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    \begin{document}

    \title{Tutorial 5 homework}
    \subtitle {Szymon Antoniak 394197}



    \maketitle
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\section{Exercise 4 - general method of moments}


\begin{enumerate}
    \item The Rayleigh distribution
    density is given as:
    \[p(x,\theta)= \frac{x}{\theta^2} exp[-\frac{x^2}{\theta^2}] \;, \;\;x>0 \,, \; \theta >0\]
    in this case, $\eta=\theta^2, \; A(\eta)=log(\eta)$, so $\mathrm{E}_\eta [T(X)]=\dot{A}(\eta)=\frac{1}{\eta}$ \\
    hence we obtain that \[ \hat{\theta}=\frac{1}{\sqrt{\hat{\mu_1}}} \]
    where $\hat{\mu_1}=\frac{1}{n}\sum_{i=1}^{n}-X^2$
    \\

    \item Beta distribution with first parameter fixed has the following density:
    \[p(x,\lambda) = \frac{1}{\Gamma(\alpha)}\lambda^\alpha x^{\alpha -1} e^{-\lambda x}                  \]
    this time $\eta=\lambda$ and $A(\eta)=-\alpha \, log(\eta)$ so, again $\mathrm{E}_\eta[T(X)]=\dot{A}(\eta)=-\frac{\alpha}{\eta}$
    so
    \[ \hat{\lambda} = -\frac{\alpha}{\hat{\mu_1}}    \]
    This time I was not certain whether the fact that $\alpha$ is fixed means that we know its value - although it was the only meaning I could attach to that phrase
    .

\end{enumerate}






\end{document}