\frac{{x}^{2}}{2}+\theta \frac{{x}^{3}}{3}\right]}_{-1}^{1}$
${\mu }_{1}^{1}=\frac{1}{2}\left[1-\left(1\right)+\frac{\theta }{3}\left(1-\left(-1\right)\right)\right]=\frac{\theta }{3}$ Maximum likelihood estimator for $\theta$
$f\left(x\right)=\frac{\left(1+\theta x\right)}{2}$
$L\left(\theta \right)=\prod _{1=1}^{n}f\left({x}_{1}\right)=\frac{\left(1+\theta {x}_{1}\right)\left(1+\theta {x}_{2}\right)\left(1+{\theta }_{3}\right)\dots \cdot \left(1+\theta {x}_{n}\right)}{2}$
$L\left(\theta \right)=\prod _{1=1}^{n}\left(1+\theta {x}_{1}\right)$
Take log, log $L\left(\theta \right)=-n\mathrm{log}2+\sum _{i=1}^{n}\left({x}_{i}\right)\frac{1}{\left(1+\theta {x}_{i}\right)=0}$
Now $\frac{\delta }{\delta \theta }L\left(x,\theta \right)=0+\sum _{i=1}^{n}\left({x}_{i}\right)\frac{1}{\left(1+\theta {x}_{i}\right)}=0$ Given $\mu ,\sum _{i=1}^{n}$

Do you have a similar question?