Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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
add comment

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.