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.

Suppose that $X = (x_{ij})n*2$ follows a bivariate normal distribution $\mathcal{N}(\mu, \sigma^2I)$, where I is the $2\times 2$ identity matrix. How to find the maximum likelihood estimates of $\mu$ and $\sigma^2$? Specifically, how to deal with the determinant part in the density formula of bivariate normal distribution? Thanks!

share|improve this question
1  
Can you please explain what you want to mean by $(x_{ij})n*2$? Does it mean a $n\times 2$ matrix $X$? –  Samrat Mukhopadhyay Feb 19 at 18:28
    
it's a n*2 matrix, has n rows and 2 columns –  user2350622 Feb 19 at 18:37
add comment

1 Answer 1

If $X$ is a $m\times n$ matrix with $n$ random vectors distributed identically as $\mathcal{N}(\mu,\sigma^2 I_{m\times m})$, and if the random vectors are independent then you can write the joint distribution of the vectors as $$p(x_1,x_2,\cdots,\ x_n)=\prod_{i=1}^np(x_i)\\=\frac{1}{(2\pi)^{mn/2} \sigma^{mn}}\exp\left(-\frac{\sum_{i=1}^n\sum_{j=1}^m(x_{ji}-\mu_{j})^2}{2\sigma^2}\right)$$ So, to find the MLE of $\mu$ and $\sigma^2$ find the simultaneous solutions of $$\nabla_{\mu}p(x_1,x_2,\cdots,x_n)=0\\ \frac{\partial p(x_1,x_2,\cdots,x_n)}{\partial \sigma^2}=0$$

share|improve this answer
    
I solved the equations but I ended up getting a strange MLE for sigma^2. I got 1/2*variance of X. –  user2350622 Feb 19 at 19:07
    
If by variance you mean the sample variance of $X$, then it is fine. –  Samrat Mukhopadhyay Feb 19 at 19:16
    
okay! It just seemed to be wired to me because usually MLE variance does not have a 1/2 in front of it –  user2350622 Feb 20 at 0:04
    
If you do the math, then you'll see that for the case where you have $m$ samples, a factor of $1/m$ is coming in front of the sample variance. –  Samrat Mukhopadhyay Feb 20 at 6:47
    
yes but I tried in R and the variance of the samples should be really close to the true sigma. –  user2350622 Feb 24 at 1:57
add comment

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.