I'm trying to find the maximum likelihood estimator of a random sample $X=(X_1,\dots,X_n)$ from a distribution with pdf $$f(x;\lambda)=\frac{2}{\lambda x \sqrt{2 \pi}} \exp\left({\frac{[ -\log(x)]^2}{2 \lambda^2}}\right), \lambda >0, 0 \leq x \leq 1$$

Until now i have calculated the joint distribution function $$f_X(\underline{x};\lambda)=\left(\frac{2}{\lambda \sqrt{2 \pi}}\right)^n \prod^n_{i=1}\frac{1}{x_i} \exp \left({\frac{\sum_{i=1}^n[-\log(x_i)]^2}{2 \lambda^2}}\right)$$

The likelihood function $$L(\lambda,\underline{x}) \propto \left(\frac{2}{\lambda \sqrt{2 \pi}}\right)^n\exp \left({\frac{\sum_{i=1}^n[-\log(x_i)]^2}{2 \lambda^2}}\right)$$

The log-likelihood function $$l(\lambda,\underline{x})=(n \log(2) - n\log(\lambda)-\frac{1}{2}\log(2 \pi) + \frac{1}{2 \lambda^2} \sum_{i=1}^n[-\log(x_i)]^2$$ To find the maximum of the log-likelihood function i want to find the point $\hat{\lambda}$ where $l'(\lambda,\underline{x})=0$ and then prove that $l''(\lambda,\underline{x})|_{\lambda = \hat{\lambda}}<0$. I have calculated the following derivatives: $$l'(\lambda;\underline{x}) = -\frac{n}{\lambda} - \frac{1}{\lambda^3}\sum_{i=1}^n[-\log(x_i)]^2$$

$$l''(\lambda,\underline{x})=\frac{1}{\lambda^2}\left( n + \frac{1}{\lambda^2} \sum_{i=1}^n[-\log(x_i)]^2\right)$$

I find

$$\hat{\lambda}= \pm \sqrt{-\frac{1}{n}\sum_{i=1}^n[-\log(x_i)]^2}$$

Which would result in a complex estimator. Are my calculations off in any way?

  • $\begingroup$ If $l'(\lambda)$ is continuous and never changes sign then the maximum likelihood would be at one end of the range of its domain $\endgroup$
    – Henry
    Jan 13, 2022 at 17:41
  • $\begingroup$ The issue is that your claimed PDF is not a PDF. $\endgroup$ Jan 13, 2022 at 17:44
  • 2
    $\begingroup$ @JoseAvilez it is presumably supposed to be a lognormal with parameters $\mu =0$ and $\sigma^2=\lambda^2$. If so, then presumably $\hat\lambda = \sqrt{\frac1n \sum (\log x_i)^2}$ $\endgroup$
    – Henry
    Jan 13, 2022 at 17:47

1 Answer 1


The problem is that $$f(x;\lambda)=\frac{2}{\lambda x \sqrt{2 \pi}} \exp{\frac{[ -\log(x)]^2}{2 \lambda^2}}, \lambda >0, 0 \leq x \leq 1$$ is not a probability density function; however, $$f(x;\lambda)=\frac{2}{\lambda x \sqrt{2 \pi}} \exp{\frac{-[ \log(x)]^2}{2 \lambda^2}}, \lambda >0, 0 \leq x \leq 1$$ is. Presumably, this was intended density. Once you repeat your computations with it, you should get an admissible estimator.

  • 1
    $\begingroup$ Thank you so much for the support, i did not question the exercise but the function clearly does not integrate to 1. $\endgroup$
    – Ciodar
    Jan 13, 2022 at 18:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .