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2021-08-16

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

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Skilled2021-08-17Added 94 answers

Given: $f\left(x\right)=C(1+\theta x)-1\le x\le 1$

a)A valid PDF satisfies${\int}_{x}f\left(x\right)dx=1$

${\int}_{-1}^{1}c(1+\theta x)dx=1$

$C{[x+\theta \frac{{x}^{2}}{2}]}_{-1}^{1}=1$

$c=\frac{1}{2}$

b) Moment estimator for$\theta ,{\mu}_{1}^{1}={\int}_{-1}^{1}xf\left(x\right)dx$

$\mu}_{1}^{1}=s\frac{1}{2}{\int}_{1}^{-1}(x+\theta {x}^{2})dx=\frac{1}{2}{[\frac{{x}^{2}}{2}+\theta \frac{{x}^{3}}{3}]}_{-1}^{1$

$\mu}_{1}^{1}=\frac{1}{2}[1-\left(1\right)+\frac{\theta}{3}(1-(-1))]=\frac{\theta}{3$
Maximum likelihood estimator for $\theta$

$f\left(x\right)=\frac{(1+\theta x)}{2}$

$L\left(\theta \right)=\prod _{1=1}^{n}f\left({x}_{1}\right)=\frac{(1+\theta {x}_{1})(1+\theta {x}_{2})(1+{\theta}_{3})\dots \cdot (1+\theta {x}_{n})}{2}$

$L\left(\theta \right)=\prod _{1=1}^{n}(1+\theta {x}_{1})$

Take log, log$L\left(\theta \right)=-n\mathrm{log}2+\sum _{i=1}^{n}\left({x}_{i}\right)\frac{1}{(1+\theta {x}_{i})=0}$

Now$\frac{\delta}{\delta \theta}L(x,\theta )=0+\sum _{i=1}^{n}\left({x}_{i}\right)\frac{1}{(1+\theta {x}_{i})}=0$
Given $\mu ,\sum _{i=1}^{n}$

a)A valid PDF satisfies

b) Moment estimator for

Take log, log

Now

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