# Find the MLE of bivariate normal

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!

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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 '14 at 18:28
it's a n*2 matrix, has n rows and 2 columns – user2350622 Feb 19 '14 at 18:37

## 1 Answer

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 $$\nabla_{\mu}p(x_1,x_2,\cdots,x_n)=0\\ \frac{\partial p(x_1,x_2,\cdots,x_n)}{\partial \sigma^2}=0$$

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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 '14 at 19:07
If by variance you mean the sample variance of $X$, then it is fine. – Samrat Mukhopadhyay Feb 19 '14 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 '14 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 '14 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 '14 at 1:57