# Maximum-likelihood estimation for continuous random variable with unknown parameter

Let $X$ be a random variable with the unknown parameter $\lambda$ and the following pdf $$f(t)=2\lambda t\cdot\mathrm e^{-\lambda t^2}\cdot\textbf{1}_{[0,\infty)}(t)$$ where $\textbf{1}_A(x)$ is an indicator function with $$\textbf{1}_A(x)=\begin{cases}1,&\text{if }x\in A,\\0,&\text{else.}\end{cases}$$ Let $\vec x=(x_1,\ldots,x_n)$ be a sample of $X$. Determine the maximum-likelihood estimator $\widehat{\lambda}$ such that the following is true for the likelihood-function $\mathcal L(\vec x;\lambda)$: $$\forall \lambda\;:\;\mathcal L(\vec x;\lambda)\leq \mathcal L(\vec x;\widehat\lambda)$$

For the sake of simplicity my first thoughts were to get the log-likelihood this way: $$\mathcal L(\vec x;\lambda)=\prod\limits_{i=1}^nf(x_i)\implies \ln(\mathcal L(\vec x;\lambda))=\sum\limits_{i=1}^n\ln(f(x_i))$$ This is the point where I'm stuck: i don't know how to compute the derivative to maximize the function $$\frac{\mathrm d \ln(\mathcal L(\vec x;\lambda))}{\mathrm d\lambda}\overset{!}{=}0.$$ Any hints on how to derive the sum would be appreciated.

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As an aside, we might recognize this as the pdf of $\sqrt Y$ when $Y$ has an exponential distribution with mean $\lambda^{-1}$. So if you know what the MLE for the exponential distribution is, you know what to expect to get here. – guy Jul 7 '12 at 17:41

## 1 Answer

We have $$L(\vec x,\lambda)=\prod_{j=1}^n(2\lambda x_j)\cdot e^{-\lambda x_j^2}=2^n\left(\prod_{k=1}^nx_k\right)\lambda ^n\exp\left(-\lambda \lVert x\rVert^2\right)\chi_{x_j\geq 0\forall j},$$ hence assuming the $x_j> 0$. $$\log L(\vec x,\lambda)=n\log 2+\sum_{j=1}^n\log x_j+n\log\lambda-\lambda\lVert x\rVert^2.$$ Now taking the derivative with respect to $\lambda$, we get $$\partial_{\lambda}\log L(\vec x,\lambda)=\frac n{\lambda}-\lVert x\rVert^2.$$ I let you finish the computation.

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I have issues following your transformation to the exp function. And could you explain how you define your norm $\|x\|$? – Christian Ivicevic Jul 7 '12 at 19:37
It's the Euclidian norm ($\lVert x\rVert=\sum_{j=1}^nx_j^2$). I used the fact that $\prod_{j=1}^n\exp(a_j)=\exp(\sum_{j=1}^na_j)$ for real numbers $a_1,\dots,a_n$ and $n$ integer. – Davide Giraudo Jul 7 '12 at 19:39
Did i miss something? I thought that $\|x\|=\sqrt{\sum\limits_{j=1}^nx_j^2}$. – Christian Ivicevic Jul 7 '12 at 20:03
You didn't miss anything since I made a typo. – Davide Giraudo Jul 7 '12 at 20:07
Sorry, but in the first equation I still cannot see why the $x_j$ from $(2\lambda x_j)$ disappeared (assuming your $\chi$ is my $\textbf 1_A(x)$). I still get a $\left(\prod\limits_{j=1}^nx_j\right)$ term left to be multiplied with the rest. – Christian Ivicevic Jul 7 '12 at 20:43