MLE of $\lambda$ Given $f(x;\lambda)=1-\dfrac{2}{3}\lambda+\lambda\sqrt{x}$

Question

$f(x;\lambda)=1-\dfrac{2}{3}\lambda+\lambda\sqrt{x}\ \ \ ; 0\le x\le1 $

$0$ otherwise

What is the maximum likelihood estimate of the parameter $\lambda$ based on two independent observations $x_1=\dfrac{1}{4}$ and $x_2=\dfrac{9}{16}$

My input

Liklihood function of the sample is given by

$L(\lambda)=\prod_{i=1}^{2}f(x_i;\lambda)=(1-\dfrac{2}{3}\lambda+\lambda\sqrt{x_i})$

=$(1-\dfrac{2}{3}\lambda+\lambda\sqrt{x_1})(1-\dfrac{2}{3}\lambda+\lambda\sqrt{x_2})$

$\log(L(\lambda)=\log(1-\dfrac{2}{3}\lambda+\lambda\sqrt{x_1})(1-\dfrac{2}{3}\lambda+\lambda\sqrt{x_2})$

$=\log(1-\dfrac{2}{3}\lambda+\lambda\sqrt{x_1})+\log(1-\dfrac{2}{3}\lambda+\lambda\sqrt{x_2})$

Am I following correct path ? Can someone tell me ?

Answer 1

You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So \begin{align}L(\lambda)&=\left(1+\left(\sqrt{x_1}-\frac23\right)\lambda\right)\left(1+\left(\sqrt{x_2}-\frac23\right)\lambda\right)\\&=\left(\sqrt{x_1}-\frac23\right)\left(\sqrt{x_2}-\frac23\right)\lambda^2+\left(\sqrt{x_1}+\sqrt{x_2}-\frac43\right)\lambda+1\\ \implies L'\left(\hat\lambda\right)&=2\left(\sqrt{x_1}-\frac23\right)\left(\sqrt{x_2}-\frac23\right)\hat\lambda+\sqrt{x_1}+\sqrt{x_2}-\frac43=0\end{align} There you can simply solve for $\hat\lambda$ and check whether it is a maximum or minimum.